rof. Anchordoqui roblems set # hysics 69 May 5, 5. A semicircular conductor of radius.5 m is rotated about the axis A at a constant rate of rev/min (Fig. ). A uniform magnetic field in all of the lower half of the figure is directed out of the plane of rotation and has a magnitude of.3 T. (i) alculate the maximum value of the emf induced in the conductor. (ii) What is the value of the average induced emf for each complete rotation? (iii) How would the answers to (i) and (ii) change if B were allowed to extend a distance above the axis of rotation? Sketch the emf versus time (iv) when the field is as drawn in Fig. and (v) when the field is extended as described in (iii). Solution (i) ε max BAω Bπ ω/.3 T π (.5 m) 4.π rad/s.6. (ii) ε π π ε dθ BAω π π sin θ dθ. (iii) The maximum and average ε would remain unchanged. (iv) See Fig.. (v) See Fig... An A power supply produces a maximum voltage of. This power supply is connected to a 4. Ω resistor, and the current and resistor voltage are measured with an ideal A ammeter and an ideal A voltmeter, as shown in Fig. 3 What does each meter read? ecall that an ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance. Solution The meters measure the rms values of potential difference and current. rms 7.7 and rms rms.95 A. These are 3. (i) For the series connection of Fig. 4, draw a phasor diagram for the voltages. The amplitudes of the voltage drop across all the circuit elements involved should be represented with phasors. (ii) An circuit consists of a 5-Ω resistor, a -µf capacitor and a 46-mH inductor, connected in series with a -, 6-Hz power supply. What is the phase angle between the current and the applied voltage? (iii) Which reaches its maximum earlier, the current or the voltage? Solution (i) For the series connection, the instantaneous voltage across the system is equal to the sum of voltage across each element. The phase angle between the voltage (across the system) and the current (through the system) is: φ arctan X X. n Fig. 5 the phasor diagram for a series circuit is shown for both the inductive case X > X and the capacitive case X < X. On the one hand, in the inductive case,, >,c, we see that leads by a phase φ. On the other hand, in the capacitance case,, >,, we have that leads by a phase φ. (ii) From the definition, the inductive reactance of the inductor is X ω π 6 Hz.46 H 73 Ω. From the definition, the capacitive reactance of the capacitor is X ω 6 Ω. π 6 Hz 6 F The phase angle between the voltage (across the system) and the current (through the system) is: φ arctan X X arctan 74 Ω 6 Ω 5 Ω 7.4 (iii) The voltage leads the current. 4. Figure 6 shows a parallel circuit. The instantaneous voltage (and rms voltage) across each of the three circuit elements is the same, and each is in phase with the current through the
resistor. The currents in the capacitor and the inductor lead or lag behind the current in the resistor. (i) Show that the rms current delivered by the source is rms rms + ( ] [ ω /. ω) (ii) Show that the phase angle between the voltage and the current is tan ϕ ( X X ). (iii) Draw a phasor diagram for the currents. The amplitudes of the currents across all the circuit elements involved should be represented with phasors. Solution We are asked to find the impedance and the phase angle for this system of elements connected in parallel. t will be easier to analyze the complex impedance. For elements connected in parallel the equivalent complex impedance of the system is equal to the sum of inverse impedance of each element Z eq Z + Z + Z + iω + iω + i ( ω ω i(ω ω). (i) Hence ( ) +(ω ω) rms Z eq ( Z eq rms (ii) tan ϕ mzeq ez eq (ω ω) ), or Zeq +i(ω ω) ) + ( ω ω), and. (iii) The phasor diagram for a parallel circuit is shown in Fig. 7, for both the inductive case X > X and the capacitive case X < X. 5. Draw to scale a phasor diagram showing Z, X, X, and ϕ for an A series circuit for which 3 Ω, µf,. H, and f 5/π Hz. Solution Before drawing the diagram, we find the complex impedance of each element. The complex impedance of a resistor is equal to its resistance, Z 3 Ω. t is therefore a real number which we marked in the complex plane on the e axis of Fig. 8. The complex impedance of an inductor is an imaginary number dependent on the angular frequency of the current and the inductance of the inductor Z iω π 5 π s. H i Ω. This number is on the m axis. Finally, the complex impedance of the capacitor is also an imaginary number dependent on the angular frequency of the current and the capacitance of the capacitor Z i ω i 5 π π s 6 F 9.9i Ω. This number is also on the m axis. Since the elements are connected in series the equivalent complex impedance is equal to the sum of the complex impedance of each element: Z Z + Z + Z 3 Ω + i Ω 9.9i Ω (3 + 9.i) Ω. The complex impedance of each element is indicated in Fig. 8. One tic corresponds to Ω. The diagram also illustrates how to find the equivalent complex impedance. Note that the phase of the complex impedance of the resistor is, the phase of the complex impedance of the capacitor is 9 and the phase of the complex impedance of the inductor is 9. 6. Draw to scale a phasor diagram showing the relationship between the current (common for all elements) and the voltages in an A series circuit for which 3 Ω, µf,. H, f 5/π Hz, and ma. Solution The complex voltage across the resistor can be found by multiplying the complex current (through the resistor) by the complex impedance of the resistor (t) (t)z. onsistent with the given values, the absolute value of this voltage is, ma 3 Ω 6. Since the phase of the complex impedance of the resistor is zero, the phase of the voltage across the resistor agrees with the phase of the current. The complex voltage across the inductor can be
found by multiplying the complex current (through the inductor) by the complex impedance of the inductor (t) (t)z. onsistent with the given values, the absolute value of this voltage is, X ma ω 4. Since the phase of the complex impedance of the inductor is 9, the voltage across the inductor leads with the current by a 9 phase angle. Finally, the complex voltage across the capacitor can be found by multiplying the complex current (through the capacitor) by the complex impedance of the capacitor (t) (t)z. onsistent with the given values, the absolute value of this voltage is, X ma 9 Ω.8. Since the phase of the complex impedance of the capacitor is 9, the voltage across the capacitor lags behind the current by 9. f we want to find the complex voltage across the entire system, we can add the complex voltages across all three elements (they are connected in series). indicated this operation on the phasor diagram. That voltage can also be found by multiplying the current by the equivalent impedance of the system. Notice that the phase difference between the voltage across the system and the current through the system is equal to the phase angle of the system (the phase of the equivalent impedance). The absolute value of the voltage is Z ma (3 Ω) + (9 Ω) 6.4. 7. The resistor in the Fig. represents the midrange speaker in a three-speaker system. Assume its resistance to be constant at 8 Ω. The source produces signals of uniform amplitude in al all frequencies. The inductor and capacitor are to function as a bandpass filter with out / in / at Hz and 4, Hz. (i) Determine the required values of and. (ii) Find the maximum value of the ratio out / in. (iii) Find the frequency f at which the ratio has its maximum value. (iv) Find the phase shift between in and out at Hz at f, and at 4, Hz. (v) Find the average power transferred to the speaker at Hz, at f, and at 4, Hz. (vi) Treating the filter as a resonant circuit, find its quality factor. Solution omplex current in the series circuit is determined by the input complex voltage and the complex impedance of the circuit in /Z in in +iω++ iω +i(ω ω ). omplex voltage across the speaker is related to the complex current through the speaker, out i(ω ω ) +i(ω ω ) in v +(ω ω ) in. The number relating complex output voltage to complex input voltage is called the complex gain of the circuit. (i) From the relation between complex voltages, the relation between the amplitudes is out out +(ω ω ) in. The number relating the amplitudes of voltage is called the gain of the circuit. Gain G is a function of frequency G +(ω ω ). t assumes a particular value at specific frequencies, determined by the difference between capacitive reactance and inductive reactance ω ω and ω G ω. Solving simultaneously for the unknowns G G ω ω G π(f f ) 4 Hz Hz π 8Ω Hz 4 Hz 4 8Ω 4 π(4 Hz Hz 58 µh and ω ω ω ω f f G πf f G 54.6 µf. (ii) The maximum gain occurs when the sum of inductive reactance and capacitive reactance is zero ω ω from which G. (iii) The + sum of inductive reactance and capacitive reactance is zero at the resonant frequency of this circuit f the phase relation. ω π π 5.8 4 H 5.46 5 F 894 Hz. (iv) omplex gain also includes The phase difference α, between the output voltage and the input volt-
ages is α arctan (ω ω ) arctan πf πf. Therefore at Hz its value is α Hz arctan π Hz 5.8 4 H π Hz 5.46 5 H 8 Ω 6. (The output voltage leads the input voltage.) At 894 Hz its value is α 894 Hz arctan(/8 Ω). (The output voltage is in phase with the input voltage.) At 4, Hz its value is α 4 Hz arctan π 4 Hz 5.46 5 H π 4 Hz 5.46 5 H 8 Ω 6. (The output voltage lags behind the input voltage.) (v) The average power delivered to an element depends on the rms values of the voltage across the element and the current through the element. Since the speaker acts like a resistor, Ohm s law can be used to relate the rms current (through the speaker) to the voltage applied to the speaker. Knowing the gain of the circuit the power can be directly related to the amplitude of the input voltage rms rms cos ϕ out out cos (G in) (.5 ). Therefore, (.5 ) 8ω.56 W, 894 ( ) 8ω 6.5 W, 4 8ω.56 W. (vi) arameters of the circuit (resistance, capacitance and inductance) affect the quality factor Q ω πf π 894 Hz 5.8 4 H 8 Ω.4. 8. A person is working near the secondary coil of a transformer as shown in Fig.. The primary voltage is at 6 Hz. The capacitance s, which is the stray capacitance between the hand and the secondary winding, is pf. Assuming the person has a body resistance to ground b 5 kω, determine the rms voltage across the body. Solution The secondary circuit consists of a capacitor with capacitance s µf, a resistor with resistance b 5 kω and a source of electromotive force (the secondary coil of the transformer). The source produces sinusoidal potential difference with rms value of 5 and frequency of 6 Hz. At this frequency the impedance of the capacitor is Z πf s and the impedance of the resistor is equal to its resistance Z b. Since the elements are connected in series, the equivalent resistance of the system (of the two elements) is Z b + (πf s). The rms current in the resistor (and the circuit) is therefore rms rms Z rms ). From Ohm s law, the rms voltage across the body (resistor) is rms,b rms,b b b rms ( ) 5 k ( + + πfs b π 6 Hz pf 5 Ω ).88. ( b + πfs rms ( ) b + πfs 9. A is a device that allows current to pass in only one direction (indicated by the arrowhead). Find, in terms of rms and, the average power dissipated in the circuit shown in Fig.. Solution When the left side of the generator has a higher potential, the top conducts (forward bias), and the bottom does not (reverse bias). The remaining circuit of three resistors acts like a single resistor. The resistors are connected in series and the pair is connected in parallel to the resistor, see Fig. 3. Therefore the equivalent resistance of the system is ( eq + +) +. Knowing the rms value of the voltage we could find the power in this circuit if the was not present. However, the presence of the reduces the power to half its value (because only half of the time current would flow in the circuit). Therefore
change if B were allowed to extend a distance above the axis of rotation? Sketch the emf versus time (d) when the field is as drawn in Figure 3.4 and (e) when the field is extended as described in (c). A A B t B 3. T f a5. m f b4. rad sg out Figure 3.4 Figure : roblem. 6. hapter 3 7 Figure 4. The rotating loop in an A generator is a square. cm on a side. t is rotated at 6. Hz in a uniform field of.8 T. alculate (a) the flux through the loop as a function of time, (b) the t emf induced in the loop, (c) the BA z d sind t F H G K J ximum and average would Figure remain Figure : Solution of problem. ged. re at the right. re at the right. before, the power dissipated at this polarity is.75 The overall (time) average power is n + + ( ) rms Figure + 4 7 rms 4. t cos BA cos t 8. T. m cos 6. t 8. mtm cos 377t FG. 3.4 B a3. fsina377tf a j f aa ff e j a f Solution (i) n order to send out power of 5 MW at 5 k potential difference, the current in the grid (transmission lines) must be rms,a rms,a A. The resistance of the two wires of the transmission lines (connected is series) is λ 58 Ω. Hence the loss of power in the lines (dissipated in the lines) is a9. Wfsin a377tf a rms ( A) 58 Ω 58 kw. (ii) A relative small fraction is lost the lines a 58 kw 5 MW. %. (iii) f the power were send at 4.5 k potential difference, the current in the grid (transmission lines) should be much larger rms,c rms,c 5 MW 4.5 k. ka. But even if nothing were connected at the end of the lines, with the potential difference of 4.5 k, the current in the lines would be a4. mn mfsin a377 rms,max tf rms,c 4.5 k 58 Ω 77 A. t would be impossible to send 5 MW of power into the lines. m 3. cos A sin 6 377t. t 8. mtm cos 377t so Figure FG. 3.4 hapter 3 7 + rms. When the polarity of the emf is reversed, the top is in reverse bias and the bottom is in forward bias. This t time the equivalent resistance of the system of resistors is different + ( ) + + 7 4; see Fig. 3. With the same arguments as a fe j a f e j a f. A transmission line that has a resistance per unit length of 4.5 4 Ω/m is to be used to transmit 5 MW over 4 miles (6.44 5 m); see Fig. 4. The output voltage of the generator is 4.5 k. (i) What is the line loss if a transformer is used to step up the voltage to 5 k? (ii) What fraction of the input power is lost to the line under these circumstances? (iii) What difficulties would be encountered in attempting to transmit the 5 MW at the generator voltage of 4.5 k. rms rents
Figure 33.3. What does each meter read? Note that an ideal ammeter has zero resistance and that an ideal voltmeter has infinite resistance. max 4. Ω Figure 33.3 Figure 3: roblem. 4. n the simple A circuit shown in Figure 33., 7. and v max sin t. (a) f v.5 max for the first time at t. s, what is the angular frequency of the source? (b) What is the next value of t for which v.5 max? 5. The current in the circuit shown in Figure 33. equals 6.% of the peak current at t 7. ms. What is the smallest frequency of the source that gives this current? 6. Figure 33.6 shows three lamps connected to a - A (rms) household supply voltage. amps and have 5-W bulbs; lamp 3 has a -W bulb. Find the rms current and resistance of each bulb. A Section 33.3 nduct 8. An inductor is con produces a 5.- to keep the insta 8. ma? 9. n a purely induct max. (a 5. Hz. alculate angular frequency. An inductor has a the maximum cur 5.-Hz source tha. For the circu 65. rad/s, a in the inductor at t. A.-mH induct outlet ( rms energy stored in t that this energy is z 3. eview problem. D through an induc outlet ( rms Figure 4: roblem 3. Figure.8. hasor diagram for the series circuit for (a) X X and (b) Figure 5: hasor diagram for the series circuit for X > X (left) and X < X (right). X X. From Figure.8.(a), we see that in the inductive case and leads by a phase. On the other hand, in the capacitive case shown in Figure.8.(b),
arallel ircuit ider the parallel circuit illustrated in Figure.6.. The A voltage sour sint. e the series circuit, the instantaneous voltages across all three circuit elem and are the same, and each voltage is in phase with the current through or. However, the currents through each element will be different. alyzing this circuit, we make use of the results discussed in Sections. urrent in the resistor is Figure.8. hasor diagram for the parallel circuit for (a) X X and (b) Figure t () X7: hasor X. diagram for the parallel circuit for X () t sint > X (left) and X sint < X (right). e / m. The voltage across the inductor is Figure.8. hasor diagram for the series circuit for (a) X X and (b) X X. From Figure.8.(a), we see that in the inductive case and leads by a phase. On the other hand, in the capacitive case shown in Figure.8.(b), and leads by a phase. 4. When, or, the circuit is at resonance. The corresponding resonant frequency is /, and the power delivered to the resistor is a maximum. 5. For parallel connection, draw a phasor diagram for the currents. The amplitudes of the currents across Figure 6: roblem 4. Figure all the circuit.6. elements arallel involved should be circuit. represented with phasors. n Figure.8. the phasor diagram for a parallel circuit is shown for both the inductive case X X and the capacitive case X X. ( From Figure.8.(a), we see that in the inductive case and leads by a phase. On the other hand, in the capacitive case shown in Figure.8.(b), and leads by a phase. gives Z Z d () () sin -5 ( t t t ϕ e dt Z Z t The t dt diagram also illustrates t how to t find the equivalent t Note that the phase of the complex ( t ( ) sin ' ' cos Figure 8: roblem sin5. sin complex impedance. X impedance of the resistor is, the phase of the complex impedance of the capacitor is -9 and the phase of the complex impedance of the inductor is 9.
roblem 33.49 elements) and the voltages in the circuit. et's assume that current has value of ma and at a certain instant t it represented by the phasor indicated in the following figure. the figure one tic corresponds to ma. m The complex volta across the resistor can found by multiplying complex current (through Z resistor) by the comp Z ϕ impedance of the resistor The resistor in the Figure 33.49 Z represents the midrange speaker in a three-speaker system. Assume its resistance to be constant at 8 Ω. The source represent an audio amplifier producing signals of uniform amplitude in al all frequencies. The inductor and capacitor are to function as a bandpass filter with out / in ½ at Hz and 4, Hz. (a) Determine the required values of and. (b) Find the maximum value of the ratio out / in. (c) Find the frequency f at which the ratio has its maximum value. (d) Find the phase shift between in and out at Hz, at f, and at 4 Hz. (e) Find the average power transferred to the speaker at Hz, at f, and at 4 Hz. (f) Treating the filter as a resonant circuit, find its quality factor. Figure 9: roblem 6. roblem 33.3 in ϕ Z e 5 out () t () t Z onsistent with the gi values, the absolute value this voltage is A person is working mear the secondary coil of a transformer as shown in Figur 33.5. The primary voltage is at 6 Hz. The capacitance s, which is th stray capacitance between the hand and the secondary winding, is pf. Assuming the person has a body resistance to ground b 5 kω, determine the rms voltag across the body. omplex current in Figure the series : roblem circuit 7. is determined by the input complex voltage and the complex impedance of the circuit in Z in in + i ω + + i ω i ω ω omplex voltage across the speaker is related to the complex current through the speaker. 5, 8 Figure : roblem 8. The secondary circuit consists of a capacitor with capacitance s µf, a resistor with resistance b 5kΩ
t allows current to n (indicated by the ws current to rms dicated of by the rms and, ipated f rms and in, the d in the 33.35. 5. of the higher side of the ias), a and higher When the polarity of the emf is oes not reversed, the top is in emaining top reverse bias and the bottom tors bias), acts is roblem in forward 33.43 bias. This and tor. The time the equivalent + - resistance of - + e ected does in the not system of resistors is onnected different e remaining 3). Therefore the equivalent resistance Figure 3: Solution esistors acts of problem 9. 7 + + esistor. The + + 4-5 k or connected in 4.5 k λ 4.5 polarity is -4 Ω/m 5 MW is connected rms 4) esistor. Therefore 75. the equivalent resistance value of the voltage we could find the 6.44 5 m f the was not present. However, the The overall (time) average power is e reduces the power to half its Figure value 4: roblem + rms rms + + 4. a) n order to send out power of 5MW at 5 k potential difference, the current in the grid 7(transmission 4 lines) must be 4 (because only half of the Figure time : current roblem would 9. flow in the circuit). Therefore ) + rms With the same arguments as before, the power dissipated at this A transmission line that has a resistance per unit length of 4.5-4 Ω/m is to be used to transmit 5 MW over 4 miles (6.44 5 m). The output voltage of the generator is 4.5 k. (a) What is the line loss if a transformer is used to step up the voltage to 5 k? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered in attempting to transmit the 5 MW at the generator voltage of 4.5 k rms,a rms,a A