27 CHAPTER Direct-Current Circuits
Answers to Understanding the Concepts Questions
1.
Tap water is an excellent conductor, and if the appliance falls into the tub there is a danger that a large current will flow through the body of the person in the tub, causing bums and sometimes heart failure. The voltmeter measures the terminal voltage V across the battery, and that depends on the current I drawn and the internal resistance r of the battery: V = Ir. While and r do not change appreciably, I can. (Even r can increase, if the battery gets warmer.) The current that flows into the battery is the same as the current flowing out of the battery. Whether there is a potential drop Ir just before the current reaches the battery or whether the drop occurs just after the current leaves the battery is irrelevant, since either way there will be the same contribution to the loop rule, and that is all that counts. It makes sense when the resistance of the wire is negligible in comparison with those of the resistors. This is impossible, since by choosing the direction in which the emf is positive, one could create a situation in which one would be creating energy. Such "perpetual motion" machines violate energy conservation. The battery could be all right, of course. Keep in mind, however, that it has to provide the right terminal voltage when hooked up to a working load, not just a voltmeter with presumably a very large resistance. As the resistance of the load decreases the current flowing in the battery increases, and the terminal voltage, V = Ir, could drop appreciably. Here r is the internal resistance of the battery. An unfair question. The circuit in a flash does not give the falling exponential characteristic of a pure RC circuit. Rather it uses solid-state devices to tailor the release of energy from a capacitor, typically of size 1000 F, so that a current that is basically flat for a period of about 0.01 s results. If we work backwards and ask what value of R in a pure RC circuit would give a time constant = RC of 0.01 s with a capacitor or 1000 F, we would find R = /C = (0.01 s)/(103 F) = 10 , a very reasonable value. The time constant of an RC circuit is = RC. To make the discharge time as short as possible we need to minimize , which means we want the lowest possible value of C out of the three capacitors. This can be done by connecting the three capacitors in series. There is no disaster. With the new choice of positive direction for the current, I is related to Q as I = dQ/dt. Thus from IR Q/C = 0 we get R(dQ/dt) Q/C = 0, which again leads to the finite solution Q = Q0 e t/RC.
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3.
4. 5.
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7.
8.
9.
10. Let's construct a potential energy diagram with a fluid analogy in mind. The batteries "raise" the liquid, increasing its potential energy. A resistance corresponds to a drop of potential energy given by IR. When the current goes through resistors in series, it is as if it cascaded down several downward slopes. If current goes through two resistors in parallel, it splits up so "at the bottom" the two currents are reunited at the same potential. Consider now this diagram turned upside down. The batteries (with reversed polarities) now lower the potential, and IR must raise them. Since R is unchanged by the reversal, I must change sign.
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Page 27-1
Chapter 27: Direct-Current Circuits
11. The maximum possible emf you can get out of two emf sources is the sum of the two emf values, which is obtained when you connect the two sources in series. To ensure that each light bulb gets that maximum emf we can hook up the light bulbs in parallel and apply the combined emf across each of them simultaneously. This ensures maximum power consumption for each light bulb and therefore maximum brightness assuming, of course, that the emf does not exceed the maximum value allowed by each light bulb (so none of them would burn out). 12. The effective net emf is the difference between the two emf values: eff = 1 2 , when they are connected + to + and to . The larger emf wins, of course, and the resulting current in the circuit is I = eff/Req . 13. The effective emf that drives the current in the circuit is now current increases.
eff = 1 + 2 , and the magnitude of the
14. Technically speaking this is certainly true. However, it is useless information. The value of the current itself depends on the rest of the circuit as well as on the value of the internal resistance, and so does the value of the "shifted" emf This is not a very useful way to think about a circuit. In contrast, the original emf is a constant which at least for an ideal battery does not vary with current. 15. In the circuit diagram depicted in Fig. 27-8(a), R2 and R3 are in series, as are R5 and R6 , and the two branches are in parallel with R 4 . The equivalent resistance of this three-branch combination is R = [1/(R2 + R3) + 1/R4 + 1/(R5 + R6)]1. Put this in series with R1 and we get a single-loop circuit. In general, such reduction is not possible for the circuit diagram depicted in Fig. 27-8(b), unless R2/R3 = R5/R6, in which case the voltage difference across R4 is zero so that it can be removed from the circuit. 16. The teenagers provide a path for the current parallel to the wire. If the wire has no resistor along it, the resistance is low compared to that of the teenager, and most of the current flows through the wire. With the resistor, more of the current flows through the teenager, with more serious consequences. They are both dumb, but the one with the resistor is dumber. 17. The steady-state value of I1 decreases as R3 is increased. 18. When the lights are on, a current runs through the battery that powers the lights. Heat is generated as the current flows through the battery due to its internal resistance, and the battery warms up.
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Page 27-2
Fishbane, Gasiorowicz, and Thornton
Solutions to Problems
1. Because the internal resistance of the battery is the only resistance in the single-loop circuit, we have I = /r; 80 A = (12 V)/r, which gives r = 0.15 . The solar panel is a source of emf, so the power output is P = Vterm I; 1200 W = Vterm(40 A), which gives V term = 30 V . For this single-loop circuit, we have I = /(R + r); 1.99 A = (3 V)/[(1.50 + r)], which gives r = 0.0075 . The energy contained in the battery is the total energy output: U = IVt = (It)V = (30 A h)(3600 s/h)(30 V) = 3.2 106 J. For this single-loop circuit, we have I = Vterm/(R + r) = (5000 V)/(230 + 20 ) = 20 A. For this single-loop circuit, we have I = /(R + r); 170 103 A = /(15 + 0.06 ), which gives = 2.56 V . The terminal voltage of the battery is V = Ir = 2.56 V (170 103 A)(0.06 ) = 2.55 V .
2.
3.
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5.
6.
7. The terminal voltage of the battery is V = Ir; 9.0 V = 12 V (100 A)r, which gives r = 0.030 . The power dissipated within the battery is P = I 2r = (100 A)2 (0.030 ) = 300 W . 8. The terminal voltage of the battery is the voltage drop across the starter: V term = Ir = IR; 8 V = 12 V Ir = I(0.11 ), which gives I = 73 A, and r = 0.05. The rate at which heat is produced is the power dissipated within the battery. For 10 s we have W = I 2rt = (73 A)2 (0.05 )(10 s)/(4.185 J/cal) = 0.6 103 cal 0.6 kcal. This would raise the temperature of a liter of electrolyte (water) by 0.6C, which may decrease the internal resistance slightly. It is more important to raise the temperature of the oil. The terminal voltage of the battery is V term = Ir = I( + I). The power dissipated within the battery is P = I 2r = I 2( + I). For I = 1.0 A, we have V 1 = (12.0 V) (1.0 A)[0.15 + (0.018 /A)(1.0 A)] = 11.8 V. P1 = (1.0 A)2 [0.15 + (0.018 /A)(1.0 A)] = 0.17 W. For I = 10.0 A, we have V 1 0 = (12.0 V) (10.0 A)[0.15 + (0.018 /A)(10.0 A)] = 8.7 V. P1 0 = (10.0 A)2 [0.15 + (0.018 /A)(10.0 A)] = 33 W. Page 27-3
9.
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Chapter 27: Direct-Current Circuits
10. The resistance of each bulb can be found from its rating: R1 = 2.5 V/0.5 A = 5.0 , and R2 = (110 V)2/10 W = 1.21 k (as P2 = V22/R2). When connected in series the equivalent resistance of the two bulbs is R = R 1 + R2 , and when hooked up to a power supply of = 110 V the current through each bulb is I = /( R1 + R2) = 110 V/(5.0 + 1.21 k) = 0.0905 A. The power consumed on each bulb is then P1 = I2 R1 = (0.0905 A)2(5.0 ) = 0.041 W and P2 = I2 R2 = (0.0905 A)2(1.21 k) = 9.9 W .
11. If we go around the single loop in the direction shown, starting at point a, we have V = IR2 + IR1 + = 0, which gives I = 2/(R1 + R2 ) . If we go from a to b, we have V a IR2 + = Vb = Va , or = IR2 . When we combine this with the expression for the current, we get = 2 R2 /(R1 + R2 ), which gives R2 = R1 . From the expression for the current, we see that I 0 when R2 .
a +
I
R2
R1
I
+
b
12. ( a ) Without the series resistor in the single-loop circuit, we have batt rbatt I = ( gen N batt)/(rgen + Nrbatt ) + ... + d c = [(110 V) 20(2.2 V)]/[(0.50 ) + 20(0.06 )] = 38.8 A. The terminal voltage of the generator is V a Vb = gen Irgen = (110 V) (38.8 A)(0.50 ) = 91 V . (b) The terminal voltage of the bank of batteries is V d Vc. R Because the batteries are being charged, this must be greater than the total emf of the bank: + V d Vc = N( batt + Irbatt) a b I = 20[2.2 V + (38.8 A)(0.06 )] = 91V . gen rgen (c) With the series resistor in the single-loop circuit, we have I = ( gen N batt)/(rgen + Nrbatt + R); 15 A = [110 V 20(2.2 V)]/[0.50 + 20(0.06 ) + R], which gives R = 2.7 . (d) The power dissipated in all the resistors is P = I 2(R) = (15 A)2 [0.50 + 2.7 + 20(0.06 )] = 9.9 102 W.
13. Because there are no resistors between points f and g, they must be at the same potential: V gf = 0. If we choose a path between two points, the potential difference is the sum of the potential differences of the segments of the path. Thus, we have V ag = Vaf = Vab + Vbc + Vcd + Vdf = Vba Vcb + Vcd + Vdf = 2 V 3.5 V + 2 V 0.5 V = 4.0 V ; V ca = Vcb + Vba = 3.5 V + 2 V = 5.5 V .
b a c +
+
g
f
e
d
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Page 27-4
Fishbane, Gasiorowicz, and Thornton
14. ( a ) If we assume initially there is no internal resistance, we have I 0 = 2/R = 2(1.5 V)/(10 ) = 0.30 A. The power delivered to the bulb is the power dissipated in the bulb: P0 = I0 2 R = (0.30 A)2 (10 ) = 0.90 W . (b) If the power delivered to the bulb, which is also the power dissipated in the bulb, decreases by one-third, we have P = I 2R; %(0.90 W) = I 2(10 ), which gives I = 0.25 A. For the single-loop circuit, we have I = 2/(R + 2r); 0.25 A = 2(1.5 V)/[(10 ) + 2r], which gives r = 1.0 /battery. 15. For the conservation of current, we have Iin = 0; I 1 + I2 + I3 + I4 + I5 + I6 = 0; (2 A) + (0.5 A) (3 A) 0.5I6 I6 + I6 = 0, which gives I6 = 1.0 A. Thus we have I 4 = +0.5 A, I5 = +1.0 A, I6 = 1.0 A.
I1 I2 I6 I3 I5 I4
16. From symmetry, the current will be the same in bulbs 1, 2, 4, and 5. Thus there will be no current through bulb 3, which will have zero brightness.
1 3 4
2
5
17. (a) With the switch open, the circuit consists of two branches in parallel, one with a resistance of R 1 = (4 + 12 ) = 16 , and the other with R2 = (8 + 6 ) = 14 . The equivalent resistance is then R eq = R1R 2/(R1 + R2) = (16 )(14 )/(16 + 14 ) = 7.5 . (b) We assume the current directions shown in the diagram. We use conservation of current at points a and b: Iin = 0; I I 1 I3 I4 = 0; I2 + I3 I5 = 0. 8 4 We apply the loop rule for the two loops indicated in the diagram: I1 I2 loop 1: I1 (4 ) I2 (8 ) + I3 (5 ) = 0; 1 5 b loop 2: I3 (5 ) I5 (6 ) + I4 (12 ) = 0. a I3 2 Also, when an emf is applied across the top and bottom of the I5 I4 circuit 12 6 = I1 (4 ) + I4 (12 ) = I2 (8 ) + I5 (6 ). I These equations yield I1 = 0.084/ and I2 = 0.059/. Thus R eq = /(I1 + I2 ) = /(0.084/ + 0.059/) = 7.0 .
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Page 27-5
Chapter 27: Direct-Current Circuits
18. Let the current in the 8- resistor be I1, to the right, and that in the 10- resistor be I2, to the right. Then the current in the remaining two resistors is I1 + I2, to the left. Apply the loop rule to the loop containing I1 and I2 : I1 (8 ) 6 V + I2 (10 ) 10 V = 0. Now apply the loop rule to the circumference of the circuit, starting clockwisely from point a: I1 (8 ) 6 V (I1 + I2 )(4 ) + 20 V (I1 + I2 )(16 ) = 0. Solve the two equations above to obtain I 1 = 0.409 A, I2 = + 1.27 A. The current in the 16- resistor is then I 1 + I2 = 0.409 A + 1.27 A = + 0.86 A, to the left. The voltage difference between a and b is V b Va = 10 V (1.27 A)(10 ) = 2.7 V , with V b < Va . 19. We assume the current directions shown in the diagram. We use conservation of current at point a: R1 Iin = 0; I 1 I2 + I3 = 0. I1 We apply the loop rule for the two loops indicated in the diagram: loop 1: I2 R 2 2 + 1 I1 R 1 = 0; + 1 I2 (10 ) 6 V + 12 V I1 (5 ) = 0; loop 2: + I3 R 3 3 + I2 R 2 + 2 = 0; + I3 (12 ) 9 V + I2 (10 ) + 6 V = 0. When we combine these equations, we get I1 = + 0.45 A, I2 = + 0.38 A, I3 = 0.068 A. 20. We assume the current directions shown in the diagram. We use conservation of current at point a: Iin = 0; I 1 I2 I3 = 0. I1 We apply the loop rule for the two loops indicated in the diagram: + loop 1: 2 I2 R 2 + 1 I1 R 1 = 0; 1 5 V I2 (3 ) + 3 V I1 (2 ) = 0; loop 2: I3 R 3 + I2 R 2 2 = 0; I3 (4 ) + I2 (3 ) 5 V = 0. When we combine these equations, we get I1 = + 1.577 A, I2 = + 1.616 A, and I 3 = 0.039 A. The negative sign means that the current is up. 21. We combine the three resistors, which are in parallel: 1/Req = 1/R1 + 1/R2 + 1/R3 = 1/(250 ) + 1/(420 ) + 1/(510 ), which gives Req = 120 . The potential difference across the equivalent resistance is V ab = IReq = (0.020 A)(120 ) = 2.4 V . Because this is the potential difference across each resistor, we have I 1 = Vab/R1 = (2.4 V)/(250 ) = 9.6 10 3 A = 9.6 mA; I 2 = Vab/R2 = (2.4 V)/(420 ) = 5.7 10 3 A = 5.7 mA; I 3 = Vab/R3 = (2.4 V)/(510 ) = 4.7 10 3 A = 4.7 mA. Note that we have conservation of current: I = I1 + I2 + I3 = (9.6 10 3 A) + (5.7 10 3 A) + (4.7 10 3 A) = 0.020 A.
a R2
2 1
R3 I3
2
+ b
I2
3
+
R1
a +
2
I2
2
I3 R3
1
R2 b
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Page 27-6
Fishbane, Gasiorowicz, and Thornton
R1 22. Because no two resistors have the same current and no two resistors have the same potential difference across them, we cannot combine them in series or parallel. We assume the current directions shown in the diagram. We use conservation of current at point a: Iin = 0; I 1 I2 + I3 = 0. We apply the loop rule for the two loops indicated in the diagram: loop 1: + I1 R 1 + I2 R 2 = 0; loop 2: I2 R 2 I3 R 3 + = 0. When we combine these equations, we get I 1 = R3 /(R1 R 2 + R1 R 3 + R2 R 3 ) , I 2 = (R1 + R3 )/(R1 R 2 + R1 R 3 + R2 R 3 ) , I 3 = R1 /(R1 R 2 + R1 R 3 + R2 R 3 ) .
1
I1
+ a +
R2 b
2
I2 I3
R3
23. With identical batteries, the terminal voltage and the current through each battery are the same. When the batteries are connected in parallel, the terminal voltage is the voltage across the resistance, so we have Ii = NIi = Ia; V ab = Iir = IaR. If we eliminate Ii from these equations, we get I a = /[R + (r/N)]. When the batteries are connected in series, the current through each battery is the current through the resistance, so we have V i = NVi = Vcd; ( Ibr) = IbR, which gives I b = /[r + (R/N)]. In general, R will be much greater than r, so Ib will be greater than Ia.
24. We assume that the current going to point C is negligible. If I is the current through the resistors, which are in series, we have V AB = I(R1 + R2 ), and VCD = IR2 . If we eliminate I, we get V CD = [R 2 /(R1 + R2 )]V AB . 25. We combine R2 and RL , which are in parallel: 1/R = 1/R2 + 1/RL, which gives R = R2 R L/(R2 + RL) . We now have a single-loop circuit, so the current is + I = /(R1 + R). I The voltage across the load is the voltage across R: V L = IR = R/(R1 + R). When we use the expression for R, we get V L = R2 R L/[(R1 R 2 + R1 R L + R2 R L)] = (10 V)(3.3 k)R L/[(3.3 k)(3.3 k) + (3.3 k)R L + (3.3 k)R L] = 33RL/(10.9 + 6.6RL) V, with R L in k. For the given loads, we have V 20 k = (33)(20 k)/[10.9 + 6.6(20 k)] = 4.62 V, V = 0.38 V; V 200 k = (33)(200 k)/[10.9 + 6.6(200 k)] = 4.96 V, V = 0.04 V ; V 2 M = (33)(2 103 k)/[10.9 + 6.6(2 103 k)] = 4.996 V, V = 0.004 V.
R2
RL
VL
R1
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Page 27-7
Chapter 27: Direct-Current Circuits
26. From the diagram, we have I 1 = Vac/R1 = /R; 55 10 3 A = (2.8 V)/R, which gives R = 51 .
I
a
1 2
b
3
+
R1 I1 c d
R2 I2
R3 I3
27. We can consider point a to be along the top and point b to be along the bottom, so the conservation of current gives Iin = 0; junction a: I1 I2 I3 I4 = 0; junction b: I2 + I3 + I4 I1 = 0. Thus there is only one independent junction. For the three loops indicated on the diagram, we have loop 1: I1 R I2 R I1 R + = 0; loop 2: + I2 R I3 R = 0; loop 3: + I3 R I4 R = 0. The solution of these four equations gives I 1 = 3/7R, I2 = I3 = I4 = /7R.
a
3 2
R
1
+ I1
R I4
R I3
R I2 b R
28. For the conservation of current, we have a I1 I junction a: I I1 I2 = 0; R junction b: I1 I3 I4 = 0; R 2 1 I2 junction d: I4 + I5 I = 0. R For the three loops indicated on the diagram, we have b c + loop 1: I1 R I4 R + = 0; I3 loop 2: I2 R + I3 R + I1 R = 0; 3 R I5 loop 3: I3 R I5 R+ I4 R = 0. R When we solve these equations, we get I4 I I 3 = 0, I1 = I2 = I4 = I5 = /2R, I = /R. d Because the current through the battery must be I = /Req , we get Req = R. From the symmetry of the resistance network, we know that there can be no current through the central resistor, which could be removed. Then we would have two pairs of series resistors in parallel, which gives Req = R.
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Page 27-8
Fishbane, Gasiorowicz, and Thornton
29. ( a ) For the conservation of current at point a, we have R2 R1 a Iin = 0; I 1 + I2 I3 = 0. For the two loops indicated on the diagram, we have 2 loop 1: 2 I2 R 1 I3 R 3 I2 R 4 = 0; 1 + + + 9 V I2 (100 ) I3 (50 ) I2 (200 ) = 0; 2 R3 1 loop 2: 1 I1 R 2 I3 R 3 I2 R 5 = 0; I3 I1 + 6 V I1 (150 ) I3 (50 ) I1 (250 ) = 0. I2 When we solve these equations, we get b R5 I 1 = 0.0106 A, I2 = 0.0242 A, I3 = 0.0348 A. R4 The power dissipated in the 50- resistor is P3 = I3 2 R 3 = (0.348 A)2 (50 ) = 0.0605 W = 60.5 mW. (b) If the terminals on the 6-V battery are reversed, we have the same equations, except for the sign of 1 : I 1 + I2 I3 = 0. loop 1: 2 I2 R 1 I3 R 3 I2 R 4 = 0; + 9 V I2 (100 ) I3 (50 ) I2 (200 ) = 0; loop 2: 1 I1 R 2 I3 R 3 I2 R 5 = 0; 6 V I1 (150 ) I3 (50 ) I1 (250 ) = 0. When we solve these equations, we get I 1 = 0.0165 A, I2 = 0.0281 A, I3 = 0.0116 A. The power dissipated in the 50- resistor is P3 = I3 2 R 3 = (0.0116 A)2 (50 ) = 0.0068 W = 6.8 mW. The total resistance of the circuit is R = 20(2 ) + 80 = 120 , while the total emf is = 20(12 V) = 240 V. The current is then I = /R = 240 V/120 = 2.0 A. (b) Now the total resistance of the circuit is R = (2 )/20 + 20 = 20.1 , while the total emf is = 12 V. The current is then I = /R = 12 V /20.1 = 0.60 A. (c) Now the total resistance of the circuit is R = 5(2 )/4 + 40 = 42.5 , while the total emf is = 5(12 V) = 60 V. The current is then I = /R = 60 V /42.5 = 1.4 A.
30. (a)
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Page 27-9
Chapter 27: Direct-Current Circuits
31. ( a ) We can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R3 and R4 , which are in parallel: 1/R5 = 1/R3 + 1/R4 = 1/(100 ) + 1/(50 ), which gives R 5 = 33.3 . We combine R1 and R2 + R5 , which are in parallel: 1/R6 = 1/R1 + 1/(R2 + R5 ) = 1/(100 ) + 1/(20 + 33.3 ), which gives R 6 = 34.8 . Because Vab = 6 V, we find I2 from I 2 = Vab/(R5 + R2 ) = (6 V)/(33.3 + (0 ) = 0.113 A. We can now find Vcb from V cb = I2 R 5 = (0.113 A)(33.3 ) = 3.75 V. The current through the 50- resistor is I 4 = Vcb/R4 = (3.75 V)/(50 ) = 0.075 A. (b) For the conservation of current, we have junction a: I I1 I2 = 0; junction c: I 2 I3 I4 = 0. For the three loops indicated on the diagram, we have loop 1: I1 R 1 = 0; 6 V I1 (100 ) = 0; loop 2: I 1 R 1 I2 R 2 I3 R 3 = 0; I 1 (100 ) I2 (20 ) I3 (100 ) = 0; loop 3: I3 R 3 + I4 R 4 = 0; I3 (100 ) + I4 (50 ) = 0. When we solve these equations, we get I 1 = 0.060 A, I2 = 0.113 A, I3 = 0.038 A, I4 = 0.075 A. Thus, the current through the 50- resistor is 0.075 A.
I2 c
3
R2
a
2
1
+
R4 I4
R3 I3
R1 I1 b
I
I2 c
R2
a
+ R5 I2 b a R1 I1 I
+ R6 I b I
32. For the conservation of current at point a, we have Iin = 0; I 1 I2 I3 = 0; R2 R1 a I 1 I2 0.1 A = 0. For the two loops indicated on the diagram, we have loop 1: 1 I1 R 1 I3 R 3 = 0; +3 V I1 (5 ) (0.1 A)R3 = 0; 2 1 + loop 2: 2 + I3 R 3 I2 R 2 = 0; 1 R3 +6 V + (0.1 A)R3 I2 (20 ) = 0. I3 When we solve these equations, we get I1 I 1 = 0.44 A, I2 = 0.34 A, and R3 = 8 . b If I 3 = 0.1 A, the equations become I 1 = I2 0.1 A; +3 V I1 (5 ) ( 0.1 A)R3 = 0; +6 V + ( 0.1 A)R3 I2 (20 ) = 0. When we solve these equations, we get I1 = 0.28 A, I2 = 0.38 A, and R3 = 16 . Because we cannot have a negative resistance, it is not possible to have I3 = 0.1 A.
+ I2
2
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Page 27-10
Fishbane, Gasiorowicz, and Thornton
33. For the conservation of current at point b, we have Iin = 0; I I1 I2 = 0; For the two loops indicated on the diagram, we have loop 1: 1 I1 r1 IR = 0; I1 +12 V I1 (0.1 ) I(5 ) = 0; b loop 2: 2 + I2 r2 IR = 0; +10 V I2 (10 ) I(5 ) = 0. I2 When we solve these equations, we get I 1 = 2.52 A, I2 = 0.17 A, and I = 2.35 A. The current through the load resistor is 2.35 A. 1 supplies 2.52 A; 2 supplies no current, it is being charged by 1 . 34. On the diagram, we show the potential difference applied between points A and B. Because all of the resistors are the same, symmetry means that the three currents leaving point A must be the same three currents entering point B. This means that there is no current in the resistor between points C and D, which can be removed without changing the currents. When we redraw the circuit, we see that we have three parallel branches between points A and B. The currents are I 1 = VAB/R = (4 V)/(1 ) = 4 A; I 2 = I3 = VAB/(R + R) = (4 V)/(1 + 1 ) = 2 A. The power dissipated in each of the resistors is PAB = I1 2 R = (4 A)2 (1 ) = 16 W; PCD = 0; Pall others = I2 2 R = (2 A)2 (1 ) = 4 W. 35. In the original configuration, there are no series or parallel combinations; however, from the symmetry of the resistors, we know that the current that goes into point A must split equally to go through the cube. The three points on the other side of the three resistors must be at the same potential, so we can connect them with a wire without changing the currents. In the same way, the other three corners of the cube must be at equal potentials, so we can connect them with a wire. From the redrawn circuit, we see that we have three parallel combinations, two of which are the same: 1/R1 = 1/R3 = (1/R) + (1/R) + (1/R), which gives R 1 = R3 = R/3; 1/R2 = (1/R) + (1/R) + (1/R) + (1/R) + (1/R) + (1/R), which gives R 2 = R/6; We combine these three in series to get R eq = (R/3) + (R/6) + (R/3) = 5R/6. The current in the equivalent resistor is I = V/(5R/6) = 6V/5R. The current in a resistor connected to point A or point B is I 1 = I3 = I/3 = 2V/5R . The current in each of the other resistors is I 2 = I/6 = V/5R.
R
I
r1
1
+
1
a
2
r2
+
2
C
R I3 R I2 A I1 R R I3 I2 A I1 R R
R D
R I3 R
I2 B R R
B
B +
A R R A I I1 A I R1 R2 I2 B R3 I3 A I B R eq R I
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Page 27-11
Chapter 27: Direct-Current Circuits
36. ( a ) For n = 1, we have two resistors in series: R 1 = R + R = 2R . (b) For n = 2, we have a resistor in series with a parallel combination of a resistor and P1 resistance R1 : R 2 = R + R1 R/(R1 + R) = R + 2RR/(2R + R) = 5R/3. (c) For n = 3, we have a resistor in series with P2 a parallel combination of a resistor and resistance R2 : R 3 = R + R2 R/(R2 + R) 1 2 = R + (5RR/3)/(5R/3 + R) = 13R/8. (d) For n rungs, we have a resistor in series with a parallel combination of a resistor and resistance Rn 1: R n = R + Rn 1R/(Rn 1 + R). In the limit of n , we have R n = Rn 1 = Req , which gives R eq = R + ReqR/(Req + R), which reduces to Req2 RReq R2 = 0. The solutions to this quadratic equation are R eq = !(1 5)R. Because the resistance cannot be negative, we have Req = !(1 + 5)R = 1.618R .
n1
n
37. The equivalent resistance of the circuit is R = R1R V/(RV + R1) + R2 , where RV is the internal resistance of the voltmeter. The current in the circuit is then I = / R = /[ R1R V/(RV + R1) + R2] . The voltage across R2 is V 2 = IR2 , so the voltage across R1 is V1 = V2 = IR2 = R2/[ R1R V/(RV + R1) + R2]. Plug in = 6 V, R1 = 1400 , R2 = 10 k, and RV = 200 k to obtain V 1 = 0.732 V (for RV = 200 k). If R V is changed to 10 M, then from the formula above we get V 1 = 0.737 V (for RV = 10 M). 38. We have a single-loop circuit of the battery and the voltmeter. Because the terminal voltage of the battery is the voltage across the voltmeter, we have V terminal = IRV; 1.45V = I(60 103 ), which gives I = 2.4 10 5 A. For the battery, we have V terminal = Ir; 1.45V = 1.5 V (2.4 10 5 A)r, which gives r = 2.1 k . 39. Without the ammeter in the circuit, we have I = /R. With the ammeter in the circuit, we have I = /(RA + R). When we combine these two equations, we get I /I = R/(R + RA) = 1/(1 + RA/R) 1 (RA/R), if RA/R << 1. To get the desired accuracy, we want I /I 0.999, or RA/R 0.001. The maximum allowable value of RA is determined by the smallest value of R: R A = 0.001(10 ) = 0.010 .
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Page 27-12
Fishbane, Gasiorowicz, and Thornton
40. The voltmeter is placed in parallel with the resistor, so the equivalent resistance is 1/Req = 1/R + 1/RV , which becomes R eq/R = RV/(R + RV) = 1/(1 + R/RV) 1 R/RV , if R/RV << 1. To get the desired accuracy, we want Req/R 0.999, or R/RV 0.001. The minimum allowable value of RV is determined by the largest value of R: R V = R/0.001 = (5 103 )/0.001 = 5 106 . 41. We find the equivalent resistance for R and RV , which are in parallel, from 1/Req = 1/R + 1/RV , which gives Req = RRV/(R + RV) . ( a ) R eq = (10 )(105 )/(10 + 105 ) 10 . (b) R eq = (105 )(105 )/(105 + 105 ) = 5 104 . (c) R eq = (100 106 )(105 )/(100 106 + 105 ) 105 . The equivalent resistance has the value of the resistor when RV >> R. 42. The resistance of the voltmeter is RV = R + RA. The maximum current through the voltmeter is the maximum current through the ammeter, so we have V max = Imax(R + RA) ; 3 V = (5 10 3 A)(R + 1.8 104 ), which gives R = 0.6 k . 43. The equivalent resistance of the circuit is R + r, where we assumed that the internal resistance of the voltmeter is nearly infinity. The current in the circuit is then I = /(R + r). The voltage across R, i.e., the reading of the voltmeter, is then V = Ir = r/( R + r) = R/( R + r). For R = 20 we have V = 23 V, and for R = 5 we have V = 16 V; so 23 V = (20 )/(20 + r); 16 V = (5 )/(5 + r). Solve these two equations to obtain = 26.9 V and r = 3.41 . Thus for R = 50 V = R/( R + r) = (26.9 V)(50 )/(50 + 3.41 ) = 25 V. 44. To be able to accommodate 2 10 3 A, which is 10 times the current that causes the galvanometer to fully deflect, we need to add a resistor of resistance R in parallel to it to route 90% of the current, leaving only 10% of 2 10 3 A, or 2 10 4 A, to flow through the galvanometer. If the full-deflection voltage is V then V = IGR G = IRR; (2 10 4 A)(20 ) = [(90%)(2 10 3 A)]R; which yields R = 2.2 . So one needs to connect a 2.2- resistor in parallel with the galvanometer. To change the full-deflection potential to V'= 0.2 V, place a resistor of resistance R' in series with the galvanometer. Upon full deflection the current in both of them is IG = 2 10 4 A, so V'= IG (RG + R'); 0.2 V = (2 10 4 A)(20 + R'); which yields R' = 0.98 k. So one needs to connect a 0.98-k resistor in series with the galvanometer.
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Page 27-13
Chapter 27: Direct-Current Circuits
45. Because the shunt resistor is in parallel with the galvanometer, we have V meter = IGR G = IsR s , which gives RG/Rs = Is/IG. We use the junction at one side of the meter to find the total current through the meter: I = IG + Is = IG(1 + Is/IG) = IG(1 + RG/Rs) . 46. We combine R2 and RV , which are in parallel: 1/R3 = 1/R2 + 1/RV , which gives R3 = R2 R V/(R2 + RV) . I For the resulting single-loop circuit, we have I = /(R1 + R3 ) = /[R1 + R2 R V/(R2 + RV)]. + The output voltage is V out = IR3 = {/[R1 + R2 R V/(R2 + RV)]}[R 2 R V/(R2 + RV)] = R2 R V/(R1 R 2 + R1 R V + R2 R V) = /(1 + R1 /R2 + R1 /RV) . For the two voltmeters, we have R V = 500 k: V out = (1200 V)/[1 + (30 k)/(50 k) + (30 k)/(500 k)] = 723 V . R V = 100 M: V out = (1200 V)/[1 + (30 k)/(50 k) + (30 k)/(100 103 k)] = 750 V . 47. When there is no current through the galvanometer, we have V BC = 0, a current I1 through R and R1 , and a current I2 through R2 and R3 . Thus we have V AD = I1 (R + R1 ) = I2 (R 2 + R3 ), and V AB = VAC or I1 R = I2 R 2 . When we divide these two equations, we get (R + R1 )/R = (R2 + R3 )/R2 , or 1 + (R1 /R) = 1 + (R3 /R2 ), which gives R 1 /R = R3 /R2 . The unknown resistance is R = R1 R 2 /R3 .
R1
R 2 Vout
RV
B R A G R2 C + R3 R1 D
S
48. The voltage read by the voltmeter is also the voltage across Rx and the ammeter: V = I(RA + Rx ), which gives R x = V/I RA . We will have R x = V/I when RA << V/I (when RA << R x ) .
Rx I
RA
A
V IT
RV
49. Because the voltmeter and Rx are in parallel, their equivalent resistance is R eq = RVR x /(RV + Rx ) . Rx The voltage read by the voltmeter is also the voltage across Req: V = IReq = IRVR x /(RV + Rx ), which gives V R x = (V/I)/(1 V/IRV) . We will have R x = V/I when V/IRV << 1, or I RV A RV >> V/I (when RV >> R x ) . R
A
Ix IV
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Page 27-14
Fishbane, Gasiorowicz, and Thornton
50. We find the resistance from time constant = RC; 5 10 4 s = R(16 10 6 F), which gives R = 31 . 51. We find the capacitance from time constant = RC; 2.0 s = (105 )C, which gives
C = 2.0 10 5 F = 20 F .
52. We use the definitions of R and C to find the units of RC: RC = (V/I)(Q/V) = Q/I = coulomb/(coulomb/second) = second. For the given data, we have R 1 C 1 = (5 106 )(30 10 6 F) = 150 s ; R 2 C 2 = (8 103 )(3 10 6 F) = 24 10 3 s = 24 ms; R 3 C 3 = (20 )(50 10 1 2 F) = 1 10 9 s = 1 ns. 53. With the time constant as the flash time, we have time constant = RC; (1/500) s = R(600 10 6 F), which gives R = 3.3 . 54. From Q = C(1 e t/RC), we obtain dQ/dt = C[ ( 1/RC)e t/RC] = (/R)e t/RC. We substitute these equations into R(dQ/dt) Q/C; R(/R)e t/RC (C/C)(1 e t/RC) = e t/RC + e t/RC = 0, so Eq. (2721) is satisfied. 55. From Q = Q0 e t/RC, we obtain dQ/dt = ( Q0 /RC)e t/RC. We substitute these equations into R(dQ/dt) + Q/C; (RQ0 /RC)e t/RC) + (Q0 /C)e t/RC = 0, so Eq. (2725) is satisfied. 56. ( a ) The time constant is RC = (3 106 )(350 10 6 F) = 1050 s. (b) For the charge on the capacitor, we have Q = Q0 (1 e t/RC) = 0.90Q0 , which gives e t/RC = 0.10. For the charging current, we have I = (/R)e t/RC = [(50 V)/(3 106 )](0.10) = 1.7 10 6 A. 57. Because there is no internal resistance in the battery, the potential difference across R2 and across the capacitor branch is . The current in R 2 is constant: I 2 = /R2 . The charging current in the capacitor branch is I 1 = (/R1)e t/R 1C. From the junction, the current in the battery is I battery = I 1 + I2 = ( /R1 )e t/R1C + (/R 2).
+ S
C I1 R1
I2
R2
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Page 27-15
Chapter 27: Direct-Current Circuits
58. The time constant of the circuit is RC = (350 103 )(20 10 6 F) = 7.0 s. The charging current in the resistor is I = (/R)e t/RC, so the voltage across the resistor is V = IR = e t/RC, = (200 V)e (4.0 s)/(7.0 s) = 113 V . The charge on the capacitor is Q = C(1 e t/RC) = [(20 10 6 F)(200 V)][1 e (4.0 s)/(7.0 s)] = 1.7 10 3 C = 1.7 mC.
59. The possible capacitance values that we have are C 1 = C2 = 5 F; C parallel = C1 + C2 = 5 F + 5 F = 10 F; C series = C1 C 2 /(C1 + C2 ) = (5 F)(5 F)/[(5 F) + (5 F)] = 2.5 F. We need to combine the resistors to produce one of the following resistance values: R a = RC/C1 = (1 10 3 s)/(5 10 6 F) = 200 ; R b = RC/Cparallel = (1 10 3 s)/(10 10 6 F) = 100 ; R c = RC/Cseries = (1 10 3 s)/(2.5 10 6 F) = 400 ; If we connect the 300- resistors in parallel, we get R 3 = R2 R 2 /(R2 + R2 ) = (300 )(300 )/[(300 ) + (300 )] = 150 . We see that we can produce Rc by putting this combination in series with the 250- resistor: R c = R1 + R3 = 250 + 150 = 400 . Thus we connect the 300- resistors in parallel with each other and in series with the 250- resistor and the two capacitors.
60. For a parallel-plate capacitor with separation d, we have C = 0 A/d. For the resistance of the dielectric, we have R = d/A. The time constant is RC = (d/A)(0 A/d) = 0 , which is independent of the area and separation.
61. We use the result of Problem 62 to find the time constant: RC = 0 = (3.2)(8.85 10 1 2 F/m)(2 101 4 m) = 5.66 103 s. As the capacitor discharges, when 70% of the charge on the plates has leaked away, we have Q = Q0 e t/RC = 0.30Q0 , which gives e t/RC = 0.30; t/RC = 1.2. The time for 70% of the charge to leak away is t = 1.2RC = (1.2)(5.66 103 s) = 6.8 103 s (1.9 h).
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Page 27-16
Fishbane, Gasiorowicz, and Thornton
62. When the switch is closed, charge will move from C1 to C2 ; C1 with a variable current I in the single loop or series circuit: I = dQ1 /dt = dQ2 /dt. + With the same current in the two capacitors, they are connected in series. Q0 We find the equivalent capacitance of the circuit from I R S 1/C = 1/C1 + 1/C2 , which gives C = C2 C 1 /(C1 + C2 ) . The time constant of the circuit is RC. C2 Q1 , the charge on C1 , will decrease and Q2 , the charge on C2 , will increase until the potential difference will be the same across both capacitors, at which point the current becomes 0. The charges will have reached their final value, which we find from V 1 = V2 ; or Q1f/C1 = Q2f/C2 . Because there has been no loss of charge, we have Q0 = Q1f + Q2f . When we combine these two equations, we get Q1f = Q0 C 1 /(C1 + C2 ), and Q2f = Q0 C 2 /(C1 + C2 ) . Capacitor C 2 will charge just like a single capacitor from 0 to its final value Q2f , so we have Q2 = [Q0 C 2 /(C1 + C2 )](1 e t/RC) . At any time the total charge is conserved and equal to the initial charge Q0 . We find the charge on C1 as a function of time from Q1 = Q0 Q2 , which gives Q1 = [Q0 C 1 /(C1 + C2 )] + [Q0 C 2 /(C1 + C2 )]e t/RC . Note that at t = 0, this gives Q 0 and after a long time it gives Q1f . [It is also possible to solve the circuit equation, obtained by adding the potential drops around the loop: Q1 /C1 IR Q2 /C2 = 0. When the current is put in terms of the rate of change of the charge and the conservation of charge is used, a differential equation is obtained. The solution is the given equations.]
63. The power used by each appliance is Pi = IiV, so the currents draw from the main supply are I 1 = P1 /V = (50 W)/(120 V) = 0.417 A ; I 2 = P2 /V = (60 W)/(120 V) = 0.500 A ; I 3 = P3 /V = (20 W)/(120 V) = 0.167 A . 20 64. ( a ) The current in the circuit will be clockwise. For the single loop, we have R I I = (1 2 )/R + + = (12 V 6 V)/(20 ) = 0.30 A. 12 V 2 6 V 1 (b) The rate at which energy is being stored in the smaller battery is P = I2 = (0.30 A)(6 V) = 1.8 W. (c) The rate of energy dissipation in the resistor is P = I 2 R = (0.30 A)2 (20 ) = 1.8 W . Note that the sum of the answers to parts (b) and (c) is the rate at which energy is being provided by the larger battery, 3.6 W.
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Page 27-17
Chapter 27: Direct-Current Circuits
65. Because the power dissipated in the effective resistance is the sum of the powers dissipated in the individual resistors, we have Peq = Pi = nPi ; 30 W = n(5 W), which gives n = 6 resistors. If we connect the six resistors in parallel, we have 1/Req = (1/Ri) = n/Ri ; 1/(100 ) = 6/Ri , which gives Ri = 600 . We can connect six 600- resistors in parallel. If we connect the six resistors in series, we have R eq = Ri = nRi ; 100 = 6Ri , which gives Ri = 16.7 . We can connect six 16.7- resistors in series. To have the same power dissipated in each resistor requires that the currents be the same, which means there must be symmetry in the arrangement of the resistors. If we connect two series resistors in parallel with two more sets of two series resistors, we have 1/Req = 1/2Ri + 1/2Ri + 1/2Ri = 3/2Ri ; 1/(100 ) = 3/2Ri , which gives Ri = 150 . We can connect two series 150- resistors in parallel with two more sets of two series 150- resistors. If we connect three series resistors in parallel with a set of three series resistors, we have 1/Req = 1/3Ri + 1/3Ri = 2/3Ri ; 1/(100 ) = 2/3Ri , which gives Ri = 66.7 . We can connect three series 66.7- resistors in parallel with three series 66.7- resistors. 66. The current for this single loop is I = /(R + r). The power delivered to the external resistor is the power dissipated in the resistor: Pext = I 2 R = 2 R/(r + R)2 . To find the value of R that maximizes the power, we set dPext/dR = 0: dP/dR = [2 /(r + R)2 ] [22 R/(r + R)3 ] = 2 (r R)/(r + R)3 = 0, which gives r = R. 67. (a) The short-circuit current (when there is no load) provided by the battery is I = /r = 12.6 V/0.05 = 0.25 kA. (b) The voltage V across the battery terminal during recharging must overcome both and the internal resistance of the battery: V = + Ir = 12.6 V + (2.5 A)(0.05 ) = 12.7 V . (c) Energy stored = It = (12.6 V)(2.5 A)(10 h)(3600 s/h) = 1.1 106 J . 68. If the batteries are in parallel then the equivalent resistance of the circuit is R + r/n, and the emf is . The current is I parallel = /(R + r/n) = n/(nR + r). If the batteries are in series then the equivalent resistance of the circuit is R + nr, and the emf is n. The current is I series = n/(R + nr). If we compare Iparallel with Iseries, it is clear that I parallel > Iseries if nR + r < R + nr, or R < r. Similarly, if R > r then Iseries > Iparallel. Therefore, to maximize the current, put the batteries in series if R > r and in parallel if R < r .
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Page 27-18
Fishbane, Gasiorowicz, and Thornton
69. For this single-loop circuit, we have I = /(R + r) = /[R + ( + I )], which is a quadratic equation for I: I 2 + (R + )I = 0. The positive solution is I = { (R + ) + [(R + )2 + 4]1/2}/2 . The ratio of power delivered to the load to the power dissipated in the battery is PR /Pr = I 2 R/[I 2 ( + I )] = R/( + I ). PR/Pr I (A) 25 30 20 15 10 5 0 R () 1 2 3 4 5 25 20 15 10 5 0 1 2 3 4 5 R ()
70. Using P = V2 /R, we find the two resistances: R 1 = V2 /P1 and R2 = V2 /P2 . The equivalent resistance for the series connection is R s = R1 + R2 = (V2 /P1 ) + (V 2 /P2 ) = V2 (P 1 + P2 )/P1 P2 . The power generated is Ps = V 2 /Rs = P1 P2 /(P1 + P2 ) . The equivalent resistance for the parallel connection is R p = R1 R 2 /(R1 + R2 ) = (V 4 /P1 P2 )/(V 2 /P1 + V2 /P2 ) = V 2 /(P1 + P2 ) . The power generated is Pp = V2 /Rp = V 2 /[V2 /(P1 + P2 )] = P1 + P2 . 71. The two heating elements in the furnace are initially connected in parallel. The power dissipated in a resistor is P = I2 R = V 2 /R. The resistances are R 1 = V2 /P1 = (110 V)2 /(1000 W) = 12.1 . R 2 = V2 /P2 = (110 V)2 /(2000 W) = 6.05 . The power can be reduced by increasing the resistance, which means connecting them in series: P = V2 /(R1 + R2 ) = (110 V)2 /(12.1 + 6.05 ) = 667 W . 72. ( a ) For the series arrangement, we have I s = ( + )/(R + 2r), which gives
I s = /(!R + r). (b) For the parallel arrangement, we use symmetry to see that the current in each battery is the same. For the junction, we have I p I I = 0, or Ip = 2I. For one loop, we have Ip R Ir = 0 = Ip R !Ip r, which gives
+
Is R
r + r Series
+ I
r
+ I
Ip R
r
Parallel
I p = /(R + !r). For large R: Is 2/R, Ip /R; so Is is larger. For small R: Is /r, Ip 2/r; so Ip is larger.
2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Page 27-19
Chapter 27: Direct-Current Circuits
73. The devices are connected in parallel, so we have I = PV; I mixer = (800 W)/(120 V) = 6.67 A. I vacuum = (600 W)/(120 V) = 5.00 A. I chandelier = 10(60 W)/(120 V) = 5.00 A. If all three devices are used at the same time, the fuse will blow. Each bulb draws 0.50 A. To find the number of bulbs that can be used without blowing the fuse, we have I max = Imixer + Ivacuum + NIbulb ; 15 A = 6.67 A + 5.00 A + N(0.50 A), which gives N = 6.7 bulbs. Thus 7 bulbs will blow the fuse.
74. We find the maximum current through a resistor from R B A P1max = Imax2 R; R R R 2 W = Imax2 (30 ), which gives Imax = 0.26 A. R When the three are connected in series, circuit A, Imax 3Imax R the maximum current goes through each resistor, so we have PAmax = 3P1max = 3(2 W) = 6 W. Imax C When the three are connected in parallel, circuit B, D R R R the maximum current goes through each resistor, R so we have R 1I PBmax = 3P1max = 3(2 W) = 6 W . 2 max Imax R 3I 1I In circuit C, the resistor in series has the maximum current. max max 2 2 From symmetry the current in the other two resistors will Imax be !Imax. The maximum power is PCmax = P1max + 2(!Imax) 2 R = P1max + !P1max = *(2 W) = 3 W . In circuit D, the branch with the single resistor has the maximum current. Because the total resistance in the other branch is 2R, the current in the other branch will be !Imax. The maximum power is PDmax = P1max + 2(!Imax) 2 R = P1max + !P1max = *(2 W) = 3 W .
75. Normally, this circuit would have six currents, one for each branch. We have used the symmetry of the circuit to reduce the number of currents to four, as shown in the diagram. For the junction equations, we have junction A (or D): I I1 I2 = 0; (1) junction B (or C): I1 + I3 I2 = 0. (2) For the loop equations, we have loop ACDA: I2 R I1 R + = 0; (3) loop BDCB: I2 R + I1 R I3 R = 0; (4) When we combine Eq. (2) and Eq. (4), we find I3 = 0 (as suggested by symmetry) and I1 = I2 . From Eq. (3), we get I1 = I2 = /2R. From Eq. (1), we get I = 2I1 = /R.
A I1 R
I I2
+ I2 D I1 R
R B I3 R
R C
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Page 27-20
Fishbane, Gasiorowicz, and Thornton
76. For a cylindrical wire, we have R = L/A, I = V/R = VA/L, J = I/A = V/L = E/ = V/L, and P = I 2 R = I 2 L/A. The two wires have the same length and area. ( a ) When the wires are in series, the currents must be the same, so we have JAl/JCu = IAl/ICu = 1. EAl/ECu = JAlAl/JCuCu = Al/Cu = (2.82 10 8 m)/(1.72 10 8 m) = 1.64. PAl/PCu = (I 2 AlL/A)/(I 2 CuL/A) = Al/Cu = 1.64. (b) When the wires are in parallel, the potential difference is the same, so we have JAl/JCu = IAl/ICu = (VA/AlL)/(VA/CuL) = Cu/Al = 0.61. EAl/ECu = JAlAl/JCuCu = (JAl/JCu)( Al/Cu) = 1. PAl/PCu = (IAl2 AlL/A)/(ICu2 CuL/A) = (IAl/ICu) 2 (Al/Cu) = (Cu/Al) 2 (Al/Cu) = Cu/Al = 0.61. (c) Because the currents and areas are the same, all wires have the same current density. For constant current, the electric field and the power loss is proportional to the resistivity. Thus silver, with the smallest resistivity, has the weakest field and the least power loss.
77. If there is no current through the ammeter, we find the current in the source loop from V S IS(R 1 + R2 ) = 0, which gives IS = VS/(R1 + R2 ) . For the loop with the unknown, we have V x ISR 2 = 0. When we use the expression for IS , we get V x = VSR 2 /(R1 + R2 ) .
+ Vs
R1 R2 + Vx
A
78. ( a ) Immediately after the switch is closed, there is no charge R2 C B R1 2 on the capacitors and thus no potential difference across them. A + For the loop we have IiR1 IiR2 = 0; 9 V Ii(300 ) Ii(1000 ) = 0, which gives Ii = 0.0069 A. + Between b and a we have S (V B VA) i = + IiR 2 = (0.0069 A)(1000 ) = 6.9 V . (b) After a long time, the current will be 0. The two capacitors are in series, with an equivalent capacitance: 1/Ceq = 1/C1 + 1/C2 = 1/(5 F) + 1/(2 F), which gives Ceq = 1.43 F. The final charge on either capacitor is Q = Ceq = (1.43 F)(9 V) = 12.9 C. Between B and A we have (V B VA) f = + Q/C2 = (12.9 C)/(1.43 F) = 6.4 V . (c) The time constant of the circuit is time constant = ReqC eq = (1 k + 0.3 k)(1.43 F) = 1.86 ms. As a measure of how fast the circuit reaches a steady state, we use 10 time constants: t = 10ReqC eq = 10(1.86 ms) = 19 ms.
C1 C + I
2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Page 27-21
Chapter 27: Direct-Current Circuits
79. ( a ) With the switch open, we have a series circuit of the three resistors and the capacitor. For the time constant we have time constant = ReqC = (r + R1 + R2 )C = [(0.04 ) + (0.1 ) + (2 )](10 F) = 21.4 s. (b) After a long time, there will be no current in the circuit. The battery emf will be across the capacitor: Q = C = (10 F)(240 V) = 2400 C = 2.4 mC. 80. ( a ) After a long time there will be a steady state; there will be no current in the capacitor branch: I 5 = 0; I1 = I3 , and I2 = I4 . For the two resistor branches we have V f Vd = = I2 (R 2 + R4 ) ; 6 V = I2 (180 + 60 ), which gives I2 = 0.025 A; V c Va = = I1 (R 1 + R3 ) ; 6 V = I1 (35 + 25 ), which gives I1 = 0.10 A. Because Va = Vd , we can find the relative potentials of b and e: V b Ve = (Vb Va) (V e Vd ) = I1 R 3 I2 R 4 = (0.10 A)(25 ) (0.025 A)(60 ) = + 1.0 V. The charge on the capacitor is Q = C(Vb Ve) = (5 F)(1.0 V) = 5 C. Because Vb > Ve , the top plate is positive. (b) The current through the 35- resistor is I1 = 0.10 A.
r
+ S R1 C
R2 R3 I3 I5 R4 d I4 b
a
R1 I1 C R2
c
e + I2
f
I
81. Because the emf has negligible resistance, the terminal voltage, which is the voltage across the capacitor, is the emf V 0 . The resistance of the material in the capacitor is R = L/A = L/A = d/r2 . ( a ) We find the electric field between the capacitor plates from the potential gradient: E = V0 /d. (b) The current density depends on the electric field: J = E = V0 /d. The current is I = JA = (V0 /d)r2 = r2 V 0 /d = V0 /R. 82. When we add another rung, as shown in the diagram, we have the resistance R* in parallel with a resistor, which has an equivalent resistance of R eq = RR*/(R + R*). This equivalent resistance is in series with two resistors. Because the resistance does not change, we have R* = 2R + [RR*/(R + R*)], which gives R* 2 2RR* 2R2 = 0. The solutions to this quadratic equation are R* = (1 3)R. Because the resistance cannot be negative, we have R* = (1 + 3)R = 2.732R .
R R R R R
a
R b R (a)
R R
R R
R R R R
R*
=
R
R*
(b)
R
2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Page 27-22