Chapter 29 Alternating-Current Circuits

hapter 9 Alternating-urrent ircuits onceptual Problems A coil in an ac generator rotates at 6 Hz. How much time elapses between successive emf values of the coil? Determine the oncept Successive s are one-half period apart. Hence the elapsed time between the s is T 8.33 ms. - f 6 s ( ) f the voltage in an ac circuit is doubled, the voltage is (a) doubled, (b) halved, (c) increased by a factor of, (d) not changed. Picture the Problem We can use the relationship between and to decide the effect of doubling the voltage on the voltage. Express the initial voltage in te of the voltage: Express the doubled voltage in ' te of the new voltage ' : Divide the second of these equations by the first and simplify to obtain: ' ' Solving for ' yields: ' (a) is correct. 3 [SSM] f the frequency in the circuit shown in Figure 9-7 is doubled, the inductance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the oncept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. (b) is correct. 4 f the frequency in the circuit shown in Figure 9-7 is doubled, the inductive reactance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple. 739

74 hapter 9 Determine the oncept The inductive reactance of an inductor varies with the frequency according to ω. Hence, doubling ω will double. (a) is correct. 5 f the frequency in the circuit in Figure 9-8 is doubled, the capacitive reactance of the circuit will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the oncept The capacitive reactance of an capacitor varies with the frequency according to ω. Hence, doubling ω will halve. (c) is correct. 6 (a) n a circuit consisting solely of a ac generator and an ideal inductor, are there any time intervals when the inductor receives energy from the generator? f so, when? Explain your answer. (b) Are there any time intervals when the inductor supplies energy back to the generator? f so when? Explain your answer. Determine the oncept Yes to both questions. (a) While the magnitude of the current in the inductor is increasing, the inductor absorbs power from the generator. (b) When the magnitude of the current in the inductor decreases, the inductor supplies power to the generator. 7 [SSM] (a) n a circuit consisting of a generator and a capacitor, are there any time intervals when the capacitor receives energy from the generator? f so, when? Explain your answer. (b) Are there any time intervals when the capacitor supplies power to the generator? f so, when? Explain your answer. Determine the oncept Yes to both questions. (a) While the magnitude of the charge is accumulating on either plate of the capacitor, the capacitor absorbs power from the generator. (b) When the magnitude of the charge is on either plate of the capacitor is decreasing, it supplies power to the generator. 8 (a) Show that the S unit of inductance multiplied by the S unit of capacitance is equivalent to seconds squared. (b) Show that the S unit of inductance divided by the S unit of resistance is equivalent to seconds. Determine the oncept (a) Substitute the S units of inductance and capacitance and simplify to obtain: s A s s s

Alternating-urrent ircuits 74 (b) Substitute the S units of inductance divided by resistance and simplify to obtain: s A Ω s A A s 9 [SSM] Suppose you increase the rotation rate of the coil in the generator shown in the simple ac circuit in Figure 9-9. Then the current (a) increases, (b) does not change, (c) may increase or decrease depending on the magnitude of the original frequency, (d) may increase or decrease depending on the magnitude of the resistance, (e) decreases. Determine the oncept Because the current through the resistor is given by NBA ω, is directly proportional to ω. (a) is correct. f the inductance value is tripled in a circuit consisting solely of a variable inductor and a variable capacitor, how would you he to change the capacitance so that the natural frequency of the circuit is unchanged? (a) triple the capacitance, (b) decrease the capacitance to one-third of its original value, (c) You should not change the capacitance.(d) You cannot determine how to change the capacitance from the data given. Determine the oncept The natural frequency of an circuit is given by f π. Express the natural frequencies of the circuit before and after the inductance is tripled: Divide the second of the these equations by the first and simplify to obtain: f and π f ' π '' f π f ' '' π '' Because the natural frequency is unchanged: ' '' '' ' When the inductance is tripled: ' 3 3 ( b) is correct. [SSM] onsider a circuit consisting solely of an ideal inductor and an ideal capacitor. How does the maximum energy stored in the capacitor compare to the maximum value stored in the inductor? (a) They are the same and each equal to the total energy stored in the circuit. (b) They are the same and each

74 hapter 9 equal to half of the total energy stored in the circuit. (c) The maximum energy stored in the capacitor is larger than the maximum energy stored in the inductor. (d) The maximum energy stored in the inductor is larger than the maximum energy stored in the capacitor. (e) You cannot compare the maximum energies based on the data given because the ratio of the maximum energies depends on the actual capacitance and inductance values. Determine the oncept The maximum energy stored in the electric field of the Q capacitor is given byu e and the maximum energy stored in the magnetic field of the inductor is given by U m. Because energy is conserved in an circuit and oscillates between the inductor and the capacitor, U e U m U total. is correct. ( a) True or false: (a) A driven series circuit that has a high Q factor has a narrow resonance curve. (b) A circuit consists solely of a resistor, an inductor and a capacitor, all connected in series. f the resistance of the resistor is doubled, the natural frequency of the circuit remains the same. (c) At resonance, the impedance of a driven series combination equals the resistance. (d) At resonance, the current in a driven series circuit is in phase with the voltage applied to the combination. (a) True. The Q factor and the width of the resonance curve at half power are related according to Q ω Δω ; i.e., they are inversely proportional to each other. (b) True. The natural frequency of the circuit depends only on the inductance of the inductor and the capacitance of the capacitor and is given by ω. (c) True. The impedance of an circuit is given by + ( ) resonance and so. (d) True. The phase angle δ is related to and according to δ tan. At resonance and so δ.. At

Alternating-urrent ircuits 743 3 True or false (all questions related to a driven series circuit): (a) Near resonance, the power factor of a driven series circuit is close to zero. (b) The power factor of a driven series circuit does not depend on the value of the resistance. (c) The resonance frequency of a driven series circuit does not depend on the value of the resistance. (d) At resonance, the current of a driven series circuit does not depend on the capacitance or the inductance. (e) For frequencies below the resonant frequency, the capacitive reactance of a driven series circuit is larger than the inductive reactance. (f) For frequencies below the resonant frequency of a driven series circuit, the phase of the current leads (is ahead of) the phase of the applied voltage. (a) False. Near resonance, the power factor, given by is close to. cosδ, ( ) + (b) False. The power factor is given by cosδ. ( ) + (c) True. The resonance frequency for a driven series circuit depends only on and and is given by ω res (d) True. At resonance and so and the current is given by app,. (e) True. Because the capacitive reactance varies inversely with the driving frequency and the inductive reactance varies directly with the driving frequency, at frequencies well below the resonance frequency the capacitive reactance is larger than the inductive reactance. (f) True. For frequencies below the resonant frequency, the circuit is more capacitive than inductive and the phase constant φ is negative. This means that the current leads the applied voltage. 4 You may he noticed that sometimes two radio stations can be heard when your receiver is tuned to a specific frequency. This situation often occurs when you are driving and are between two cities. Explain how this situation can occur.

744 hapter 9 Determine the oncept Because the power curves received by your radio from two stations he width, you could he two frequencies overlapping as a result of receiving signals from both stations. 5 True or false (all questions related to a driven series circuit): (a) At frequencies much higher than or much lower than the resonant frequency of a driven series circuit, the power factor is close to zero. (b) The larger the resonance width of a driven series circuit is, the larger the Q factor for the circuit is. (c) The larger the resistance of a driven series circuit is, the larger the resonance width for the circuit is. (a) True. Because the power factor is given by cosδ ω ω +, for values of ω that are much higher or much lower than the resonant frequency, the term in parentheses becomes very large and cosδ approaches zero. (b) False. When the resonance curve is reasonably narrow, the Q factor can be approximated by Q ω Δω. Hence a large value for Q corresponds to a narrow resonance curve. (c) True. See Figure 9-. 6 An ideal transformer has N turns on its primary and N turns on its secondary. The erage power delivered to a load resistance connected across the secondary is P when the primary voltage is. The current in the primary windings can then be expressed as (a) P /, (b) (N /N )(P / ), (c) (N /N )(P / ), (d) (N /N )(P / ). Picture the Problem et the subscript denote the primary and the subscript the secondary. Assuming no loss of power in the transformer, we can equate the power in the primary circuit to the power in the secondary circuit and solve for the current in the primary windings. Assuming no loss of power in the transformer: Substitute for P and P to obtain: P P,,,,

Alternating-urrent ircuits 745 Solving for, and simplifying yields:,,,, ( a) is correct. P, 7 [SSM] True or false: (a) A transformer is used to change frequency. (b) A transformer is used to change voltage. (c) f a transformer steps up the current, it must step down the voltage. (d) A step-up transformer, steps down the current. (e) The standard household wall-outlet voltage in Europe is, about twice that used in the United States. f a European treler wants her hair dryer to work properly in the United States, she should use a transformer that has more windings in its secondary coil than in its primary coil. (f) The standard household wall-outlet voltage in Europe is, about twice that used in the United States. f an American treler wants his electric razor to work properly in Europe, he should use a transformer that steps up the current. (a) False. A transformer is a device used to raise or lower the voltage in a circuit. (b) True. A transformer is a device used to raise or lower the voltage in a circuit. (c) True. f energy is to be conserved, the product of the current and voltage must be constant. (d) True. Because the product of current and voltage in the primary and secondary circuits is the same, increasing the current in the secondary results in a lowering (or stepping down) of the voltage. (e) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the voltage in order to make her hair dryer work properly. (f) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the current (and decrease the voltage) in order to make his razor work properly. Estimation and Approximation 8 The impedances of motors, transformers, and electromagnets include both resistance and inductive reactance. Suppose that phase of the current to a large industrial plant lags the phase of the applied voltage by 5 when the plant

746 hapter 9 is under full operation and using.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 6 Hz line voltage at the plant is 4 k. The resistance of the transmission line from the substation to the plant is 5. Ω. The cost per kilowatt-hour to the company that owns the plant is $.4, and the plant pays only for the actual energy used. (a) Estimate the resistance and inductive reactance of the plant s total load. (b) Estimate the current in the power lines and the voltage at the substation. (c) How much power is lost in transmission? (d) Suppose that the phase that the current lags the phase of the applied voltage is reduced to 8º by adding a bank of capacitors in series with the load. How much money would be sed by the electric utility during one month of operation, assuming the plant operates at full capacity for 6 h each day? (e) What must be the capacitance of this bank of capacitors to achieve this change in phase angle? Picture the Problem We can find the resistance and inductive reactance of the plant s total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the voltage at the substation by applying Kirchhoff s loop rule to the substation-transmission wires-load circuit. The power lost in transmission can be found from P trans trans. We can find the cost sings by finding the difference in the power lost in transmission when the phase angle is reduced to 8. Finally, we can find the capacitance that is required to reduce the phase angle to 8 by first finding the capacitive reactance using the definition of tanδ and then applying the definition of capacitive reactance to find. 5. Ω trans substation f 6 Hz 4 k δ 5 (a) elate the resistance and inductive reactance of the plant s total load to and δ: Express in te of the current in the power lines and the voltage at the plant: cosδ and sinδ

Alternating-urrent ircuits 747 Express the power delivered to the plant in te of, solve for :, and δ and P cosδ and P () cosδ Substitute to obtain: evaluate : cos P ( 4 k) δ cos 5 63Ω.3MW evaluate and : ( 63Ω).57 kω and ( 63Ω).7 kω cos 5 57Ω sin 5 66Ω (b) Use equation () to find the current in the power lines: ( 4k) 63A.3MW 63.4A cos 5 Apply Kirchhoff s loop rule to the circuit: sub trans tot Solve for sub : sub ( trans + tot ) evaluate sub : sub ( 63.4 A)( 5.Ω + 63Ω) 4.3 k (c) The power lost in transmission is: P ( 63.4A)( 5.Ω) trans trans.9 kw kw (d) Express the cost sings Δ in te of the difference in energy consumption (P 5 P 8 )Δt and the per-unit cost u of the energy: ( P P ) Δtu Δ 5 8

748 hapter 9 Express the power lost in transmission when δ 8 : P 8 8 trans Find the current in the transmission lines when δ 8 : ( 4 k) Evaluate : 8.3MW cos8 8 6.5A P P ( 6.5A) ( 5.Ω) 9. kw evaluate Δ: 8 h d $.4 Δ d month kw h (.9 kw 9. kw) 6 3 $ 8 (e) The required capacitance is given by: elate the new phase angle δ to the inductive reactance, the reactance due to the added capacitance, and the resistance of the load : πf tan δ tanδ Substituting for yields: evaluate : π ( 6 s ) 66Ω ( 57Ω) πf tanδ ( tan8 ) ( ) 33 μf - Alternating urrent in esistors, nductors, and apacitors 9 [SSM] A -W light bulb is screwed into a standard -- socket. Find (a) the current, (b) the current, and (c) the power. Picture the Problem We can use, and P P to find. P to find, to find (a) elate the erage power delivered by the source to the voltage across the bulb and the current through it: P P

Alternating-urrent ircuits 749 evaluate : W.8333 A.833 A (b) Express in te of : Substitute for and evaluate : (.8333A).8A.785 A (c) Express the maximum power in te of the maximum voltage and maximum current: evaluate : P P P (.785A) ( ) W A circuit breaker is rated for a current of 5 A at a voltage of. (a) What is the largest value of the current that the breaker can carry? (b) What is the maximum erage power that can be supplied by this circuit? Picture the Problem We can breaker can carry and P circuit. (a) Express in te of : to find the largest current the to find the erage power supplied by this ( ) 5 A A (b) elate the erage power to the current and voltage: P ( 5A)( ).8kW [SSM] What is the reactance of a.-μh inductor at (a) 6 Hz, (b) 6 Hz, and (c) 6. khz? Picture the Problem We can use any frequency. ω to find the reactance of the inductor at Express the inductive reactance as a function of f: ω πf (a) At f 6 Hz: ( 6s )(. mh).38ω π

75 hapter 9 (b) At f 6 Hz: ( 6s )(. mh) 3.77Ω (c) At f 6. khz: ( 6. khz)(.mh) 37.7Ω π π An inductor has a reactance of Ω at 8 Hz. (a) What is its inductance? (b) What is its reactance at 6 Hz? Picture the Problem We can use at any frequency. ω to find the inductance of the inductor (a) elate the reactance of the inductor to its inductance: ω πf πf Solve for and evaluate : Ω.99 H π ( 8s ).H (b) At 6 Hz: ( 6s )(.99 H).kΩ π 3 At what frequency would the reactance of a -μf capacitor equal the reactance of a.-μh inductor? Picture the Problem We can equate the reactances of the capacitor and the inductor and then solve for the frequency. Express the reactance of the inductor: Express the reactance of the capacitor: ω πf ω π f Equate these reactances to obtain: π f f πf π evaluate f: π ( μf)(.mh) f.6 khz 4 What is the reactance of a.-nf capacitor at (a) 6. Hz, (b) 6. khz, and (c) 6. MHz? Picture the Problem We can use capacitor at any frequency. ω to find the reactance of the

Alternating-urrent ircuits 75 Express the capacitive reactance as a function of f: ω π f (a) At f 6. Hz: π ( 6.s )(. nf).65 MΩ (b) At f 6. khz: π ( 6. khz)(. nf) 6.5 kω (c) At f 6. MHz: π ( 6. MHz)(. nf) 6.5Ω 5 [SSM] A -Hz ac generator that produces a emf of is connected to a -μf capacitor. Find (a) the current and (b) the current. Picture the Problem We can use / and /ω to express as a function of, f, and. Once we ve evaluate, we can use / to find. Express and : in te of Express the capacitive reactance: ω π f Substitute for and simplify to obtain: πf (a) evaluate : π ( s )( μf)( ) 5.mA 5mA (b) Express in te of : 5.mA 8mA 6 At what frequency is the reactance of a -μf capacitor (a). Ω, (b) Ω, and (c). mω?

75 hapter 9 Picture the Problem We can use the capacitor to the frequency. ω πf to relate the reactance of The reactance of the capacitor is given by: f ω π f π (a) Find f when. Ω: f π ( μf)(.ω) 6 khz (b) Find f when Ω: f π ( μf)( Ω).6 khz (c) Find f when. mω: f π ( μf)(. mω).6 MHz 7 A circuit consists of two ideal ac generators and a 5-Ω resistor, all connected in series. The potential difference across the terminals of one of the generators is given by (5. ) cos(ωt α), and the potential difference across the terminals of the other generator is given by (5. ) cos(ωt + α), where α π/6. (a) Use Kirchhoff s loop rule and a trigonometric identity to find the current in the circuit. (b) Use a phasor diagram to find the current in the circuit. (c) Find the current in the resistor if α π/4 and the amplitude of is increased from 5. to 7.. Picture the Problem We can use the trigonometric identity cos θ + cosφ cos ( θ + φ) cos ( θ φ) to find the sum of the phasors and and then use this sum to express as a function of time. n (b) we ll use a phasor diagram to obtain the same result and in (c) we ll use the phasor diagram appropriate to the given voltages to express the current as a function of time. (a) Applying Kirchhoff s loop rule to the circuit yields: Solve for to obtain: + +

Alternating-urrent ircuits 753 Use the trigonometric identity cos θ cosφ cos ( θ + φ) cos ( θ φ) to find + : + + ( 5.) [ cos( ωt α ) + cos( ωt + α )] ( 5) [ cos ( ωt ) cos ( α )] π 6 ( ) cos cosωt ( 8.66) cosωt Substitute for + and to obtain: ( 8.66) cosωt 5Ω (.35A) cosωt where.35 A (.346 A) cosωt (b) Express the magnitude of the current in : The phasor diagram for the voltages is shown to the right. r r 3 3 r r Use vector addition to find r : cos3 ( 5. ) r r 8.66 cos3 Substitute for r and to obtain: 8.66.346 A 5Ω and (.35A) cosωt where.35 A

754 hapter 9 (c) The phasor diagram is shown to the right. Note that the phase angle between r and r is now 9. r 9 o α δ α r Use the Pythagorean theorem to find r : r 8.6 r r r + ( 5. ) + ( 7. ) Express as a function of t: r cos( ω t + δ ) where δ 45 9 α α 45 ( ) 7. tan 45 5. 9.46.65rad evaluate : 8.6 cos 5Ω ( ωt +.65rad) (.34A) cos( ωt +.7 rad) Undriven ircuits ontaining apacitors, esistors and nductors 8 (a) Show that has units of inverse seconds by substituting S units for inductance and capacitance into the expression. (b) Show that ω / (the expression for the Q-factor) is dimensionless by substituting S units for angular frequency, inductance, and resistance into the expression. Picture the Problem We can substitute the units of the various physical quantitities in / and Q ω to establish their units.

Alternating-urrent ircuits 755 (a) Substitute the units for and in the expression and simplify to obtain: H F ( Ω s) s Ω s s (b) Substitute the units for ω,, and in the expression Q ω and simplify to obtain: s s A A s s A A no units 9 [SSM] (a) What is the period of oscillation of an circuit consisting of an ideal.-mh inductor and a -μf capacitor? (b) A circuit that oscillates consists solely of an 8-μF capacitor and a variable ideal inductor. What inductance is needed in order to tune this circuit to oscillate at 6 Hz? Picture the Problem We can use T π/ω and ω f) to and. to relate T (and hence (a) Express the period of oscillation of the circuit: π T ω For an circuit: ω Substitute for ω to obtain: evaluate T: T T π () (.mh)( F).3ms π μ (b) Solve equation () for to obtain: T 4π 4π f evaluate : 4π ( 6s ) ( 8 μf) 88mH 3 An circuit has capacitance and inductance. A second circuit has capacitance and inductance, and a third circuit has capacitance and inductance. (a) Show that each circuit oscillates with the same frequency. (b) n which circuit would the current be greatest if the voltage across the capacitor in each circuit was the same?

756 hapter 9 Picture the Problem We can use the expression f π for the resonance frequency of an circuit to show that each circuit oscillates with the same frequency. n (b) we can use ωq, where Q is the charge of the capacitor at time zero, and the definition of capacitance Q to express in te of ω, and. Express the resonance frequency for an circuit: f π (a) Express the product of and ircuit :,, and ircuit 3: 33 ( )( ) for each circuit: ircuit : ( )( ) Because the same. (b) Express 33, the resonance frequencies of the three circuits are in te of the ωq charge stored in the capacitor: Express Q in te of the capacitance of the capacitor and the potential difference across the capacitor: Substituting for Q yields: Q ω or, for ω and constant,. Hence, the circuit with capacitance has the greatest current. 3 A 5.-μF capacitor is charged to 3 and is then connected across an ideal -mh inductor. (a) How much energy is stored in the system? (b) What is the frequency of oscillation of the circuit? (c) What is the current in the circuit? Picture the Problem We can use U to find the energy stored in the electric field of the capacitor, ω πf to find f, and ωq and Q to find.

Alternating-urrent ircuits 757 (a) Express the energy stored in the system as a function of and : U evaluate U: U ( 5. F)( 3).3mJ μ (b) Express the resonance frequency of the circuit in te of and : ω πf f π evaluate f : π ( mh)( 5. μf) f.7khz 7Hz (c) Express in te of the ωq charge stored in the capacitor: Express Q in te of the capacitance of the capacitor and the potential difference across the capacitor: Q Substituting for Q yields: evaluate : ω π ( 7s )( 5. μf)( 3).67 A 3 A coil with internal resistance can be modeled as a resistor and an ideal inductor in series. Assume that the coil has an internal resistance of. Ω and an inductance of 4 mh. A.-μF capacitor is charged to 4. and is then connected across coil. (a) What is the initial voltage across the coil? (b) How much energy is dissipated in the circuit before the oscillations die out? (c) What is the frequency of oscillation the circuit? (Assume the internal resistance is sufficiently small that has no impact on the frequency of the circuit.) (d) What is the quality factor of the circuit?

758 hapter 9 Picture the Problem n Part (a) we can apply Kirchhoff s loop rule to find the initial voltage across the coil. (b) The total energy lost via joule heating is the total energy initially stored in the capacitor. (c) The natural frequency of the circuit is given by f π. n Part (d) we can use its definition to find the quality factor of the circuit. (a) Application of Kirchhoff s loop rule leads us to conclude that the initial voltage across the coil is 4.. (b) Because the ideal inductor can not dissipate energy as heat, all of the energy initially stored in the capacitor will be dissipated as joule heat in the resistor: U μ.576 mj (. F)( 4. ) (c) The natural frequency of the circuit is: π π ( 4 mh)(.μf) f 78 Hz (d) The quality factor of the circuit is given by: ω Q Substituting for ω and simplifying yields: Q evaluate Q: Q. Ω 4 mh. μf 447 33 [SSM] An inductor and a capacitor are connected, as shown in Figure 9-3. nitially, the switch is open, the left plate of the capacitor has charge Q. The switch is then closed. (a) Plot both Q versus t and versus t on the same graph, and explain how it can be seen from these two plots that the current leads the charge by 9º. (b) The expressions for the charge and for the current are given by Equations 9-38 and 9-39, respectively. Use trigonometry and algebra to show that the current leads the charge by 9º. Picture the Problem et Q represent the instantaneous charge on the capacitor and apply Kirchhoff s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect

Alternating-urrent ircuits 759 to time, an expression for the current as a function of time. We ll use a spreadsheet program to plot the graphs. Apply Kirchhoff s loop rule to a clockwise loop just after the switch is closed: Because dq dt : The solution to this equation is: Because Q() Q, δ and: Q d + dt d Q Q d Q + or + Q dt dt Q t) Q cos t ( where ω ( ω δ ) Q t) Q cosωt ( The current in the circuit is the [ Q cos t] Q sin t derivative of Q with respect to t: (a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time.,, and Q were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 9. dq dt d dt ω ω ω...6 harge urrent.6 Q (m).. (ma) -.6 -.6 -. -. 4 6 8 t (s) (b) The equation for the current is: ωq sinωt () The sine and cosine functions are related through the identity: π sinθ cos θ +

76 hapter 9 Use this identity to rewrite equation π (): ωq sinωt ωq cos ωt + Thus, the current leads the charge by 9. Driven ircuits 34 A circuit consists of a resistor, an ideal.4-h inductor and an ideal 6-Hz generator, all connected in series. The voltage across the resistor is 3 and the voltage across the inductor is 4. (a) What is the resistance of the resistor? (b) What is the emf of the generator? Picture the Problem We can express the ratio of to and solve this expression for the resistance of the circuit. n (b) we can use the fact that, in an circuit, leads by 9 to find the ac input voltage. (a) Express the potential differences across and in te of the common current through these components: and ω Divide the second of these equations by the first to obtain: ω ω ω 3 4 ( 6s )(.4H).4kΩ evaluate : π (b) Because leads by 9 in an circuit: evaluate : + ( 3 ) + ( 4 ) 7 35 [SSM] A coil that has a resistance of 8. Ω has an impedance of Ω when driven at a frequency of. khz. What is the inductance of the coil? Picture the Problem We can solve the expression for the impedance in an circuit for the inductive reactance and then use the definition of to find.

Alternating-urrent ircuits 76 Express the impedance of the coil in te of its resistance and inductive reactance: + Solve for to obtain: Express in te of : πf Equate these two expressions to obtain: πf πf evaluate : ( Ω) ( 8.Ω) π (. khz) 9. mh 36 A two conductor transmission line simultaneously carries a superposition of two voltage signals, so the potential difference between the two conductors is given by +, where (. ) cos(ω t) and (. ) cos(ω t), where ω rad/s and ω rad/s. A. H inductor and a. kω shunt resistor are inserted into the transmission line as shown in Figure 9-3. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) What is the voltage ( out ) at the output of the transmission line? (b) What is the ratio of the low-frequency amplitude to the high-frequency amplitude at the output? Picture the Problem We can express the two output voltage signals as the product of the current from each source and. kω. We can find the currents due to each source using the given voltage signals and the definition of the impedance for each of them. (a) Express the voltage signals observed at the output side of the transmission line in te of the potential difference across the resistor:, out and, out

76 hapter 9 Evaluate and : and (.) cost - (. kω) + ( s )(.H) [ ] 4 (.) cos t 4 (. kω) + ( s )(.H) [ ] ( 9.95 ma) cos t Substitute for and to obtain:, out 4 (.995mA) cos t (. kω)( 9.95mA) ( 9.95) cost where ω rad/s and ω rad/s. and. kω.995 ma, out ( )( ) 4 (.995 ) cos t cost cos 4 t (b) Express the ratio of,out to,out :, out, out 9.95.995 : 37 A coil is connected to a -, 6-Hz line. The erage power supplied to the coil is 6 W, and the current is.5 A. Find (a) the power factor, (b) the resistance of the coil, and (c) the inductance of the coil. (d) Does the current lag or lead the voltage? Explain your answer. (e) Support your answer to Part (d) by determining the phase angle. Picture the Problem The erage power supplied to coil is related to the power factor by P cosδ. n (b) we can use P to find. Because the inductance is related to the resistance and the phase angle δ according to ω tanδ, we can use this relationship to find the resistance of the coil. Finally, we can decide whether the current leads or lags the voltage by noting that the circuit is inductive. (a) Express the erage power supplied to the coil in te of the power factor of the circuit: P cosδ cosδ P 6 W cos δ.333 evaluate cosδ: ( )(.5A).33

Alternating-urrent ircuits 763 (b) Express the power supplied by the source in te of the resistance of the coil: P P evaluate : (.5A) 6 W 6.7Ω 7Ω (c) elate the inductive reactance to the resistance and phase angle: ω tanδ [ ] Solving for yields: tanδ tan cos (.333) ω πf evaluate : ( 6.7Ω) tan( 7.5 ) π ( 6s ).H (d) Evaluate : ( 6.7 Ω) tan( 7.5 ) 75. 4Ω Because the circuit is inductive, the current lags the voltage. (e) From Part (a): δ cos (.333) 7 38 A 36-mH inductor that has a resistance of 4 Ω is connected to an ideal ac voltage source whose output is given by (345 ) cos(5πt), where t is in seconds. Determine (a) the current in the circuit, (b) the and voltages across the inductor, (c) the erage power dissipation, and (d) the and erage magnetic energy stored in the inductor. + ω and, ω to find the current in the circuit and the voltage across the inductor. (b) Once we ve found we can find using Picture the Problem (a) We can use ( ),,,,. (c) We can use P to find the erage power dissipation, and (d) U, to find the and erage magnetic energy stored in the inductor. The erage energy stored in the magnetic field of the inductor can be found using U P dt,.

764 hapter 9 (a) Apply Kirchhoff s loop rule to the circuit to obtain: + ( ω ) evaluate : and ( 345) cos( 5πt ) ( 4Ω) + ( 5π s )( 36mH) ( 7.94A) cos( 5πt ) 7.9 A. [ ] (b) Because (345 ) cos(5πt) :, 345 Find from :,, 345,, 44 (c) elate the erage power dissipation to and : evaluate : P P P ( 7.94 A) ( 4Ω).3kW (d) The maximum energy stored in the magnetic field of the inductor is: U,.J ( 36mH)( 7.94A) The definition of U, is: U, T T U ()dt t U(t) is given by: U () t [ () t ] Substitute for U(t) to obtain: U, T T [ () t ] dt Evaluating the integral yields: U, T T 4 evaluate : U, U, 4 ( 36mH)( 7.94A).57 J

Alternating-urrent ircuits 765 39 [SSM] A coil that has a resistance and an inductance has a power factor equal to.866 when driven at a frequency of 6 Hz. What is the coil s power factor it is driven at 4 Hz? Picture the Problem We can use the definition of the power factor to find the relationship between and when the coil is driven at a frequency of 6 Hz and then use the definition of to relate the inductive reactance at 4 Hz to the inductive reactance at 6 Hz. We can then use the definition of the power factor to determine its value at 4 Hz. Using the definition of the power cosδ () factor, relate and : + Square both sides of the equation cos δ to obtain: + ( ) Solve for 6Hz : Substitute for cosδ and simplify to obtain: 6 cos δ ( Hz) ( Hz) (.866) 6 3 Use the definition of to obtain: ( f ) 4πf and ( f' ) 4πf' Dividing the second of these equations by the first and simplifying yields: or ( f' ) 4πf' f' ( f ) 4 π f f ( f' ) ( f ) f' f Substitute numerical values to 4s 6s 6 3 obtain: ( 4Hz) ( 6Hz) 6 3

766 hapter 9 Substitute in equation () to obtain: ( cosδ ) 4 Hz 6 + 3 3 9.397 4 A resistor and an inductor are connected in parallel across an ideal ac voltage source whose output is given by cosωt as shown in Figure 9-3. Show that (a) the current in the resistor is given by / cos ωt, (b) the current in the inductor is given by / cos(ωt 9º), and (c) the current in the voltage source is given by + cos(ωt δ), where max /. Picture the Problem Because the resistor and the inductor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through the resistor and the current through the inductor. Because these two currents are not in phase, we ll need to use phasors to calculate their sum. The amplitudes of the applied voltage and the currents are equal to the magnitude of the phasors. That is r, r,, r, and r,. (a) The ac source applies a voltage given by cosωt. Thus, the voltage drop across both the load resistor and the inductor is: cosωt The current in the resistor is in phase with the applied voltage:, cosωt Because, : t cosω (b) The current in the inductor lags the applied voltage by 9 : ( ) cos 9, ω t Because, : cos( ωt 9 ) (c) The net current is the sum of the currents through the parallel branches: +

Draw the phasor diagram for the circuit. The projections of the phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, and the current in the inductor lags the applied voltage by 9. The net current phasor is the sum of the r r r branch current phasors ( + ). The current through the parallel combination is equal to, where is the impedance of the combination: From the phasor diagram we he: Solving for yields: Alternating-urrent ircuits 767 r δ cos t where r ωt ( ω δ ),, r 9 ωt +, + + where +. where r ω t δ + From the phasor diagram: cos t where ( ω δ ) tanδ,, 4 [SSM] Figure 9-33 shows a load resistor that has a resistance of. Ω connected to a high-pass filter consisting of an inductor that has inductance 3.-mH and a resistor that has resistance 4.-Ω. The output of the ideal ac generator is given by ( ) cos(πft). Find the currents in all three branches of the circuit if the driving frequency is (a) 5 Hz and (b) Hz. Find the fraction of the total erage power supplied by the ac generator that is delivered to the load resistor if the frequency is (c) 5 Hz and (d) Hz.

768 hapter 9 Picture the Problem +, where is the voltage drop across and is r r r the voltage drop across the parallel combination of and. + is the r r r relation for the phasors. For the parallel combination +. Also, is in phase with and is in phase with. First draw the phasor diagram for the currents in the parallel combination, then add the phasors for the voltages to the diagram. The phasor diagram for the currents in the circuit is: r r δ Adding the voltage phasors to the diagram gives: r r r r δ ωt r δ r The maximum current in the inductor,,, is given by: r,, () + where () tan δ is given by: tanδ,,,, ω πf Solve for δ to obtain: δ tan πf (3)

Alternating-urrent ircuits 769 Apply the law of cosines to the triangle formed by the voltage phasors to obtain: or + + cosδ,, + +,, cosδ Dividing out the current squared yields: Solving for yields: + + cosδ + + cosδ (4) The maximum current circuit is given by: in the (5) is related to to: according (6) (a) Substitute numerical values in equation (3) and evaluate δ : δ tan tan π ( 5Hz)( 3. mh).ω.53ω.ω 63.3 Solving equation () for yields: + evaluate : ( ).Ω + (.53Ω ) 8.98Ω evaluate : ( 4.Ω) + ( 8.98Ω) + ( 4.Ω)( 8.98Ω) cos63.3.36ω Substitute numerical values in equation (5) and evaluate :.36Ω 8.86 A

77 hapter 9 Substitute for in equation (6) and evaluate : The maximum and values of are given by: ( 8.86 A) 6.3A, and, ( 8.86 A)( 8.98 Ω) 79.95, ( 79.95 ) 55.99 The values of are:,, and 55.99.Ω,, and 55.99.53Ω,,.8A 5.53A (b) Proceed as in (a) with f Hz to obtain: 4. Ω, δ 6. 4, 7. 9Ω,. 6Ω, 4.64A, and 3.8A, 83., 58.7,,max,.94A, and.46a,, (c) The power delivered by the ac source equals the sum of the power dissipated in the two resistors. The fraction of the total power delivered by the source that is dissipated in load resistor is given by: P P P + + + P P, Substitute numerical values for f 5 Hz to obtain: P P + P f 5 Hz + ( 6.3A) ( 4.Ω) (.8A) (.Ω).5 5.%

Alternating-urrent ircuits 77 (d) Substitute numerical values for f Hz to obtain: P P + P f Hz + ( 3.8A) ( 4.Ω) (.94A) (.Ω).8 8.% 4 An ideal ac voltage source whose emf is given by ( ) cos(πft) and an ideal battery whose emf is 6 are connected to a combination of two resistors and an inductor (Figure 9-34), where Ω, 8. Ω, and 6. mh. Find the erage power delivered to each resistor if (a) the driving frequency is Hz, (b) the driving frequency is Hz, and (c) the driving frequency is 8 Hz. Picture the Problem We can treat the ac and dc components separately. For the dc component, acts like a short circuit. et, denote the value of the voltage supplied by the ac voltage source. We can use P to find the power dissipated in the resistors by the current from the ideal battery. We ll apply Kirchhoff s loop rule to the loop including,, and to derive an expression for the erage power delivered to each resistor by the ac voltage source. (a) The total power delivered to and is: P + () P, dc P, ac and P P + (), dc P, ac The dc power delivered to the resistors whose resistances are and is: P,dc and P,dc Express the erage ac power delivered to : P,ac,, Apply Kirchhoff s loop rule to a clockwise loop that includes,, and : Solving for yields:,,

77 hapter 9 Express the erage ac power delivered to : Substituting in equations () and () yields: P, ac P and P,, + +,,, evaluate P : evaluate P : P ( 6) ( ) + Ω ( Ω) 46 W P ( 6 ) ( ) ( 8.Ω) + 8.Ω ( 8.Ω) + π { s }{ 6. mh} [ ( ) ] 5 W (b) Proceed as in (a) to evaluate P and P with f Hz: P 5.6W +.W P 3.W + 3.W 46W 45W (c) Proceed as in (a) to evaluate P and P with f 8 Hz: P 5.6W +.W P 3.W +.64W 46W 34W 43 An ac circuit contains a resistor and an ideal inductor connected in series. The voltage drop across this series combination is - and the voltage drop across the inductor alone is 8. What is the voltage drop across the resistor? Picture the Problem We can use the phasor diagram for an circuit to find the voltage across the resistor.

Alternating-urrent ircuits 773 The phasor diagram for the voltages in the circuit is shown to the right: r Use the Pythagorean theorem to express : r evaluate : ( ) ( 8) 6 Filters and ectifiers 44 The circuit shown in Figure 9-35 is called an high-pass filter because it transmits input voltage signals that he high frequencies with greater amplitude than it transmits input voltage signals that he low frequencies. f the input voltage is given by in in cos ωt, show that the output voltage is out H cos(ωt δ) where ( ) H in + ω. (Assume that the output is connected to a load that draws only an insignificant amount of current.) Show that this result justifies calling this circuit a high-pass filter. Picture the Problem The phasor diagram for the high-pass filter is shown to the right. r app and r are the phasors for in and out, respectively. Note that tanδ. That δ is negative follows from the fact that r app lags r by δ. The projection of r app onto the horizontal axis is app in, and the projection of r onto the horizontal axis is out. δ ωt r r ωt δ r app Express app : Because δ < : app where app, cosωt app, and + () ωt + δ ωt δ

774 hapter 9 is given by: where cos t ( ω δ ),, H Solving equation () for and substituting for yields: + ω () out ( ) out, cos ωt δ cos( ωt δ ) Because : Using equation () to substitute for in in cos ( ωt δ ) in yields: out cos( ωt δ ) + ω Simplify further to obtain: As ω : or out out where H + in ( ω) cos t H + cos ( ω δ ) in ( ω) ( ) ( ωt δ ) in H in showing that + the result is consistent with the highpass name for this circuit. 45 (a) Find an expression for the phase constant δ in Problem 44 in te of ω, and. (b) What is the value of δ in the limit that ω? (c) What is the value of δ in the limit that ω? (d) Explain your answers to Parts (b) and (c).

Alternating-urrent ircuits 775 Picture the Problem The phasor diagram for the high-pass filter is shown below. r app and r are the phasors for in and out, respectively. The projection of r app onto the horizontal axis is app in, and the projection of r onto the horizontal axis is out. (a) Because r app lags r by δ. tan δ δ ωt r r r app Use the definition of to obtain: tanδ ω ω Solving for δ yields: δ tan ω (b) As ω : δ 9 (c) As ω : δ (d) For very low driving frequencies, >> and so r effectively lags r in by 9. For very high driving frequencies, << and so r is effectively in phase with r in. 46 Assume that in Problem 44, kω and 5 nf. (a) At what frequency is H? This particular frequency is known as the 3 db in frequency, or f 3dB for the circuit. (b) Using a spreadsheet program, make a graph of log ( H ) versus log (f), where f is the frequency. Make sure that the scale extends from at least % of the 3-dB frequency to ten times the 3-dB frequency. (c) Make a graph of δ versus log (f) for the same range of frequencies as in Part (b). What is the value of the phase constant when the frequency is equal to the 3-dB frequency? Picture the Problem We can use the results obtained in Problems 44 and 45 to find f 3 db and to plot graphs of log( out ) versus log(f) and δ versus log(f).

776 hapter 9 (a) Use the result of Problem 44 to express the ratio out / in : out in + in ( ω) in + ( ω) When / : out in + ( ω) Square both sides of the equation and solve for ω to obtain: ω ω f 3 db π evaluate f 3 db : B ( f π kω)( 5nF) 3 db.53khz (b) From Problem 44 we he: out + in ( ω) n Problem 45 it was shown that: δ tan ω ewrite these expressions in te in of f 3 db to obtain: out f 3 db + + πf f and f 3dB δ tan tan πf f A spreadsheet program to generate the data for a graph of out versus f and δ versus f is shown below. The formulas used to calculate the quantities in the columns are as follows: ell Formula/ontent Algebraic Form B.E+3 B.5E 8 B3 in B4 53 f 3 db A8 53.f 3 db

8 $B$3/SQT(+(($B$4/A8))^) Alternating-urrent ircuits 777 in f 3 db + f D8 OG(8) log( out ) E8 ATAN( $B$4/A8) f tan 3dB f F8 E8*8/P() δ in degrees A B D E F.E+4 ohms.5e 8 F 3 in 4 f 3 db 53 Hz 5 6 7 f log(f) out log( out ) delta(rad) delta(deg) 8 53.7.99.3.47 84.3 9 63.8.8.98.453 83. 73.86.36.865.434 8. 83.9.55.8.46 8. 55 53.7.7.54.793 45.4 56 533.73.79.5.783 44.9 57 543.73.75.46.774 44.3 53 583 3.7.995.. 5.7 53 593 3.7.995.. 5.7 533 533 3.7.995.. 5.7 534 533 3.73.995.. 5.7 The following graph of log( out ) versus log(f) was plotted for in.. -. log( out ) -.4 -.6 -.8 -..5..5 3. 3.5 4. log(f )

778 hapter 9 A graph of δ (in degrees) as a function of log(f) follows. - - -3 delta (deg) -4-5 -6-7 -8-9.5..5 3. 3.5 4. log(f ) eferring to the spreadsheet program, we see that when f f 3 db, δ 44.9. This result is in good agreement with its calculated value of 45.. 47 [SSM] A slowly varying voltage signal (t) is applied to the input of the high-pass filter of Problem 44. Slowly varying means that during one time constant (equal to ) there is no significant change in the voltage signal. Show that under these conditions the output voltage is proportional to the time derivative of (t). This situation is known as a differentiation circuit. Picture the Problem We can use Kirchhoff s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. Because the voltage drop across the resistor is small compared to the voltage drop across the capacitor, we can express the voltage drop across the capacitor in te of the input voltage. Apply Kirchhoff s loop rule to the input side of the filter to obtain: ( t) where is the potential difference across the capacitor. Substitute for () t Because Q : and to obtain: dq in cosω t c dt dq dt d dt [ ] d dt Substitute for dq/dt to obtain: d cosω t dt the differential equation describing the potential difference across the capacitor.

Alternating-urrent ircuits 779 Because there is no significant change in the voltage signal during one time constant: d d dt dt Substituting for d dt yields: in cosω t and cosωt in onsequently, the potential difference across the resistor is given by: d dt d dt [ cosωt] in 48 We can describe the output from the high-pass filter from Problem 44 using a decibel scale: β ( db) log ( H in ), where β is the output in decibels. Show that for H, β 3. db. The frequency at which in H in is known as f 3dB (the 3-dB frequency). Show that for f << f 3dB, the output β drops by 6 db if the frequency f is halved. Picture the Problem We can use the expression for H from Problem 44 and the definition of β given in the problem to show that every time the frequency is halved, the output drops by 6 db. From Problem 44: H or H in + in ( ω) + ( ω) Express this ratio in te of f and f 3 db and simplify to obtain: H + f f 3 db f 3 db f + f f 3 db For f << f 3dB : B H f 3 db f + f f 3 db f f 3 db From the definition of β we he: β log H

78 hapter 9 Substitute for H / to obtain: β log f f 3 db Doubling the frequency yields: The change in decibel level is: f β ' log f Δβ β' β log log 3 db f f log ( ) 6dB 3 db f f 3 db 49 [SSM] Show that the erage power dissipated in the resistor of the in high-pass filter of Problem 44 is given by P e + ω ( ) Picture the Problem We can express the instantaneous power dissipated in the resistor and then use the fact that the erage value of the square of the cosine function over one cycle is ½ to establish the given result.. The instantaneous power P(t) dissipated in the resistor is: P( t) out The output voltage is: cos( ωt δ ) out out H From Problem 44: H + in ( ω) Substitute in the expression for P(t) H to obtain: P( t) cos ( ωt δ ) Because the erage value of the square of the cosine function over in [ + ( ω) ] e one cycle is ½: + ( ω) P cos in [ ] ( ωt δ )

Alternating-urrent ircuits 78 5 One application of the high-pass filter of Problem 44 is a noise filter for electronic circuits (a filter that blocks out low-frequency noise). Using a resistance value of kω, find a value for the capacitance for the high-pass filter that attenuates a 6-Hz input voltage signal by a factor of. That is, so H in. Picture the Problem We can solve the expression for H from Problem 44 for the required capacitance of the capacitor. From Problem 44: H + in ( ω) We require that: or H in + + ( ω ) ( ω) Solving for yields: 99ω π 99f π evaluate : 99 ( kω )( 6Hz ) 3nF 5 [SSM] The circuit shown in Figure 9-36 is an example of a lowpass filter. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) f the input voltage is given by in in cos ωt, show that the output voltage is out cos(ωt δ) where in + ( ω). (b) Discuss the trend of the output voltage in the limiting cases ω and ω. Picture the Problem n the phasor diagram for the low-pass filter, r app and r are the phasors for in and out, respectively. The projection of r onto the horizontal axis is app app in, the projection of r onto the horizontal axis is out, r app, and φ is the angle between ωt φ δ r r r app r and the horizontal axis.

78 hapter 9 (a) Express app : app where in in cosωt and + () out is given by: f we define δ as shown in the phasor diagram, then: out out, cosφ cosφ cos in cos ( ωt δ ) ( ωt δ ) Solving equation () for and substituting for yields: + ω () Using equation () to substitute for ω in and substituting for yields: out cos( ωt δ ) + ω Simplify further to obtain: or out out where in cos t + ( ω) cos t ( ω δ ) in ( ω ) + ( ω δ ) (b) Note that, as ω,. This makes sense physically in that, for low frequencies, is large and, therefore, a larger input voltage will appear across it than appears across it for high frequencies. Note further that, as ω,. This makes sense physically in that, for high frequencies, is small and, therefore, a smaller voltage will appear across it than appears across it for low frequencies. emarks: n Figures 9-9 and 9-, δ is defined as the phase of the voltage drop across the combination relative to the voltage drop across the resistor.

Alternating-urrent ircuits 783 5 (a) Find an expression for the phase angle δ for the low-pass filter of Problem 5 in te of ω, and. (b) Find the value of δ in the limit that ω and in the limit that ω. Explain your answer. Picture the Problem The phasor diagram for the low-pass filter is shown to the right. r app and r are the phasors for in and out, respectively. The projection of r app onto the r r app horizontal axis is app in and the projection of r onto the horizontal axis is. r. out app ω t δ r (a) From the phasor diagram we he: tanδ Use the definition of to obtain: tan δ ω ω Solving for δ yields: δ tan ( ω) (b) As ω, δ. This behior makes sense physically in that, at low frequencies, is very large compared to and, as a consequence, is in phase with. in As ω, δ 9. This behior makes sense physically in that, at high frequencies, is very small compared to and, as a consequence, is out of phase with. in emarks: See the spreadsheet solution in the following problem for additional evidence that our answer for Part (b) is correct. 53 Using a spreadsheet program, make a graph of versus input frequency f and a graph of phase angle δ versus input frequency for the low-pass filter of Problems 5 and 5. Use a resistance value of kω and a capacitance value of 5. nf.

784 hapter 9 Picture the Problem We can use the expressions for and δ derived in Problems 5 and 5 to plot the graphs of versus f and δ versus f for the lowpass filter of Problem 5. We ll simplify the spreadsheet program by expressing both and δ as functions of f 3 db. From Problems 5 and 5 we he: ewrite each of these expressions in te of f to obtain: in + and δ tan ( ω ) ( ω) in + f and δ tan πf ( π ) ( ) A spreadsheet program to generate the data for graphs of versus f and δ versus f for the low-pass filter is shown below. Note that in has been arbitrarily set equal to. The formulas used to calculate the quantities in the columns are as follows: ell Formula/ontent Algebraic Form B.E+3 B 5.E 9 B3 in B8 $B$3/SQT(+((*P()*A8* in *$B$*$B$)^)) + ( πf) 8 ATAN(*P()*A8**$B$*$B$) tan ( πf) D8 8*8/P() δ in degrees A B D.E+4 ohms 5.E 9 F 3 in 4 5 6 f(khz) out δ(rad) δ (deg) 7... 8.954.34 7.4 9.847.56 3. 3.78.756 43.3

54 47.68.53 86. 55 48.66.55 86. 56 49.65.56 86.3 57 5.64.57 86.4 A graph of as a function of f follows: Alternating-urrent ircuits 785..8.6 ( ).4.. 3 4 5 f ( khz) A graph of δ as a function of f follows: 9 6 delta (deg) 3 3 4 5 f (khz) 54 A rapidly varying voltage signal (t) is applied to the input of the lowpass filter of Problem 5. apidly varying means that during one time constant (equal to ) there are significant changes in the voltage signal. Show that under these conditions the output voltage is proportional to the integral of (t) with respect to time. This situation is known as an integration circuit. Picture the Problem We can use Kirchhoff s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We ll then assume a solution to this equation that is a linear combination of sine and cosine te with coefficients that we can find by substitution in the differential equation. The solution to these simultaneous equations will yield the amplitude of the output voltage.

786 hapter 9 Apply Kirchhoff s loop rule to the input side of the filter to obtain: Substitute for (t) and to obtain: Because Q : Substitute for dq/dt to obtain: ( t) where is the potential difference across the capacitor. dq in cosω t dt dq dt d dt [ ] d dt d in cosω t dt the differential equation describing the potential difference across the capacitor. is given by: c ω The fact that (t) varies rapidly means that ω >> and so: and d cosω t dt Separating the variables in this differential equation and solving for yields: tdt cosω 55 [SSM] The circuit shown in Figure 9-37 is a trap filter. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) Show that the trap filter acts to reject signals in a band of frequencies centered at ω /. (b) How does the width of the frequency band rejected depend on the resistance? Picture the Problem The phasor diagram for the trap filter is shown below. r r r app and + are the phasors for in and out, respectively. The projection of r r r app onto the horizontal axis is app in, and the projection of + onto the horizontal axis is + out. equiring that the impedance of the trap be zero will yield the frequency at which the circuit rejects signals. Defining the bandwidth as Δω ω ωtrap and requiring that trap will yield an expression for the bandwidth and reveal its dependence on.

Alternating-urrent ircuits 787 r ωt r app δ ω δ t r r + r r (a) Express app : out is given by: app app, cosωt where app, and ( ) out + () cos t out, where and trap out, ( ω δ ) trap Solving equation () for yields: ( ) Because + : out + () ( ) out out, cos ωt δ cos( ωt δ ) Using equation () to substitute for trap trap cos ( ωt δ ) yields: out trap cos( ωt δ ) + trap Noting that out provided trap, set trap to obtain: Substituting for and yields: trap ω ω ω (b) et the bandwidth Δω be: Δ ω ω (3) ω trap

788 hapter 9 et the frequency bandwidth be defined by the frequency at which trap. Then: ω ω ω ω Because ω trap : ω ω trap ω For ω ω trap : Solve for ω : trap ω ω ωtrap trap ω trap ω ω ω ( ω ω )( ω + ω ) Because ω ω trap, ω + ω trap ω trap : ω ωtrap ω trap( ω ωtrap ) trap trap trap Substitute in equation (3) to obtain: Δω ω ω ω trap trap 56 A half-we rectifier for transforming an ac voltage into a dc voltage is shown in Figure 9-38. The diode in the figure can be thought of as a one-way valve for current. t allows current to pass in the forward direction (the direction of the arrowhead) only when in is at a higher electric potential than out by.6 (that is, whenever in out +.6 ). The resistance of the diode is effectively infinite when in out is less than +.6. Plot two cycles of both input and output voltages as a function of time, on the same graph, assuming the input voltage is given by in in cos ωt. Picture the Problem For voltages greater than.6, the output voltage will mirror the input voltage minus a.6 drop. But when the voltage swings below.6, the output voltage will be. A spreadsheet program was used to plot the following graph. The angular frequency and the voltage were both arbitrarily set equal to one.

Alternating-urrent ircuits 789..5 _in _out..5 in ().. out () -.5 -.5 -. 4 6 8 -. t (s) 57 The output of the rectifier of Problem 56, can be further filtered by putting its output through a low-pass filter as shown in Figure 9-39a. The resulting output is a dc voltage with a small ac component (ripple) shown in Figure 9-39b. f the input frequency is 6 Hz and the load resistance is. kω, find the value for the capacitance so that the output voltage varies by less than 5 percent of the mean value over one cycle. Picture the Problem We can use the decay of the potential difference across the capacitor to relate the time constant for the circuit to the frequency of the input signal. Expanding the exponential factor in the expression for will allow us to find the approximate value for that will limit the variation in the output voltage by less than 5 percent (or any other percentage). The voltage across the capacitor is given by: in e t Expand the exponential factor to obtain: For a decay of less than 5 percent: e t t t.5 t Because the voltage goes positive every cycle, t /6 s and: Driven ircuits. kω 6 s 33μF 58 The generator voltage in Figure 9-4, is given by ( ) cos (πft). (a) For each branch, what is the current and what is the phase of the current relative to the phase of the generator voltage? (b) At the resonance frequency there is no current in the generator. What is the angular frequency at resonance? (c) At the resonance frequency, find the current in the inductor and what is the current in the capacitor. Express your results as

79 hapter 9 functions of time. (d) Draw a phasor diagram showing the phasors for the applied voltage, the generator current, the capacitor current, and the inductor current for the case where frequency is higher than the resonance frequency. Picture the Problem We know that the current leads the voltage across and capacitor and lags the voltage across an inductor. We can use, and to find the amplitudes of these currents. The current in the, generator will vanish under resonance conditions, i.e., when. To find the currents in the inductor and capacitor at resonance, we can use the common potential difference across them and their reactances together with our knowledge of the phase relationships mentioned above. (a) Express the amplitudes of the currents through the inductor and the capacitor:, and πf, πf πf Substitute numerical values to obtain: and 5. /H, lagging by 9 4.H πf πf, ( ) ( 5. μf)( ) ω (.5m F) πf, leading by, 9 (b) Express the condition that : or ω ω ω Solve for ω to obtain: ω ω evaluate ω: ( 4.H)( 5. μf) rad/s

Alternating-urrent ircuits 79 (c) Express the current in the inductor at ω ω : Express the current in the capacitor at ω ω : 5. /H π cos ω t s ( 5mA) cos ω t where ω rad/s. π (.5 m F)( s ) cos ωt + π ( 5mA) cos ωt + where ω rad/s. π (d) The phasor diagram for the case where the inductive reactance is larger than the capacitive reactance is shown to the right. r r r max 59 A circuit consist of an ideal ac generator, a capacitor and an ideal inductor, all connected in series. The charge on the capacitor is given by Q (5 μ) cos(ωt + π 4 ), where ω 5 rad/s. (a) Find the current in the circuit as a function of time. (b) Find the capacitance if the inductance is 8 mh. (c) Write expressions for the electrical energy U e, the magnetic energy U m, and the total energy U as functions of time. Picture the Problem We can differentiate Q with respect to time to find as a function of time. n (b) we can find by using ω. The energy stored in the magnetic field of the inductor is given by U m and the energy stored in Q the electric field of the capacitor by U e. r

79 hapter 9 (a) Differentiate the charge with respect to time to obtain the current: dq d π () t ( 5μ) cos ωt + ( 5μ)( 5s ) dt dt where ω 5 rad/s π 4 ( 8.75mA) sin ωt + ( 9mA) sin ωt + 4 π 4 sin ωt + π 4 (b) elate to and ω: ω Solve for to obtain: ω evaluate : ( 5s ) ( 8mH) 3μF (c) Express and evaluate the magnetic energy U m ( t) : U m π 4.86μF () t ( 8mH)( 8.75mA) sin ωt + ( 4.9μJ) sin ωt + where ω 5 rad/s π 4 Q Use U e to find the electrical energy stored in the capacitor as a function of time: U e () t ( 5μF) π cos ωt +.86 μf 4 ( 4.9μJ) π cos ωt + 4 ( 4.9μJ) cos ωt + where ω 5 rad/s π 4 The total energy stored in the electric and magnetic fields is the sum of : U e () t U m () t and U π + 4 t where ω 5 rad/s. π 4 ( 4.9μ J) sin ωt + ( 4.9 μj) cos ω + 4.9 μj

Alternating-urrent ircuits 793 6 One method for determining the compressibility of a dielectric material uses a driven circuit that has a parallel-plate capacitor. The dielectric is inserted between the plates and the change in resonance frequency is determined as the capacitor plates are subjected to a compressive stress. n one such arrangement, the resonance frequency is MHz when a dielectric of thickness. cm and dielectric constant κ 6.8 is placed between the plates. Under a compressive stress of 8 atm, the resonance frequency decreases to 6 MHz. Find the Young s modulus of the dielectric material. (Assume that the dielectric constant does not change with pressure.) Picture the Problem We can use the definition of the capacitance of a dielectricfilled capacitor and the expression for the resonance frequency of an circuit to derive an expression for the fractional change in the thickness of the dielectric in te of the resonance frequency and the frequency of the circuit when the dielectric is under compression. We can then use this expression for Δt/t to calculate the value of Young s modulus for the dielectric material. Use its definition to express Young s modulus of the dielectric material: Y stress strain ΔP Δt t () etting t be the initial thickness of the dielectric, express the initial capacitance of the capacitor: κ t A Express the capacitance of the capacitor when it is under compression: c κ A t Δt Express the resonance frequency of the capacitor before the dielectric is compressed: ω κ t A When the dielectric is compressed: ω c c κ A t Δt Express the ratio of ω c to ω and simplify to obtain: κ A ωc t Δt ω κ A t t Δt

794 hapter 9 Expand the radical binomially to obtain: ωc Δt Δt ω t t provided Δt << t. Solve for Δt/t: Δt t ω c ω Substitute in equation () to obtain: evaluate Y: Y Y ΔP ω c ω ( 8atm)(.35kPa/atm) 6 MHz MHz. 9 N/m 6 Figure 9-4 shows an inductor in series with a parallel plate capacitor. The capacitor has a width w of cm and a gap of. mm. A dielectric that has a dielectric constant of 4.8 can be slid in and out of the gap. The inductor has an inductance of. mh. When half the dielectric is between the capacitor plates (when x w ), the resonant frequency of this combination is 9 MHz. (a) What is the capacitance of the capacitor without the dielectric? (b) Find the resonance frequency as a function of x for x w. Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel. et be the capacitance of the dielectric-filled capacitor and be the capacitance of the air-filled capacitor. We ll derive expressions for the capacitances of the parallel capacitors and add these expressions to obtain (x). We can then use the given resonance frequency when x w/ and the given value for to evaluate. n Part (b) we can use our result for (x) and the relationship between f,, and (x) at resonance to express f(x). (a) Express the equivalent capacitance of the two capacitors in parallel: Express A in te of the total area of a capacitor plate A, w, and the distance x: κ A A ( x) + + d d A A x w x A A w ()

Alternating-urrent ircuits 795 Express A in te of A and A : w x A A A A Substitute in equation () and simplify to obtain: ( ) + + x w w x w x d A w x d A w x d A x κ κ κ κ κ d A where Find (w/): + κ κ κ κ κ κ κ w w w Express the resonance frequency of the circuit in te of and (x): ( ) ( ) x x f π () Evaluate f(w/): ( ) w f + + κ π κ π Solve for to obtain: ( ) + π κ w f evaluate : ( ) ( )( ) 5.4 ff 4.8. mh cm 9 MHz + π

796 hapter 9 (b) Substitute for (x) in equation () to obtain: evaluate f(x): f ( x) π κ κ x κw f ( x) π (.mh)( 4.8)( 5.39 F) (.m) 7MHz ( 4. )x 6 4.8 m 4.8 x Driven ircuits 6 A circuit consists of an ideal ac generator, a -μf capacitor and an 8-Ω resistor, all connected in series. The output of the generator has a emf of -, and the armature of the generator rotates at 4 rad/s. Find (a) the power factor, (b) the current, and (c) the erage power supplied by the generator. Picture the Problem The diagram shows the relationship between δ,,, and. We can use this reference triangle to express the power factor for the given circuit. n (b) we can find the current from the potential difference and the impedance of the circuit. We can find the erage power delivered by the source from the current and the resistance of the resistor. ( ) + δ (a) The power factor is defined to cosδ be: ( ) + With no inductance in the circuit: and cosδ + + ω

Alternating-urrent ircuits 797 evaluate cosδ: cosδ ( 8Ω).54 + 8Ω ( 4s ) ( μf) (b) Express the current in the circuit: max max + + ω evaluate : ( 8Ω) 95.3mA 95mA + ( 4s ) ( μf) (c) The erage power delivered by the generator is given by: evaluate : P P P ( 95.3mA) ( 8Ω).73W 63 [SSM] Show that the expression P gives the correct result for a circuit containing only an ideal ac generator and (a) a resistor, (b) a capacitor, and (c) an inductor. n the expression P, P is the erage power supplied by the generator, is the root-mean-square of the emf of the generator, is the resistance, is the capacitance and is the inductance. (n Part (a),, in Part (b), and in Part (c),. Picture the Problem The impedance of an ac circuit is given by +. We can evaluate the given expression for first for ( ) and then for. P (a) For, and: P

798 hapter 9 (b) and (c) f, then: ( ) P ( ) emarks: ecall that there is no energy dissipation in an ideal inductor or capacitor. 64 A series circuit that has an inductance of mh, a capacitance of. μf, and a resistance of 5. Ω is driven by an ideal ac voltage source that has a emf of. Find (a) the resonant frequency and (b) the root-mean-square current at resonance. When the frequency is 8 rad/s, find (c) the capacitive and inductive reactances, (d) the impedance, (e) the root-mean-square current, and (f) the phase angle. Picture the Problem We can use ω to find the resonant frequency of the circuit, to find the current at resonance, the definitions of and to find these reactances at ω 8 rad/s, the definitions of and to find the impedance and current at ω 8 rad/s, and the definition of the phase angle to find δ. (a) Express the resonant frequency ω in te of and : ω ω evaluate ω : ( mh)(. μf) 7. 3 rad/s (b) elate the current at resonance to and the impedance of the circuit at resonance: 4A max ( 5.Ω) (c) Express and evaluate and at ω 8 rad/s: and ω ( 8s )(. μf) 6.5 Ω 63Ω ω ( 8s )( mh) 8Ω

(d) Express the impedance in te of the reactances, substitute the results from (c), and evaluate : Alternating-urrent ircuits 799 ( ) + ( 5.Ω) + ( 8Ω 6.5Ω) 8.Ω 8Ω (e) elate the current at ω 8 rad/s to and the impedance of the circuit at this frequency: 3.9A max ( 8.Ω) (f) δ is given by: evaluate δ: δ tan 8Ω 6.5Ω δ tan 5.Ω 74 65 [SSM] Find (a) the Q factor and (b) the resonance width (in hertz) for the circuit in Problem 64. (c) What is the power factor when ω 8 rad/s? Picture the Problem The Q factor of the circuit is given byq ω, the resonance width by Δ f f Q ω πq, and the power factor by cos δ. Because is frequency dependent, we ll need to find and at ω 8 rad/s in order to evaluate cosδ. Using their definitions, express the Q factor and the resonance width of the circuit: ω Q () and f ω Δ f () Q πq (a) Express the resonance frequency for the circuit: ω Substituting for ω in equation () yields: Q evaluate Q: Q 5.Ω mh. μf 4. 4

8 hapter 9 (b) Substitute numerical values in Δf 7.7 equation () and evaluate Δf: π ( 4.) 3 rad/s 8Hz (c) The power factor of the circuit is given by: cosδ + ( ) + ω ω evaluate cosδ: cosδ 5. Ω ( 5. Ω) + ( 8s )( mh) ( 8s )(. μf).7 66 FM radio stations typically operate at frequencies separated by. MHz. Thus, when your radio is tuned to a station operating at a frequency of. MHz, the resonance width of the receiver circuit should be much smaller than. MHz, so that you do not receive a signal from stations operating at adjacent frequencies. Assume your receiving circuit has a resonance width of.5 MHz. When tuned in to this particular station, what is the Q factor of your circuit? Picture the Problem We can use its definition, Q f Δf to find the Q factor for the circuit. The Q factor for the circuit is given by: evaluate Q: Q Q f Δ f.mhz.5 MHz. 3 67 A coil is connected to a 6-Hz ac generator with a emf equal to. At this frequency, the coil has an impedance of Ω and a reactance of 8. Ω. (a) What is the current in the coil? (b) What is the phase angle between the current and the applied voltage? (c) A capacitor is put in series with the coil and the generator. What capacitance is required so that the current is in phase with the generator emf? (d) What is the voltage measured across this capacitor?

Picture the Problem We can use Alternating-urrent ircuits 8 to find the current in the coil and the definition of the phase angle to evaluate δ. We can equate and to find the capacitance required so that the current and the voltage are in phase. Finally, we can find the voltage measured across the capacitor by using. (a) Express the current in the coil in te of the potential difference across it and its impedance: evaluate : Ω A (b) The phase angle δ is given by: δ cos sin evaluate δ: 8.Ω δ sin Ω 53 (c) Express the condition on the reactances that must be satisfied if the current and voltage are to be in phase: ω ω πf 33 μf evaluate : π ( 6s )( 8.Ω).33mF (d) Express the potential difference across the capacitor: elate the current in the circuit to the impedance of the circuit when : Substitute for to obtain: πf elate the impedance of the circuit to the resistance of the coil: +

8 hapter 9 Substituting for yields: π f evaluate : π ( 6s )( 33μF) ( Ω) ( 8.Ω).3 k 68 An ideal.5-h inductor and a capacitor are connected in series with an ideal 6-Hz generator. An digital voltmeter is used to measure the voltages across the inductor and capacitor independently. The voltmeter reading across the capacitor is 75 and that across the inductor is 5. (a) Find the capacitance and the current in the circuit. (b) What is the voltage across the series combination of the capacitor and the inductor? Picture the Problem We can find using and from the potential difference across the inductor. n the absence of resistance in the circuit, the measured voltage across both the capacitor and inductor is. (a) elate the capacitance to the potential difference across the capacitor: πf πf Use the potential difference across the inductor to express the current in the circuit: πf Substitute for to obtain: ( πf ) evaluate : [ π ( 6s )] (.5H)( 75) 9μF 5 (b) Express the measured voltage across both the capacitor and the inductor when :

Alternating-urrent ircuits 83 5 75 5 evaluate : 69 [SSM] n the circuit shown in Figure 9-4 the ideal generator produces an voltage of 5 when operated at 6 Hz. What is the voltage between points (a) A and B, (b) B and, (c) and D, (d) A and, and (e) B and D? Picture the Problem We can find the current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge of the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations. (a) Express the potential difference between points A and B in te of and : () AB Express in te of and : + ( ) Evaluate and to obtain: πf π 5.648Ω and πf π 6.Ω ( 6s )( 37mH) ( 6s )( 5μF) evaluate : ( ) 5Ω + ( 5.648Ω 6.Ω ).5556A 5 Substitute numerical values in equation () and evaluate AB : AB (.5556 A)( 5.648Ω) 8 8.344 (b) Express the potential difference between points B and in te of and : B (.5556A)( 5Ω) 77.78 78

84 hapter 9 (c) Express the potential difference between points and D in te of and : D (.5556 A)( 6.Ω) 65.5.7 k (d) The voltage across the inductor leads the voltage across the resistor as shown in the phasor diagram to the right: r AB r A Use the Pythagorean theorem to find A : A AB + B r B ( 8. ) + ( 77.78 ).58.k (e) The voltage across the capacitor lags the voltage across the resistor as shown in the phasor diagram to the right: r B Use the Pythagorean theorem to find BD : BD r D D + B r BD ( 65.5 ) + ( 77.78 ) 8.46.8 k 7 When an series circuit is connected to a -, 6-Hz line, the current in the circuit is A and this current leads the line voltage by 45º. (a) Find the erage power supplied to the circuit. (b) What is the resistance in the circuit? (c) f the inductance in the circuit is 5 mh, find the capacitance in the circuit. (d) Without changing the inductance, by how much should you change the capacitance to make the power factor equal to? (e) Without changing the capacitance, by how much should you change the inductance to make the power factor equal to? Picture the Problem We can use P cosδ to find the power supplied to the circuit and P to find the resistance. n Part (c) we can relate the capacitive reactance to the impedance, inductive reactance, and resistance of the circuit and solve for the capacitance. We can use the condition on and at

Alternating-urrent ircuits 85 resonance to find the capacitance or inductance you would need to add to the circuit to make the power factor equal to. (a) Express the power supplied to the circuit in te of,, and the power factor cosδ: P cosδ evaluate : P P ( )( A).93kW cos45 933W (b) elate the power dissipated in the circuit to the resistance of the resistor: P P evaluate : ( A) 933W 7.7Ω 7.7Ω (c) Express the capacitance of the capacitor in te of its reactance: elate the capacitive reactance to the impedance, inductive reactance, and resistance of the circuit: () ω πf + ( ) Express the impedance of the circuit in te of the emf and the current : Equating these two expressions yields: + ( ) Solve for : Note that because leads, the circuit is capacitive and >. Hence: and ( )

86 hapter 9 + πf + evaluate : π + ( 6s )( 5mH) ( ) ( 7.7Ω) ( A) 8.8 Ω + 7.7 Ω 6.6Ω Substitute in equation () and 99.9μF evaluate : π ( 6s )( 6.6Ω).mF (d) et pf represent the capacitance required to make cosδ. The necessary change in capacitance is given by: Δ pf pf 99.9 μf elate pf to : πf pf Solving for pf yields: pf πf Substitute for for Δ to obtain: pf in the expression Δ πf 99.9μF Δ evaluate Δ: π ( 6s )( 8.8Ω) 4μF 99.9μF (e) et pf represent the inductance required to make cosδ. The necessary change in inductance is given by: Δ pf pf 5mH

Alternating-urrent ircuits 87 elate pf to : π fpf Solving for pf yields: pf πf Substitute for for Δ to obtain: pf in the expression Δ πf 5 mh 6.6Ω Δ 5 mh evaluate Δ: π ( 6s ) mh 7 Plot the circuit impedance versus the angular frequency for each of the following circuits. (a) A driven series circuit, (b) a driven series circuit, and (c) a driven series circuit. Picture the Problem The impedance for the three circuits as functions of the angular frequency is shown in the three figures below. Also shown in each figure (dashed line) is the asymptotic approach for large angular frequencies. (a) (b) (c) 7 n a driven series circuit, the ideal generator has a emf equal to, the resistance is 6. Ω and the capacitance is 8. μf. The inductance can be varied from 8. mh to 4. mh by the insertion of an iron core in the solenoid. The angular frequency of the generator is 5 rad/s. f the capacitor voltage is not to exceed 5, find (a) the current and (b) the range of inductances that is safe to use. Picture the Problem We can find the maximum current in the circuit from the maximum voltage across the capacitor and the reactance of the capacitor. To find the range of inductance that is safe to use we can express for the circuit in te of and and solve the resulting quadratic equation for. (a) Express the current in te of the maximum potential difference across the capacitor and its, ω,

88 hapter 9 reactance: evaluate : (b) elate the maximum current in the circuit to the emf of the source and the impedance of the circuit: Express in te of,, and : ( 5 rad/s)( 8. F)( 5 ) μ 3. A + ( ) Substitute to obtain: + ( ) Evaluating yields: ω 5.Ω 5 rad/s ( )( 8. μf) Substitute numerical values to obtain: Solving for yields: ( ) ( 3.A) ( 6.Ω) + (( 5 rad/s) 5.Ω) 5.Ω ± 844Ω 5s Denoting the solutions as + and, find the values for the inductance: The ranges for are: 3.6mH and 8.38mH + 8. mh < < 8.38 mh and 3.6 mh < < 4. mh 73 A certain electrical device draws an current of A at an erage power of 7 W when connected to a -, 6-Hz power line. (a) What is the impedance of the device? (b) What series combination of resistance and reactance would he the same impedance as this device? (c) f the current leads the emf, is the reactance inductive or capacitive?

Alternating-urrent ircuits 89 Picture the Problem We can find the impedance of the circuit from the applied emf and the current drawn by the device. n Part (b) we can use P to find and the definition of the impedance of a series circuit to find. (a) Express the impedance of the device in te of the current it draws and the emf provided by the power line: evaluate : A Ω (b) Use the relationship between the erage power supplied to the device and the current it draws to express : P P evaluate : ( A) 7 W 7.Ω 7.Ω The impedance of a series circuit is given by: ( ) + or + ( ) Solving for yields: evaluate : ( Ω) ( 7.Ω) Ω (c) f the current leads the emf, the reactance is capacitive. 74 A method for measuring inductance is to connect the inductor in series with a known capacitance, a known resistance, an ac ammeter, and a variablefrequency signal generator. The frequency of the signal generator is varied and the emf is kept constant until the current is maximum. (a) f the capacitance is μf, the emf is, the resistance is Ω, and the current in the circuit is maximum when the driving frequency is 5 rad/s, what is the value of the inductance? (b) What is the maximum current? Picture the Problem (a) We can use the fact that when the current is a maximum,

8 hapter 9, to find the inductance of the circuit. n Part (b), we can find, max from and the impedance of the circuit at resonance. (a) elate and at resonance: Solve for to obtain: or ω ω ω evaluate : ( 5s ) ( μf) (b) Noting that, at resonance,, express, max in te of the applied emf and the impedance of, max the circuit at resonance: ( Ω ) max 4.mH 7mA 75 A resistor and a capacitor are connected in parallel across an ac generator (Figure 9-43) that has an emf given by cos ωt. (a) Show that the current in the resistor is given by ( /) cos ωt. (b) Show that the current in the capacitor branch is given by ( / ) cos(ωt + 9º). (c) Show that the total current is given by cos(ωt + δ), where tan δ / and /. Picture the Problem Because the resistor and the capacitor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through the resistor and the current through the capacitor. Because these two currents are not in phase, we use phasors to calculate their sum. The amplitudes of the applied voltage and the currents are equal to the magnitude of the phasors. That is r, r, r,, and r,. (a) The ac source applies a voltage given by cosωt. Thus, the voltage drop across both the load resistor and the capacitor is: The current in the resistor is in phase with the applied voltage: cosωt, cosωt

Because, : t cosω Alternating-urrent ircuits 8 (b) The current in the capacitor leads the applied voltage by 9 : ( t + ) cos ω 9, Because, : cos( ωt + 9 ) (c) The net current is the sum of the currents through the parallel branches: + Draw the phasor diagram for the circuit. The projections of the phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, and the current in the capacitor leads the applied voltage by 9. The net current phasor is the sum of the r r r branch current phasors ( + ). r δ ωt r r r max The current through the parallel combination is equal to, where is the impedance of the combination: From the phasor diagram we he: cos t where ( ω δ ),, +, + + where + Solving for yields: where + From the phasor diagram: cos( ω t + δ ) where

8 hapter 9 tanδ 76 Figure 9-44 shows a plot of erage power P versus generator frequency ω for a series circuit driven by an ac generator. The erage power P is given by Equation 9-56. The full width at half-maximum, Δω, is the width of the resonance curve between the two points, where P is one-half its maximum value. Show that for a sharply ed resonance, Δω / and that Q ω /Δω in this case (Equation 9-58). Hint: The half-power points occur when the denominator of Equation 9-56 is equal to twice the value it has at resonance; ω ω + ω ω. et ω and ω be the solutions of this that is, when ( ) equation. Then, show that Δω ω ω /. Picture the Problem We can use the condition determining the half-power points to obtain a quadratic equation that we can solve for the frequencies corresponding to the half-power points. etting ω be the half-power frequency that is less than ω and ω be the half-power frequency that is greater than ω will lead us to the result that Δω ω ω /. We can then use the definition of Q to complete the proof that Q ω /Δω. Equation 9-56 is: P app, ( ω ω ) + ω ω The half-power points occur when the denominator of Equation 9-56 is twice the value near resonance; that is, when: or ( ω ω ) + ω ω [( ω ω )( ω + ω )] + ω ω For a sharply ed resonance, ω + ω ω. Hence: et ω be a solution to this equation. Noting that, for a sharply ed resonance, ω ω, it follows that: [( ω ω )( ω )] + ω ω or 4ω ω ω + ω ω ( ) ( ω ω ) + ω ω 4ω or, simplifying, ( ω ω ) 4

Alternating-urrent ircuits 83 Solving for ω yields: ω ω where we ve used the minus sign because ω < ω.

84 hapter 9 Similarly for ω : ω ω + where we ve used the plus sign because ω > ω. Evaluating Δω ω ω yields: From the definition of Q: Substitute in the expression for Δω to obtain: Δω ω + ω ω Q ω ω Δ Q Q ω Δω 77 Show by direct substitution that d Q + dq + dt dt Q t τ (Equation 9-43b) is satisfied by Q Q e cos ' t, whereτ, ( ) τ ω ω', and Q is the charge on the capacitor at t. Picture the Problem We ll differentiate t τ Q Q e cos ' t twice and substitute ω this function and both its derivatives in the differential equation of the circuit. ewriting the resulting equation in the form Acosω t + Bsinω t will reveal that B vanishes. equiring that Acosω t hold for all values of t will lead to the ω' τ. result that ( ) Equation 9-43b is: Assume a solution of the form: d Q dq + + Q dt dt Q Q e cos t τ ω ' t Differentiate Q(t) twice to obtain: dq dt Q Q e Q e d dt t τ [ e cosω' t] t t τ τ d d cosω' t + cosω' t e dt dt ω'sinω' t cosω' t τ t τ

Alternating-urrent ircuits 85 and d Q Q dt Q e d dt e t τ t τ ω'sin ω' t cosω' t τ ω' τ ω' cosω't+ sin ω't τ Substitute these derivatives in the differential equation and simplify to obtain: t τ ω' t τ Qe ω' cos + sin + 'sin ' cos ' ω't ω't Q e ω ω t ω t τ τ τ t τ + Qe cosω' t Because Q and to obtain: t e τ are never zero, divide them out of the equation and simplify ω' ω' cos + sin ' sin ' cos ' + cos ' ω't ω't ω ω t ω t ω t τ τ τ ewriting this equation in the form Acosω t + Bsinω t yields: or τ + τ ( ω ' ω' ) sinω't + ω' cosω't ω' cos + ω't τ τ f this equation is to hold for all values of t, its coefficient must vanish: ω' + τ τ Solving for ω yields: ω ', the condition that must be satisfied if t Q Q e τ cosω't is the solution to Equation 9-43b. 78 One method for measuring the magnetic susceptibility of a sample uses an circuit consisting of an air-core solenoid and a capacitor. The resonant frequency of the circuit without the sample is determined and then measured

86 hapter 9 again with the sample inserted in the solenoid. Suppose you he a solenoid that is 4. cm long, 3. mm in diameter, and has 4 turns of fine wire. You he a sample that is inserted in the solenoid and completely fills the air space. Neglect end effects. (a) alculate the inductance of your empty solenoid. (b) What value for the capacitance of the capacitor should you choose that the resonance frequency of the circuit without a sample is exactly 6. MHz? (c) When a sample is inserted in the solenoid, you determine that the resonance frequency drops to 5.9989 MHz. Use your data to determine the sample s susceptibility. Picture the Problem We can use μ n Al to determine the inductance of the empty solenoid and the resonance condition to find the capacitance of the samplefree circuit when the resonance frequency of the circuit is 6. MHz. By expressing as a function of f and then evaluating df /d and approximating the derivative with Δf /Δ, we can evaluate χ from its definition. (a) Express the inductance of an aircore solenoid: μ n Al evaluate : 7 4 π ( 4π N/A ) ( 3. cm) ( 4. cm) 3.553mH 3.55mH 4. cm 4 (b) Express the condition for resonance in the circuit: Solving for yields: π f () πf 4π f evaluate : 4π ( 6. MHz) ( 3.553mH).983 3 F.98 pf (c) Express the sample s susceptibility in te of and Δ: Solve equation () for f : Δ χ () f π

Alternating-urrent ircuits 87 Differentiate f with respect to : df d d π d 4π f 4π 3 Approximate df /d by Δf /Δ: Δf Δ f or Δf f Δ Substitute in equation () to obtain: Δf χ f evaluate χ: 5.9989 MHz 6. MHz χ 6. MHz 3.7 4 The Transformer 79 [SSM] A voltage of 4 is required for a device whose impedance is Ω. (a) What should the turns ratio of a transformer be, so that the device can be operated from a - line? (b) Suppose the transformer is accidentally connected in reverse with the secondary winding across the -- line and the -Ω load across the primary. How much current will then be in the primary winding? Picture the Problem et the subscript denote the primary and the subscript the secondary. We can use N N and N N to find the turns ratio and the primary current when the transformer connections are reversed. (a) elate the number of primary and secondary turns to the primary and secondary voltages: (), N, N Solve for and evaluate the ratio N, 4 N /N : N 5, (b) elate the current in the primary to the current in the secondary and to the turns ratio: N,, N

88 hapter 9 Express the current in the primary winding in te of the voltage across it and its impedance:,, Substitute for, to obtain:, N N, evaluate, : 5 Ω 5A 8 A transformer has 4 turns in the primary and 8 turns in the secondary. (a) s this a step-up or a step-down transformer? (b) f the primary is connected to a voltage source, what is the open-circuit voltage across the secondary? (c) f the primary current is. A, what is the secondary current, assuming negligible magnetization current and no power loss? Picture the Problem et the subscript denote the primary and the subscript the secondary. We can decide whether the transformer is a step-up or step-down transformer by examining the ratio of the number of turns in the secondary to the number of te in the primary. We can relate the open-circuit voltage in the secondary to the primary voltage and the turns ratio. (a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer. (b) elate the open-circuit voltage, in the secondary to the voltage, in the primary: evaluate, : (c) Because there are no power losses: N,, N 8 4 ( ).4,,,,, and,,,, evaluate, :.4 (. A) 5. A,

Alternating-urrent ircuits 89 8 The primary of a step-down transformer has 5 turns and is connected to a - line. The secondary is to supply A at 9.. Find (a) the current in the primary and (b) the number of turns in the secondary, assuming percent efficiency. Picture the Problem et the subscript denote the primary and the subscript the secondary. We can use to find the current in the primary and N,, N,,,, to find the number of turns in the secondary. (a) Because we he percent efficiency:,,,, and,,,, evaluate, : (b) elate the number of primary and secondary turns to the primary and secondary voltages: evaluate N /N : 9. ( A).5A,, N, N N, N N 9. ( 5) 9, 8 An audio oscillator (ac source) that has an internal resistance of Ω and an open-circuit output voltage of. is to be used to drive a loudser coil that has a resistance of 8. Ω. (a) What should be the ratio of primary to secondary turns of a transformer, so that maximum erage power is transferred to the ser? (b) Suppose a second identical ser is connected in parallel with the first ser. How much erage power is then supplied to the two sers combined? Picture the Problem Note: n a simple circuit maximum power transfer from source to load requires that the load resistance equals the internal resistance of the source. We can use Ohm s law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the ser(s). (a) Express the effective loudser resistance at the primary of the transformer: eff,,

8 hapter 9 elate, to,, N, and N :,, N N Express, in te of,, N, N,, and N : N Substitute for, and, and simplify to obtain: eff,, N N N N,, N N Solve for N /N : N N, eff eff (), Evaluate N /N for eff coil : N Ω 5.8 N 8.Ω 5.8 (b) Express the power delivered to the two sers connected in parallel: P () sp, eff Find the equivalent resistance sp the two 8.-Ω sers in parallel: of sp 8.Ω + 8.Ω sp 4.Ω Solve equation () for eff to obtain: N eff N evaluate : eff Find the current supplied by the source: Substitute numerical values in equation () and evaluate the power delivered to the parallel sers: ( 4.Ω)( 5.8) Ω eff, P tot 4. ma. Ω + Ω ( 4.mA) ( Ω) 6.mW sp

Alternating-urrent ircuits 8 General Problems 83 The distribution circuit of a residential power line is operated at. This voltage must be reduced to 4 for use within residences. f the secondary side of the transformer has 4 turns, how many turns are in the primary? Picture the Problem We can relate the input and output voltages to the number of turns in the primary and secondary using, N, N. elate the output voltages, to the input voltage, and the number of turns in the primary N and secondary N : evaluate N : N N N,, N,, N 4 3 ( 4 ) 3.33 84 A resistor that has a resistance carries a current given by (5. A) sin πft + (7. A) sin 4πft, where f 6 Hz. (a) What is the current in the resistor? (b) f Ω, what is the erage power delivered to the resistor? (c) What is the voltage across the resistor? to relate the current to the current carried by the resistor and find ( by integrating. Picture the Problem We can use its definition, ( ) ) (a) Express the current in te of the ( : ) ( ) Evaluate : ) [( 5. A) sin πft + ( 7. A) sin 4πft] ( 5A ) sin πft + ( 7 A ) sin πft sin 4πft + ( 49 A ) sin 4πft Find ( by integrating from t to t T π/ω and dividing by T: π ω ω ( ) {( 5A ) sin πft + ( 7 A ) π + ( 49 A ) sin 4πft}dt sin πft sin 4πft

8 hapter 9 Use the trigonometric identity sin x ( cos x) to simplify and evaluate the st and 3 rd integrals and recognize that the middle term is of the form sinxsinx to obtain: ( ).5A + + 4.5A 37. A Substitute for ( ) and evaluate : 37.A 6.A (b) elate the power dissipated in the resistor to its resistance and the current in it: evaluate P: (c) Express the voltage across the resistor in te of and : P P ( 6.8A) ( Ω).44kW evaluate : ( 6.8A)( Ω) 73 85 [SSM] Figure 9-45 shows the voltage versus time for a squarewe voltage source. f, (a) what is the voltage of this source? (b) f this alternating weform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what is the new voltage? Picture the Problem The erage of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT. We can use this definition to find both the erage of the voltage squared, ( ) and then use the definition of the voltage. (a) From the definition of he: we ( ) Noting that :, evaluate

Alternating-urrent ircuits 83 (b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval ΔT : Express the square of the voltage during this half cycle: alculate ( ) by integrating from t to t ΔT and dividing by ΔT: ΔT Δ ( ) T dt [] t ΔT ΔT Substitute to obtain: 8.5 86 What are the erage values and values of current for the two current wefo shown in Figure 9-46? Picture the Problem The erage of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT. We can use this definition to find both the erage current, and the erage of the current squared, ( ) From the definition of we he: and ΔT Δ T dt and ( ) Weform (a) Express the current during the first half cycle of time interval ΔT: 4 A a t ΔT where is in A when t and T are in seconds. Evaluate :, a, a ΔT 4. A 4. A Δ tdt Δ tdt T T 4. A t ( ΔT ) ΔT ( ΔT ).A ΔT

84 hapter 9 Express the square of the current during this half cycle: Noting that the erage value of the squared current is the same for each time interval ΔT, calculate ( a ) by integrating a from t to t ΔT and dividing by ΔT: a ( 4. A) ( ΔT ) ΔT ( ) ( 4. A) a t ΔT ( ΔT ) ( 4. A) 3 ( ΔT ) 3 t 3 ΔT t dt 6 3 A Substitute in the expression for to obtain:, a 6 A 3, a.3a Weform (b) Noting that the current during the second half of each cycle is zero, express the current during the first half cycle of the time interval ΔT : b 4.A Evaluate, b :, b ΔT 4 ΔT [] t.a 4.A Δ dt T ΔT.A Express the square of the current during this half cycle: alculate ( b ) by integrating b from t to t ΔT and dividing by ΔT: ( 4. ) A b ( ) ( 4.A) b Δ Δ dt T ( 4.A) [] ΔT ΔT t 8.A Substitute in the expression for to obtain:, b, b 8.A.8A 87 n the circuit shown in Figure 9-47, ( ) cos πft, where f 8 Hz; 8, and 36 Ω. Find the maximum, minimum, erage, and values of the current in the resistor. Picture the Problem We can apply Kirchhoff s loop rule to express the current in the circuit in te of the emfs of the sources and the resistance of the resistor.

We can then find max and min Alternating-urrent ircuits 85 by considering the conditions under which the time-dependent factor in will be a maximum or a minimum. Finally, we can use ( ) to derive an expression for that we can use to determine its value. Apply Kirchhoff s loop rule to obtain:, cosω t + Solving for yields:, cosω t + or A cosω t + A where, A and A Substitute numerical values to obtain: cos 36Ω ( π ( 8s ) t) 8 36Ω (.556 A) cos( 3s ) t +.5A + The current is a maximum when cos( 3s ) t. Hence : max.5a +.556 A.6A Evaluate min : min.5a.556a.6a Because the erage value of cosωt :.5A The current is the square root of the erage of the squared current: [ ] () [ ] is given by: [ ] [( A cosωt + A ) ] [ A cos ωt + A A cosωt + A ] [ A cos ωt] + [ A A cosωt] + [ A ] A [ cos + A A [ cosωt] + A ω t and Because [ cos ] [ cos t] ω : [ ] A + A

86 hapter 9 Substituting in equation () yields: A + A Substitute for A and A to obtain: + evaluate : 36 Ω 8 + 36 Ω.64 A 88 epeat Problem 87 if the resistor is replaced by a.-μf capacitor. Picture the Problem We can apply Kirchhoff s loop rule to obtain an expression for charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find max and min by considering the conditions under which the time-dependent factor in to derive an will be a maximum or a minimum. Finally, we can use ( ) expression for that we can use to determine its value. Apply Kirchhoff s loop rule to obtain: Solving for q(t) yields: ( t) q cosω t + q ( t) ( cosωt + ) A cosωt + A where A and A Differentiate this expression with respect to t to obtain the current as a function of time: Substituting numerical values yields: dq d dt dt ωa sinωt ( A cosωt + A ) ( 8 Hz)(. μf) sin( ( 8 Hz) t) (.6 ma) sin( 3s )t π π The current is a minimum when sin 3s t. Hence: ( ) min.3ma The current is a maximum when sin 3s t. Hence: ( ) max.3ma

Alternating-urrent ircuits 87

88 hapter 9 Because the dc source sees the capacitor as an open circuit and the erage value of the sine function over a period is zero: The current is the square root of the erage of the squared current: [ ] () [ ] is given by: [ ] [( A cosωt + A ) ] [ A cos ωt + A A cosωt + A ] [ A cos ωt] + [ A A cosωt] + [ A ] A [ cos + A A [ cosωt] + A ω t and Because [ cos ] [ cos t] ω : [ ] A + A Substituting in equation () yields: A + A Substitute for A and A to obtain: ( ) + ( ) ( ) + ( ) evaluate : (. μf) ( ) + ( 8 ) 46 μa 89 [SSM] A circuit consists of an ac generator, a capacitor and an ideal inductor all connected in series. The emf of the generator is given by cosωt. (a) Show that the charge on the capacitor obeys the equation d Q Q + cosωt. (b) Show by direct substitution that this equation is dt satisfied by Q Q cosωt where Q can be written as cos( ωt δ ), where ( ω ω ). (c) Show that the current ω ω ω δ 9º for ω < ω and δ 9º for ω > ω, where ω is the resonance frequency. and

Alternating-urrent ircuits 89 Picture the Problem n Part (a) we can apply Kirchhoff s loop rule to obtain the nd order differential equation relating the charge on the capacitor to the time. n Part (b) we ll assume a solution of the form Q Q cosωt, differentiate it twice, and substitute for d Q/dt and Q to show that the assumed solution satisfies the differential equation provided Q ( ω ω ). n Part (c) we ll use our results from (a) and (b) to establish the result for given in the problem statement. (a) Apply Kirchhoff s loop rule to obtain: Substitute for and rearrange the differential equation to obtain: d dt Q + Q d dt max cosωt Because dq dt : d Q + dt Q max cosωt (b) Assume that the solution is: Differentiate the assumed solution twice to obtain: dq d Q Substitute for and in the dt dt differential equation to obtain: Factor cosωt from the left-hand side of the equation: f this equation is to hold for all values of t it must be true that: Q Q dq dt and cosωt ωq sinωt d Q ω Q cosωt dt ω Q ω Q Q cosωt + cosωt cosωt Q + cosωt cosωt Q ω Q +

83 hapter 9 Solving for Q yields: Q ω + Factor from the denominator and substitute for / to obtain: Q ω + ( ω ω ) (c) From (a) and (b) we he: dq dt ω ωq ( ω ω ) where cos ω ω ω sinωt sinωt ( ωt δ ) ω ω sinωt ω ω ω f ω > ω, > and the current lags the voltage by 9 (δ 9 ). f ω < ω, < and the current leads the voltage by 9 (δ 9 ).