# CHAPTER 28 ELECTRIC CIRCUITS

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1 CHAPTER 8 ELECTRIC CIRCUITS 1. Sketch a circuit diagram for a circuit that includes a resistor R 1 connected to the positive terminal of a battery, a pair of parallel resistors R and R connected to the lower-voltage end of R 1, then returned to the battery s negative terminal, and a capacitor across R. A literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is shown in sketch (b). 1 (a). 1 (b).. A circuit consists of two batteries, a resistor, and a capacitor, all in series. Sketch this circuit. Does the description allow any flexibility in how you draw the circuit? In a series circuit, the same current must flow through all elements. One possibility is shown. The order of elements and the polarity of the battery connections are not specified... Resistors R 1 and R are connected in series, and this series combination is in parallel with R. This parallel combination is connected across a battery whose internal resistance is R int. Draw a diagram representing this circuit.

2 660 CHAPTER 8 The circuit has three parallel branches: one with R 1 and R in series; one with just R ; and one with the battery (an ideal emf in series with the internal resistance). Section 8-: Electromotive Force. 4. What is the emf of a battery that delivers 7 J of energy as it moves.0 C between its terminals? From the definition of emf (as work per unit charge), E = W= q = 7 J = C = 9 V. 5. A 1.5-V battery stores 4.5 kj of energy. How long can it light a flashlight bulb that draws 0.60 A? The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an ideal battery, P = EI, therefore EIt = 4. 5 kj, or t = 4. 5 kj = ( 1. 5 V )( A ) = 5 10 s = 19. h. 7. A battery stores 50 W h of chemical energy. If it uses up this energy moving C through a circuit, what is its voltage? The emf is the energy (work done going through the source from the negative to the positive terminal) per unit charge: 4 E = ( 50 W h )( 600 s/h ) = ( 10 C ) = 6 V. (This is the average emf; the actual emf may vary with time.) 9. What resistance should be placed in parallel with a 56-kΩ resistor to make an equivalent resistance of 45 kω? The solution for R in Equation 8-a is R = R1 Rparallel= ( R1 Rparallel ) = ( 56 k Ω)( 45) = ( 56 45) = 9 k Ω. 11. In Fig. 8-49, take all resistors to be 1. 0 Ω. If a 6.0-V battery is connected between points A and B, what will be the current in the vertical resistor?

3 CHAPTER The circuit in Fig. 8-49, with a battery connected across points A and B, is similar to the circuit analysed in Example 8-4. In 8 this case, R = ( 1 Ω)( ) = ( 1 + ) = ( ) Ω, and R tot = 1 Ω + 1 Ω + Ω = Ω. The total current (that through the battery) is Itot = E= Rtot = 6 V = ( Ω ) = ( 4 ) A. The voltage across the parallel combination is Itot R = ( 4 A)( Ω ) = V, which is the voltage across the vertical 1 Ω resistor. The current through this resistor is then ( V ) = ( 1 Ω ) = 1. 5 A. 17. A partially discharged car battery can be modeled as a 9-V emf in series with an internal resistance of Ω. Jumper cables are used to connect this battery to a fully charged battery, modeled as a 1-V emf in series with a 0. 0-Ω internal resistance. How much current flows through the discharged battery? Terminals of like polarity are connected with jumpers of negligible resistance. Kirchhoff s voltage law gives E1 E IR IR = 0, or I = ( E E ) = ( R + R ) = ( 1 9) V = ( ) Ω = 0 A What is the equivalent resistance between A and B in each of the circuits shown in Fig. 8-50? Hint: In (c), think about symmetry and the current that would flow through R. 1 (a) There are two parallel pairs ( R 1) in series, so RAB = 1 R R1 = R1. (b) Here, there are two series pairs ( R 1) in parallel, so RAB = ( R1 )( R1 ) = ( R1 + R1 ) = R1. (c) Symmetry requires that the current divides equally on the right and left sides, so points C and D are at the same potential. Thus, no current flows through R, and the circuit is equivalent to (b). (Note that the reasoning in parts (a) and (b) is easily generalized to resistances of different values; the generalization in part (c) requires the equality of ratios of resistances which are mirror images in the plane of symmetry.) FIGURE

4 66 CHAPTER 8. What is the current through the -Ω resistor in the circuit of Fig. 8-51? Hint: This is trivial. Can you see why? 5.0 Ω V.0 Ω V FIGURE The current is IΩ = V Ω= RΩ = 6 V = Ω = A, from Ohm s law. The answer is trivial because the potential difference across the Ω resistor is evident from the circuit diagram. (However, if the 6 V battery had internal resistance, an argument like that in Example 8-5 must be used.). Take E = 1 V and R1 = 70 Ω in the voltage divider of Fig (a) What should be the value of R in order that 4.5 V appear across R? (b) What will be the power dissipation in R? (a) For this voltage divider, Equation 8-b gives V = RE= ( R1 + R ), or R = R1V = ( E V ) = ( 70 Ω )( 4. 5) = ( ) = 16 Ω. (b) The power dissipated (Equation 7-9b) is P = V = R = ( 4. 5 V ) = 16 Ω = 15 mw. 5. In the circuit of Fig. 8-5, R 1 is a variable resistor, and the other two resistors have equal resistances R. (a) Find an expression for the voltage across R 1, and (b) sketch a graph of this quantity as a function of R 1 as R 1 varies from 0 to 10R. (c) What is the limiting value as R 1? FIGURE (a) The resistors in parallel have an equivalent resistance of R = RR1= ( R + R1 ). The other R, and R, is a voltage divider in series with E, so Equation 8- gives V = ER = ( R + R ) = ER 1 = ( R + R 1 ). (b) and (c) If R 1 = 0 (the second resistor shorted out), V = 0, while if R 1 = (open circuit), V = 1 E (the value when R 1is removed). If R1 = 10R, V = ( 10= 1) E (as in 4).

5 CHAPTER In the circuit of Fig. 8-5 find (a) the current supplied by the battery and (b) the current through the 6-Ω resistor. (a) With reference to the solution of the next problem, the resistance of the three parallel resistors in ( 1= 11) Ω, so the current supplied by the battery is I = E= ( R1 + R ) = ( 6 V ) = ( = 11) Ω =. 87 A. (b) The voltage drop across the resistors in parallel is V = E IR1 = IR and the current through the 6 Ω resistor is I Ω = V = Ω. Thus, I 6 Ω = (. 87 A)( = 11) = 5 ma., 6 6 FIGURE 8-5 6, and In the circuit of Fig it makes no difference whether the switch is open or closed. What is E in terms of the other quantities shown? If the switch is irrelevant, then there is no current through its branch of the circuit. Thus, points A and B must be at the same potential, and the same current flows through R 1 and R. Kirchhoff s voltage law applied to the outer loop, and to the lefthand loop, gives E1 IR1 IR + E = 0, and E IR + E = E = IR E = F HG I, respectively. Therefore, E1 + E E E 1 E1 = R1 + RKJ R R R R1 + R 1 1. FIGURE What is the current through the ammeter in Fig. 8-55? If the ammeter has zero resistance, the potential difference across it is zero, or nodes C and D are at equal potentials. If I is the current through the battery, 1 I must go through each of the Ω -resistors connected at node A (because 1 A C A D V V = I( Ω ) = V V ). At node B, the Ω -resistor inputs twice the current of the 4 Ω -resistor, or I and 1 I 1 respectively (because VC VB = I( Ω) = I( 4 Ω ) = VD VB ). Therefore 1 6 I must go through the ammeter from D to C, as required by Kirchhoff s current law. To find the value of I, note that the upper pair of resistors are effectively in parallel ( V = V ) as is the lower pair. The effective resistance between A and B is R eff = Ω=( + ) + 4 Ω C D

6 664 CHAPTER ( + 4) = 1 Ω + ( ) Ω = ( ) Ω. Thus I = V= R eff, and the ammeter current is I = 6 ( 6 V ) = ( ) Ω = ( 7) A = 0.49 A. 7 FIGURE In Fig. 8-56, what is the equivalent resistance measured between points A and B? The effective resistance is determined by the current which would flow through a pure emf if it were connected between A and B: RAB = E= I. Since I is but one of six branch currents, the direct solution of Kirchhoff s circuit laws is tedious (6 6 determinants). (The method of loop currents, not mentioned in the text, involves more tractable determinants.) However, because of the special values of the resistors in Fig. 8-56, a symmetry argument greatly simplifies the calculation. The equality of the resistors on opposite sides of the square implies that the potential difference between A and C equals that between D and B, i.e., VA VC = VD VB. Equivalently, VA VD = VC VB. Since VA VC = I1 R, VA VD = I ( R), etc., the symmetry argument requires that both R-resistors on the perimeter carry the same current, I 1, and both Rresistors carry current I. Then Kirchhoff s current law implies that the current through E is I1 + I, and the current through the central resistor is I1 I (as added to Fig. 8-56). Now there are only two independent branch currents, which can be found from Kirchhoff s voltage law, applied, for example, to loops ACBA, E I1R I ( R) = 0, and ACDA, I1R ( I1 I ) R + I( R) = 0. These equations may be rewritten as I1 + I = E= R and I1 + I = 0, with solution I1 = E= 7R and I = E= 7R. Therefore, I = I1 + I = 5E= 7R, and RAB = E= I = 7R= 5. (The configuration of resistors in Fig is called a Wheatstone bridge.) FIGURE In Fig. 8-57, take E1 = 6. 0 V, E = 1. 5 V, E = 4. 5 V, R1 = 70 Ω, R = 150 Ω, R = 560 Ω, and R4 = 80 Ω. Find the current in R,and give its direction. The general expressions for the branch currents can be found from the solution to the next problem. Here, we only need ( 6 V 1. 5 V )( 80 Ω) + ( 4. 5 V 1. 5 V )( 40 Ω) I b = = ma. ( 560 Ω)( 80 Ω) + ( 40 Ω)( 180 Ω) A negative current is downward through E in Fig

7 CHAPTER FIGURE , and s 5 and A voltmeter with 00-kΩ resistance is used to measure the voltage across the 10-kΩ resistor in Fig By what percentage is the measurement in error because of the finite meter resistance? 5.0 kω V 10 kω FIGURE 8-59 s 9 and 40. The voltage across the 10 kω resistor in Fig is ( 150 V )( 10) = ( ) = 100 V (the circuit is just a voltage divider as described by Equations 8-a and b), as would be measured by an ideal voltmeter with infinite resistance. With the real voltmeter connected in parallel across the 10 kω resistor, its effective resistance is changed to R = ( 10 k Ω)( 00 kω ) ( 10 k Ω) = 9. 5 kω, and the voltage reading is only ( 150 V )( 9. 5) = ( ) = 98.4 V, or about 1.64% lower. 4. The voltage across the 0-kΩ resistor in Fig is measured with (a) a 50-kΩ voltmeter, (b) a 50-kΩ voltmeter, and (c) a digital meter with 10-MΩ resistance. To two significant figures, what does each read? FIGURE With a meter of resistance R m connected as indicated, the circuit reduces to two pairs of parallel resistors in series. The total resistance is Rtot = ( 0 k Ω) Rm= ( 0 k Ω + Rm ) + 40 kω=. The voltage reading is Vm = Rm Im = Rm ( 0 kω ) Itot ( 0 kω + R m ), where Itot = ( 100 V) = Rtot (the expression for V m follows from Equation 8-, with R 1 and R as the above pairs, or from I m as a fraction of I tot, as in the solution to 65). For the three voltmeters specified, I tot =. 58 ma,. 14 ma, and.00 ma, while V m = 48.4 V, 57. V, and 59.9 V, respectively. (After checking the calculations, round off to two figures. Of course, 60 V is the ideal voltmeter reading.) 4. In Fig what are the meter readings when (a) an ideal voltmeter or (b) an ideal ammeter is connected between

8 666 CHAPTER 8 points A and B? 10 kω A + 0 V 0 kω B FIGURE (a) An ideal voltmeter has infinite resistance, so AB is still an open circuit (as shown on Fig. 8-61) when such a voltmeter is connected. The meter reads the voltage across the 0 kω resistor (part of a voltage divider), or ( 0 V ) 0= ( ) = 0 V (see Equation 8-a or b). (b) An ideal ammeter has zero resistance, and thus measures the current through the points A and B when short-circuited (i.e., no current flows through the 0 kω resistor). In Fig. 8-61, this would be I AB = 0 V = 10 Ω = ma. (Such a connection does not measure the current in the original circuit, since an ammeter should be connected in series with the current to be measured.) 45. Show that the quantity RC has the units of time (seconds). The SI units for the time constant, RC, are ( Ω)( F) = ( V= A)( C= V) = ( s= C)( C) = s, as stated. 47. Show that a capacitor is charged to approximately 99% of the applied voltage in five time constants. 5 After five time constants, Equation 8-6 gives a voltage of VC 1 e = ' 99. % of the applied voltage. 49. Figure 8-6 shows the voltage across a capacitor that is charging through a 4700-Ω resistor in the circuit of Fig Use the graph to determine (a) the battery voltage, (b) the time constant, and (c) the capacitance.

9 CHAPTER FIGURE (a) For the circuit considered, the voltage across the capacitor asymptotically approaches the battery voltage after a long time (compared to the time constant). In Fig. 8-6, this is about 9 V. (b) The time constant is the time it takes the capacitor 1 voltage to reach 1 e = 6. % of its asymptotic value, or 5.69 V in this case. From the graph, τ ' 15. ms. (c) The time constant is RC, so C = 1. 5 ms = 4700 Ω = µ F. 50. The voltage across a charging capacitor in an RC circuit rises to 1 1=e of the battery voltage in 5.0 ms. (a) How long will it take to reach 1 1=e of the battery voltage? (b) If the capacitor is charging through a -kω resistor, what is its capacitance? (a) Equation 8-6 and the given circuit characteristics imply that the time constant is τ = RC = 5. 0 ms. Therefore, in three time constants, or 15 ms, the capacitor is charged to 1 e of the battery voltage. (b) Evidently, C = τ= R = 5 ms = kω = 0. 7 µ F. 51. A 1. 0-µ F capacitor is charged to 10.0 V. It is then connected across a 500-kΩ resistor. How long does it take (a) for the capacitor voltage to reach 5.0 V and (b) for the energy stored in the capacitor to decrease to half its initial value? A capacitor discharging through a resistor is described by exponential decay, with time constant RC (see Equation 8-8), and, 1 1 t= RC t= RC of course, UC ( t) = CV( t) = CV0e = UC( 0) e is the energy stored (see Equation 6-8b). (a) V( t) = V( 0) = 1= implies t = RC ln = ( 500 k Ω )( 1 µ F )( 0. 69) = 47 ms. (b) U ( t) = U ( 0) = 1= implies 1 t = RC ln = 17 ms. 56. In the circuit of Fig the switch is initially open and both capacitors initially uncharged. All resistors have the same value R. Find expressions for the current in R (a) just after the switch is closed and (b) a long time after the switch is closed. (c) Describe qualitatively how you except the current in R to behave after the switch is closed. c c FIGURE (a) An uncharged capacitor acts instantaneously like a short circuit (see Example 8-9), so initially ( t = 0 ) all of the current from the battery goes through R 1 and C 1, and none goes through R and R. Thus, I1( 0) = E= R, and I( 0) = I( 0) = 0. (b) A fully charged capacitor acts like an open circuit (when responding to a constant applied emf ), so after a long time ( t = ), all of the current goes through R 1 and R in series, and none goes through R. Thus I1( ) = I( ) = E= R, and I ( ) = 0. (c) One can easily guess that I 1 and I respectively decrease and increase monotonically from their initial to their

10 668 CHAPTER 8 final values, and that I first increases from, and then decreases to zero. (One can use the loop and node equations to solve for the currents. They turn out to be linear combinations of two decaying exponentials with different time constants.) 57. In the circuit for Fig the switch is initially open and the capacitor is uncharged. Find expressions for the current I supplied by the battery (a) just after the switch is closed and (b) a long time after the switch is closed. I R R + R C R FIGURE (a) Just after the switch is closed, the uncharged capacitor acts instantaneously like a short circuit and the resistors act like two parallel pairs in series. The effective resistance of the combination is ( R)( R) = ( R + R) = 4R=, and the current supplied by the battery is I( 0) = E= 4R. (b) A long time after the switch is closed, the capacitor is fully charged and acts like an open 1 circuit. Then the resistors act like two series pairs in parallel, with an effective resistance of ( ) ( R + R ) = R=. The battery current is I( ) = E= R. 75. Write the loop and node laws for the circuit of Fig. 8-71, and show that the time constant for this circuit is R R C=( R + R ). 1 1 Consider the loops and node added to Fig Kirchhoff s laws are E = I R + I R VC = I R V = q= C and I = dq= dt, the equations can be combined to yield C C,, and IC = I1 I. Since 1 1 VC E I1R1 IR E ( IC I ) R1 IR E ICR1 ( R R ) E IC R R = + = F H G I K J + = 1 1 q = 0. CR = ( R + R ) 1 This is exactly in the same form as the first equation, solved in the text, in the section The RC Circuit: Charging (with I IC, R R1 and C CR= ( R1 + R )), so the time constant for the circuit is τ = CR1 R=( R1 + R ) (the ratio of the coefficients of I C and q). FIGURE The circuit in Fig. 8-7 extends forever to the right, and all the resistors have the same value R. Show that the equivalent

11 CHAPTER resistance measured across the two terminals at left is 1 R( 1 + 5). Hint: You don t need to sum an infinite series.... FIGURE Since the circuit line is infinite, the addition or deletion of one more element leaves the equivalent resistance unchanged. Diagrammatically: 78. The right-hand picture represents R in series with the parallel combination R and R eq, therefore R = R + RR =( R + R ). 1 eq eq eq Solving for R eq, one finds Req RReq R = 0, or Req = ( 1 + 5) R (only the positive root is physically meaningful for a resistance).

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