# Chapter 29 Alternating-Current Circuits

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1 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

3 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

4 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

5 Alternating-urrent ircuits 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.

6 744 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 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,,,,

7 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

8 746 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

9 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

10 748 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 cos A 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

11 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

12 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

13 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?

14 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: + +

15 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: A 5Ω and (.35A) cosωt where.35 A

16 754 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 rad 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.

17 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?

18 756 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.

19 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?

20 758 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 [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

21 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) 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 +

22 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.

23 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

24 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 : 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

25 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,.

26 764 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

27 Alternating-urrent ircuits [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

28 766 hapter 9 Substitute in equation () to obtain: ( cosδ ) 4 Hz 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: +

29 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.

30 768 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)

31 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) cos ω Substitute numerical values in equation (5) and evaluate :.36Ω 8.86 A

32 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, ( ) The values of are:,, and Ω,, and Ω,,.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.%

33 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:,,

34 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.

35 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

36 774 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).

37 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).

38 776 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

39 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 f log(f) out log( out ) delta(rad) delta(deg) The following graph of log( out ) versus log(f) was plotted for in.. -. log( out ) log(f )

40 778 hapter 9 A graph of δ (in degrees) as a function of log(f) follows delta (deg) log(f ) eferring to the spreadsheet program, we see that when f f 3 db, δ This result is in good agreement with its calculated value of [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.

41 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

42 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 )

43 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.

44 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.

45 Alternating-urrent ircuits (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.

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