Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19 Page 19 – 1 CHAPTER 19 1. When the bulbs are connected in series, the equivalent resistance is R series = áR i = 4R bulb = 4(140 Ω) = 560 Ω. When the bulbs are connected in parallel, we find the equivalent resistance from 1/R parallel = á(1/R i ) = 4/R bulb = 4/(140 Ω), which gives R parallel = 35 Ω. 2. (a) When the bulbs are connected in series, the equivalent resistance is R series = áR i = 3R 1 + 3R 2 = 3(40 Ω) + 3(80 Ω) = 360 Ω. (b) When the bulbs are connected in parallel, we find the equivalent resistance from 1/R parallel = á(1/R i ) = (3/R 1 ) + (3/R 2 ) = [3/(40 Ω)] + [3/(80 Ω)], which gives R parallel = 8.9 Ω. 3. If we use them as single resistors, we have R 1 = 30 Ω; R 2 = 50 Ω. When the bulbs are connected in series, the equivalent resistance is R series = áR i = R 1 + R 2 = 30 Ω + 50 Ω = 80 Ω. When the bulbs are connected in parallel, we find the equivalent resistance from 1/R parallel = á(1/R i ) = (1/R 1 ) + (1/R 2 ) = [1/(30 Ω)] + [1/(50 Ω)], which gives R parallel = 19 Ω. 4. Because resistance increases when resistors are connected in series, the maximum resistance is R series = R 1 + R 2 + R 3 = 500 Ω + 900 Ω + 1400 Ω = 2800 Ω = 2.80 kΩ. Because resistance decreases when resistors are connected in parallel, we find the minimum resistance from 1/R parallel = (1/R 1 ) + (1/R 2 ) + (1/R 3 ) = [1/(500 Ω)] + [1/(900 Ω)] + [1/(1400 Ω)], which gives R parallel = 261 Ω. 5. The voltage is the same across resistors in parallel, but is less across a resistor in a series connection. We connect three 1.0-Ω resistors in series as shown in the diagram. Each resistor has the same current and the same voltage: V i = @V = @(6.0 V) = 2.0 V. Thus we can get a 4.0-V output between a and c. R R R a b c d I V
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Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 1
CHAPTER 19 1. When the bulbs are connected in series, the equivalent resistance is
Rseries = áRi = 4Rbulb = 4(140 Ω) = 560 Ω.
When the bulbs are connected in parallel, we find the equivalent resistance from
5. The voltage is the same across resistors in parallel, but is less
across a resistor in a series connection. We connect three 1.0-Ω resistors in series as shown in the diagram. Each resistor has the same current and the same voltage:
Vi = @V = @(6.0 V) = 2.0 V.
Thus we can get a 4.0-V output between a and c.
R R Ra b c d
I
V
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 2
6. When the resistors are connected in series, as shown in A, we have
RA = áRi = 3R = 3(240 Ω) = 720 Ω.
When the resistors are connected in parallel, as shown in B, we have
1/RB = á(1/Ri) = 3/R = 3/(240 Ω), so RB = 80 Ω.
In circuit C, we find the equivalent resistance of the two resistors in parallel:
1/R1 = á(1/Ri) = 2/R = 2/(240 Ω), so R1 = 120 Ω.
This resistance is in series with the third resistor, so we have
RC = R1 + R = 120 Ω + 240 Ω = 360 Ω.
In circuit D, we find the equivalent resistance of the two resistors in series:
R2 = R + R = 240 Ω + 240 Ω = 480 Ω.
This resistance is in parallel with the third resistor, so we have
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 3
8. (a) In series the current must be the same for all bulbs. If all bulbs have the same resistance, they will have the same voltage: Vbulb = V/N = (110 V)/8 = 13.8 V.
(b) We find the resistance of each bulb from
Rbulb = Vbulb/I = (13.8 V)/(0.40 A) = 34 Ω.
The power dissipated in each bulb is Pbulb = IVbulb = (0.40 A)(13.8 V) = 5.5 W.
9. For the parallel combination, the total current from the source is I = NIbulb = 8(0.240 A) = 1.92 A.
The voltage across the leads is
Vleads = IRleads = (1.92 A)(1.5 Ω) = 2.9 V.
The voltage across each of the bulbs is Vbulb = V – Vleads = 110 V – 2.88 V = 107 V.
We find the resistance of a bulb from
Rbulb = Vbulb/Ibulb = (107 V)/(0.240 A) = 450 Ω.
The power dissipated in the leads is IVleads and the total power used
is IV, so the fraction wasted is IVleads/IV = Vleads/V = (2.9 V)/(110 V) = 0.026 = 2.6%.
10. In series the current must be the same for all bulbs. If all bulbs have the same resistance, they will have
the same voltage: Vbulb = V/N = (110 V)/8 = 13.8 V.
We find the resistance of each bulb from Pbulb = Vbulb
2/Rbulb ;
7.0 W = (13.8 V)2/Rbulb , which gives Rbulb = 27 Ω.
11. Fortunately the required resistance is less. We can reduce the resistance by adding a parallel resistor,
which does not require breaking the circuit. We find the necessary resistance from 1/R = (1/R1) + (1/R2);
1/(320 Ω) = [1/(480 Ω)] + (1/R2), which gives R2 = 960 Ω in parallel.
12. The equivalent resistance of the two resistors connected in series is Rs = R1 + R2 .
We find the equivalent resistance of the two resistors connected in parallel from 1/Rp = (1/R1) + (1/R2), or Rp = R1R2/(R1 + R2).
The power dissipated in a resistor is P = V2/R, so the ratio of the two powers is Pp/Ps = Rs/Rp = (R1 + R2)
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 4
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 5
13. With the two bulbs connected in parallel, there will be 110 V across each bulb, so the total power will be 75 W + 40 W = 115 W. We find the net resistance of the bulbs from
P = V2/R;
115 W = (110 V)2/R, which gives R = 105 Ω. 14. We can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R1 and R2 , which are in series:
The current in the single loop is the current through R6 :
I6 = I = å/Req = (12 V)/(3.58 kΩ) = 3.36 mA.
For VAC we have
VAC = IR10 = (3.36 mA)(1.38 kΩ) = 4.63 V.
This allows us to find I5 and I4 ;
I5 = VAC/R5 = (4.63 V)/(2.20 kΩ) = 2.11 mA;
I4 = VAC/R9 = (4.63 V)/(3.67 kΩ) = 1.26 mA.
For VAB we have
VAB = I4R8 = (1.26 mA)(1.47 kΩ) = 1.85 V.
This allows us to find I3 , I2 , and I1 ;
I3 = VAB/R3 = (1.85 V)/(2.20 kΩ) = 0.84 mA;
I1 = I2 = VAB/R7 = (1.85 V)/(4.40 kΩ) = 0.42 mA.
From above, we have VAB = 1.85 V.
R3
+ –
R1
R2
R4
R5
R6
R3
R7
R4
R5
R6
R8
å
R4
R5
R6
R9
R5
R6
R10
R6
Req
A
B
C
D
A
A A
A A
B
C
C C
D
D D
D
DC
I I
II
II
+ –
+ –
+ – + –
+ –
I4
I5I5
I4
I3
B
I2I2
I3
I5I4
I
å
å å
å
å
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 6
15. (a) When the switch is closed the addition of R2 to the parallel set will decrease the equivalent
resistance, so the current from the battery will increase. This causes an increase in the voltage across R1 , and a corresponding decrease across R3 and R4. The voltage across R2 increases from zero.
Thus we have V1 and V2 increase; V3 and V4 decrease.
(b) The current through R1 has increased. This current is now split into three, so currents through
R3 and R4 decrease. Thus we have
I1 = I and I2 increase; I3 and I4 decrease.
(c) The current through the battery has increased, so the power output of the battery increases. (d) Before the switch is closed, I2 = 0. We find the resistance for R3 and R4 in parallel from
1/RA = á(1/Ri) = 2/R3 = 2/(100 Ω),
which gives RA = 50 Ω.
For the single loop, we have I = I1 = V/(R1 + RA)
= (45.0 V)/(100 Ω + 50 Ω) = 0.300 A. This current will split evenly through R3 and R4 :
I3 = I4 = !I = !(0.300 A) = 0.150 A.
After the switch is closed, we find the resistance for R2 , R3 , and R4 in parallel from
1/RB = á(1/Ri) = 3/R3 = 3/(100 Ω),
which gives RB = 33.3 Ω.
For the single loop, we have I = I1 = V/(R1 + RB)
= (45.0 V)/(100 Ω + 33.3 Ω) = 0.338 A. This current will split evenly through R2 , R3 , and R4 :
I2 = I3 = I4 = @I = @(0.338 A) = 0.113 A.
I4R3
+
–
R1
R2
a
b
V
I
R4
I
I3
I4R3
+
–
R1
R2
a
b
V
I
R4I3I2
S
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 7
16. (a) When the switch is opened, the removal of a resistor from the parallel set will increase the equivalent resistance, so the current from the battery will decrease. This causes a decrease in the voltage across R1 , and a corresponding increase across R2 . The voltage across R3 decreases to zero.
Thus we have V1 and V3 decrease; V2 increases.
(b) The current through R1 has decreased. The current through R2 has increased. The current through
R3 has decreased to zero. Thus we have
I1 = I and I3 decrease; I2 increases.
(c) Because the current through the battery decreases, the Ir term decreases, so the terminal voltage of the battery will increase. (d) When the switch is closed, we find the resistance for R2 and R3 in parallel from
1/RA = á(1/Ri) = 2/R = 2/(5.50 Ω),
which gives RA = 2.75 Ω.
For the single loop, we have I = V/(R1 + RA + r)
= (18.0 V)/(5.50 Ω + 2.75 Ω + 0.50 Ω) = 2.06 A. For the terminal voltage of the battery, we have
Vab = å – Ir = 18.0 V – (2.06 A)(0.50 Ω) = 17.0 V.
When the switch is opened, for the single loop, we have
I′ = V/(R1 + R2 + r)
= (18.0 V)/(5.50 Ω + 5.50 Ω + 0.50 Ω) = 1.57 A. For the terminal voltage of the battery, we have
Vab′ = å – I′r = 18.0 V – (1.57 A)(0.50 Ω) = 17.2 V.
17. We find the resistance for R1 and R2 in parallel from
+ 6.0 V – I3(1.0 Ω) – I3(18 Ω) – 12.0 V + I2(1.0 Ω) – I3(15 Ω) = 0.
When we solve these equations, we get I1 = 1.30 A, I2 = 1.12 A, I3 = 0.18 A.
36. For the conservation of current at point b, we have I = I1 + I2 .
For the two loops indicated on the diagram, we have loop 1: å1 – I1r1 – IR = 0;
+ 2.0 V – I1(0.10 Ω) – I(4.0 Ω) = 0;
loop 2: å2 – I2r2 – IR = 0;
+ 3.0 V – I2(0.10 Ω) – I(4.0 Ω) = 0.
When we solve these equations, we get I1 = 5.31 A, I2 = – 4.69 A, I = 0.62 A.
For the voltage across R we have
Vab = IR = (0.62 A)(4.0 Ω) = 2.5 V.
Note that one battery is charging the other with a significant current. 37. We find the equivalent capacitance for a parallel connection from Cparallel = áCi = 6(3.7 µF) = 22 µF.
When the capacitors are connected in series, we find the equivalent capacitance from 1/Cseries = á(1/Ci ) = 6/(3.7 µF), which gives Cseries = 0.62 µF.
38. Fortunately the required capacitance is greater. We can increase the capacitance by adding a parallel
capacitor, which does not require breaking the circuit. We find the necessary capacitor from C = C1 + C2 ;
16 µF = 5.0 µF + C2 , which gives C2 = 11 µF in parallel.
a
bc
d
e f
g
R3
r1å
1
R4 R5
r2
r3å
3
å2
R2R1
1
2
I1
I2
I3
r1
R
å1
ab
+–
I
I1
I2å
2 r2
1
+–
c
d
2
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 14
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 15
39. We can decrease the capacitance by adding a series capacitor. We find the necessary capacitor from 1/C = (1/C1) + (1/C2);
Yes, it is necessary to break a connection to add a series component. 40. The capacitance increases with a parallel connection, so the maximum capacitance is Cmax = C1 + C2 + C3 = 2000 pF + 7500 pF + 0.0100 µF = 0.0020 µF + 0.0075 µF + 0.0100 µF = 0.0195 µF.
The capacitance decreases with a series connection, so we find the minimum capacitance from 1/Cmin = (1/C1) + (1/C2) + (1/C3) = [1/(2000 pF)] + [1/(7500 pF)] + [1/(0.0100 µF)]
U = !CV2. To increase the energy we must increase the capacitance, which means adding a parallel capacitor.
Because the potential is constant, to have three times the energy requires three times the capacitance: C = 3C1 = C1 + C2 , or C2 = 2C1 = 2(150 pF) = 300 pF in parallel.
42. (a) From the circuit, we see that C2 and C3 are in series
and find their equivalent capacitance from 1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3). From the new circuit, we see that C1 and C4 are in parallel,
with an equivalent capacitance Ceq = C1 + C4 = C1 + [C2C3/(C2 + C3)]
= (C1C2 + C1C3 + C2C3)/(C2 + C3).
(b) Because V is across C1 , we have
Q1 = C1V = (12.5 µF)(45.0 V) = 563 µC.
Because C2 and C3 are in series, the charge on each is the
charge on their equivalent capacitance: Q2 = Q3 = C4V = C2C3/(C2 + C3)V
= [(12.5 µF)(6.25 µF)/(12.5 µF + 6.25 µF)](45.0 V) = 188 µC. 43. From Problem 42 we know that the equivalent capacitance is
We find the potential differences from Q1 = C1V1 ;
2.00 µC = (0.40 µF)V1 , which gives V1 = 5.0 V.
Q2 = C2V2 ;
2.00 µC = (0.50 µF)V2 , which gives V2 = 4.0 V.
(b) As we found above
Q1 = Q2 = 2.0 µC.
(c) For the parallel network, we have V1 = V2 = 9.0 V.
We find the two charges from
Q1 = C1V1 = (0.40 µF)(9.0 V) = 3.6 µC;
Q2 = C2V2 = (0.50 µF)(9.0 V) = 4.5 µC.
45. For the parallel network the potential difference is the same for all capacitors, and the total charge is the
sum of the individual charges. We find the charge on each from Q1 = C1V = (Å0A1/d1)V; Q2 = C2V = (Å0A2/d2)V; Q3 = C3V = (Å0A3/d3)V.
Thus the sum of the charges is Q = Q1 + Q2 + Q3 = [(Å0A1/d1)V] + [(Å0A2/d2)V] + [(Å0A3/d3)V].
The definition of the equivalent capacitance is Ceq = Q/V = [(Å0A1/d1)] + [(Å0A2/d2)] + [(Å0A3/d3)] = C1 + C2 + C3 .
46. The potential difference must be the same on each half of the capacitor, so we can treat the system as two capacitors in parallel:
C = C1 + C2 = [K1Å0(!A)/d] + [K2Å0(!A)/d]
= (Å0!A/d)(K1 + K2) = !(K1 + K2)(Å0A/d)
= !(K1 + K2)C0 .
47. If we think of a layer of equal and opposite charges on the interface between the two dielectrics, we see that they are in series. For the equivalent capacitance, we have
(b) We find the charge on each of the two in series:
Q1 = Q2 = Q4 = C4V = (2.1 µF)(24 V) = 50.4 µC.
We find the voltages from Q1 = C1V1 ;
50.4 µC = (7.0 µF)V1 , which gives V1 = 7.2 V;
Q2 = C2V2 ;
50.4 µC = (3.0 µF)V2 , which gives V2 = 16.8 V.
The applied voltage is across C3: V3 = 24 V.
50. Because the two sides of the circuit are identical, we find the resistance from the time constant:
τ = RC;
3.0 s = R(3.0 µF), which gives R = 1.0 MΩ.
C1
V
C2
C3
C4V
CV
C3
a
b
a
b
c
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 18
51. (a) We know from Example 19–7 that the equivalent
resistance of the two resistors in parallel is 2.7 Ω. There can be no steady current through the capacitor, so we can find the current in the series resistor circuit: I = å/(R6 + R2.7 + R5 + r)
= (9.0 V)/(6.0 Ω + 2.7 Ω + 5.0 Ω + 0.50 Ω) = 0.634 A. We use this current to find the potential difference across the capacitor:
(b) As we found above, the steady state current through
the 6.0-Ω and 5.0-Ω resistors is 0.63 A.
The potential difference across the 2.7-Ω resistor is
Vbc = IR2.7 = (0.634 A)(2.7 Ω) = 1.7 V.
We find the currents through the 8.0-Ω and 4.0-Ω resistors from Vbc = I8R8 ;
1.7 V = I8(8.0 Ω), which gives I8 = 0.21 A;
Vbc = I4R4 ;
1.7 V = I4(4.0 Ω), which gives I4 = 0.42 A.
52. (a) We find the capacitance from
τ = RC;
3.5 × 10–6 s = (15 × 103 Ω)C, which gives C = 2.3 × 10–9 F = 2.3 nF. (b) The voltage across the capacitance will increase to the final steady state value. The voltage across the resistor will start at the battery voltage and decrease exponentially:
VR = å e – t/τ ;
16 V = (24 V)e – t/(35 µs), or t/(35 µs) = ln(24 V/16 V) = 0.405, which gives t = 14 µs.
53. The time constant of the circuit is
τ = RC = (6.7 × 103 Ω)(3.0 × 10–6 F) = 0.0201 s = 20.1 ms. The capacitor voltage will decrease exponentially:
VC = V0 e – t/τ ;
0.01V0 = V0 e – t/(20.1 ms), or t/(20.1 ms) = ln(100) = 4.61,
which gives t = 93 ms.
R8
C
R6
åI R5
a c
I6
bR4
r
I8
R6
åI R5
a c
I
b R2.7
r
C
I4
å C
S
R
+
–
å C
S
R
+
–
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 19
54. (a) In the steady state there is no current through the capacitors. Thus the current through the resistors is
I = Vcd/(R1 + R2) = (24 V)/(8.8 Ω + 4.4 Ω) = 1.82 A. The potential at point a is
Va = Vad = IR2 = (1.82 A)(4.4 Ω) = 8.0 V.
(b) We find the equivalent capacitance of the two series capacitors:
which gives C = 0.16 µF. We find the charge on each of the two in series:
Q1 = Q2 = Q = CVcd = (0.16 µF)(24 V) = 3.84 µC.
The potential at point b is
Vb = Vbd = Q2/C2 = (3.84 µC)/(0.24 µF) = 16 V.
(c) With the switch closed, the current is the same. Point b must have the same potential as point a: Vb = Va = 8.0 V.
(d) We find the charge on each of the two capacitors, which are no longer in series:
Q1 = C1Vcb = (0.48 µF)(24 V – 8.0 V) = 7.68 µC;
Q2 = C2Vbd = (0.24 µF)(8.0 V) = 1.92 µC.
When the switch was open, the net charge at point b was zero, because the charge on the negative plate of C1 had the same magnitude as the charge on the positive plate of C2. With the switch
closed, these charges are not equal. The net charge at point b is
Qb = – Q1 + Q2 = – 7.68 µC + 1.92 µC = – 5.8 µC, which flowed through the switch.
55. We find the resistance on the voltmeter from
R = (sensitivity)(scale) = (30,000 Ω/V)(250 V) = 7.50 × 106 Ω = 7.50 MΩ. 56. We find the current for full-scale deflection of the ammeter from
I = Vmax/R = Vmax/(sensitivity)Vmax = 1/(10,000 Ω/V) = 1.00 × 10–4 A = 100 µA.
57. (a) We make an ammeter by putting a resistor in parallel with the galvanometer. For full-scale deflection, we have Vmeter = IGr = IsRs ;
(50 × 10–6 A)(30 Ω) = (30 A – 50 × 10–6 A)Rs ,
which gives Rs = 50 × 10–6 Ω in parallel.
(b) We make a voltmeter by putting a resistor in series with the galvanometer. For full-scale deflection, we have Vmeter = I(Rx + r) = IG(Rx + r);
1000 V = (50 × 10–6 A)(Rx + 30 Ω),
which gives Rx = 20 × 106 Ω = 20 MΩ in series.
R2
IS
R1C1
C2
a b
c
d
Vc = 24 V
Vd = 0
Rs
rI IG
G
Is
rRx
I
G
IG
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 20
58. (a) The current for full-scale deflection of the galvanometer is
I = 1/(sensitivity) = 1/(35 kΩ/V) = 2.85 × 10–2 mA = 28.5 µA. We make an ammeter by putting a resistor in parallel with the galvanometer. For full-scale deflection, we have Vmeter = IGr = IsRs ;
We find the percent inaccuracies introduced by the meter: (25.6 V – 22.1 V)(100)/(25.6 V) = 14% low; (19.4 V – 16.7 V)(100)/(19.4 V) = 14% low. 61. We find the voltage of the battery from the series circuit with the ammeter in it: å = IA(RA + R1 + R2)
= (5.25 × 10–3 A)(60 Ω + 700 Ω + 400 Ω) = 6.09 V. Without the meter in the circuit, we have å = I0(R1 + R2);
6.09 V = I0(700 Ω + 400 Ω),
which gives I0 = 5.54 × 10–3 A = 5.54 mA.
RV
+
–
R1
R2
å
I1
a
b
c
RV
+
–
R1
R2
å
I2
a
b
c
+
–
R1
R2
å
I0
a
b
c
RA
+
–
R1
R2
å
IA
a
b
c
+
–
R1
R2
å
I0
a
b
c
d
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 22
62. We find the equivalent resistance of the voltmeter in parallel with one of the resistors:
1/R = (1/R1) + (1/RV) = (1/9.0 kΩ) + (1/15 kΩ),
which gives R = 5.63 kΩ. The current in the circuit, which is read by the ammeter, is I = å/(r + RA + R + R2)
= (12.0 V)/(1.0 Ω + 0.50 Ω + 5.63 kΩ + 9.0 kΩ) = 0.82 mA. The reading on the voltmeter is
Vab = IR = (0.82 mA)(5.63 kΩ) = 4.6 V.
63. In circuit 1 the voltmeter is placed in parallel with the resistor, so we find their equivalent resistance from 1/Req1 = (1/R) + (1/RV), or Req1 = RRV/(R + RV).
The ammeter measures the current through this equivalent resistance and the voltmeter measures the voltage across this equivalent resistance, so we have R1 = V/I = Req1 = RRV/(R + RV).
In circuit 2 the ammeter is placed in series with the resistor, so we find their equivalent resistance from Req2 = R + RA .
The ammeter measures the current through this equivalent resistance and the voltmeter measures the voltage across this equivalent resistance, so we have R2 = V/I = Req2 = R + RA .
Thus both circuits give about the same inaccuracy. (c) For circuit 1 we get
R1c = (5.0 kΩ)(10.0 kΩ)/(5.0 kΩ + 10.0 kΩ)
= 3.3 kΩ. For circuit 2 we get
R2c = (5.0 kΩ + 1.00 Ω) = 5.0 kΩ.
Thus circuit 2 is better. Circuit 1 is better when the resistance is small compared to the voltmeter resistance. Circuit 2 is better when the resistance is large compared to the ammeter resistance.
RV+
–
R1
R2
å
I
a
b
c
V
A
RA
r
RV+
–
RI1
a
b
c
V
ARA
+
–
I2
RV
R
d
e
f
V
A RA
å
å
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
The voltmeter measures the voltage across this equivalent resistance, so the current in the circuit is
I = Vab/R = (2.0 V)/(2.13 kΩ) = 0.937 mA.
For the series circuit, we have
å = I(R + Req) = (0.937 mA)(7.4 kΩ + 2.13 kΩ) = 8.9 V.
65. We know from Example 19–14 that the voltage across the resistor without the voltmeter connected is 4.0 V. Thus the minimum voltmeter reading is Vab = (0.97)(4.0 V) = 3.88 V.
We find the maximum current in the circuit from
I = Vbc/R2 = (8.0 V – 3.88 V)/(15 kΩ) = 0.275 mA.
Now we can find the minimum equivalent resistance for the voltmeter and R1 :
Req = Vab/I = (3.88 V)/(0.275 mA) = 14.1 kΩ.
For the equivalent resistance, we have 1/Req = (1/R1) + (1/RV);
We have two equations for two unknowns, with the results:
R1 = 9.4 kΩ, and R2 = 6.8 kΩ.
68. The voltage is the same across resistors in parallel, but is less across a resistor in a series connection. We connect two resistors in series as shown in the diagram. Each resistor has the same current: I = V/(R1 + R2) = (9.0 V)/(R1 + R2).
When the body is connected across ab, we want very negligible current through the body, so the potential
difference does not change. This requires Rbody = 2000 Ω ª R1 . If we also do not want a large current from
the battery, a possible combination is
R1 = 2 Ω, R2 = 70 Ω.
69. Because the voltage is constant and the power is additive, we can use two resistors in parallel. For the lower ratings, we use the resistors separately; for the highest rating, we use them in parallel. The rotary switch shown allows the B contact to successively connect to C and D. The A contact connects to C and D for the parallel connection. We find the resistances for the three settings from P = V2/R;
50 W = (120 V)2/R1 , which gives R1 = 288 Ω;
100 W = (120 V)2/R2 , which gives R2 = 144 Ω;
150 W = (120 V)2/R3 , which gives R3 = 96 Ω.
As expected, for the parallel arrangement we have 1/Req = (1/R1) + (1/R2);
(b) When the capacitors are connected in series, we find the equivalent capacitance from 1/Cseries = (1/C1) + (1/C2) = [1/(0.40 µF)] + [1/(0.60 µF)], which gives Cseries = 0.24 µF.
73. The time between firings is t = (60 s)/(72 beats) = 0.833 s. We find the time for the capacitor to reach 63% of maximum from
V = V0(1 – e – t/τ) = 0.63V0 , which gives e – t/τ = 0.37, or
t = τ = RC;
0.833 s = R(7.5 µC), which gives R = 0.11 MΩ. 74. We find the required current for the hearing aid from P = IV; 2 W = I(4.0 V), which gives I= 0.50 A. With this current the terminal voltage of the three mercury cells would be
Vmercury = 3(åmercury – Irmercury) = 3[1.35 V – (0.50 A)(0.030 Ω)] = 4.01 V.
With this current the terminal voltage of the three dry cells would be
Vdry = 3(ådry – Irdry) = 3[1.5 V – (0.50 A)(0.35 Ω)] = 3.98 V.
Thus the mercury cells would have a higher terminal voltage. 75. (a) We find the current from
I1 = V/Rbody = (110V)/(900 Ω) = 0.12 A.
(b) Because the alternative path is in parallel, the current is the same: 0.12 A. (c) The current restriction means that the voltage will change. Because the voltage will be the same across both resistances, we have I2R2 = IbodyRbody ;
I2(40 ) = Ibody(900 ), or I2 = 22.5Ibody .
For the sum of the currents, we have I2 + Ibody = 23.5Ibody = 1.5 A, which gives Ibody = 6.4 × 10–2 A = 64 mA.
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
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Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
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76. (a) When there is no current through the galvanometer, we have VBD = 0, a current I1 through R1 and R2 , and
a current I3 through R3 and Rx . Thus we have
VAD = VAB ; I1R1 = I3R3 , and
VBC = VDC ; I1R2 = I3Rx .
When we divide these two equations, we get R2/R1 = Rx/R3 , or Rx = (R2/R1)R3 .
= (10.6 × 10–8 Ω á m)L/?(0.460 × 10–3 m)2, which gives L = 26.4 m. 78. (a) When there is no current through the galvanometer,
the current I must pass through the long resistor R′, so the potential difference between A and C is VAC = IR.
Because there is no current through the measured emf, for the bottom loop we have å = IR. When different emfs are balanced, the current I is the same, so we have ås = IRs , and åx = IRx .
When we divide the two equations, we get ås/åx = Rs /Rx , or åx = (Rx/Rs)ås .
(b) Because the resistance is proportional to the length, we have åx = (Rx/Rs)ås = (45.8 cm/25.4 cm)(1.0182 V) = 1.836 V.
(c) If we assume that the current in the slide wire is much greater than the galvanometer current, the uncertainty in the voltage is
®V = ± IGRG = ± (30 Ω)(0.015 mA) = ± 0.45 mV.
Because this can occur for each setting and there will be uncertainties in measuring the distances, the minimum uncertainty is ± 0.90 mV. (d) The advantage of this method is that there is no effect of the internal resistance, because there is no current through the cell. 79. (a) We see from the diagram that all positive plates are connected to the positive side of the battery, and that all negative plates are connected to the negative side of the battery, so the capacitors are connected in parallel. (b) For parallel capacitors, the total capacitance is the sum, so we have Cmin = 7(Å0Amin/d)
= 7(8.85 × 10–12 C2/N á m2)(2.0 × 10–4 m2)/(2.0 × 10–3 m) = 6.2 × 10–12 F = 6.2 pF. Cmax = 7(Å0Amin/d)
= 7(8.85 × 10–12 C2/N á m2)(12.0 × 10–4 m2)/(2.0 × 10–3 m) = 3.7 × 10–11 F = 37 pF. Thus the range is 6.2 pF ? C ? 37 pF.
A
B
D
C
S
G
+ –
R1
R2
R3 Rx
I1
I3
å
+
–
R‘
RVI
BC
S
G
AR
–
+åx
åS
or
å
V
d
+
–
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 28
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 29
80. The terminal voltage of a discharging battery is V = å – Ir. For the two conditions, we have 40.8 V = å – (7.40 A)r; 47.3 V = å – (2.20 A)r.
We have two equations for two unknowns, with the solutions: å = 50.1 V, and r = 1.25 Ω. 81. One arrangement is to connect N resistors in series. Each resistor will have the same power, so we need
N = Ptotal/P = 5 W/! W = 10 resistors.
We find the required value of resistance from Rtotal = NRseries ;
Thus we have 10 22-kΩ resistors in parallel. 82. If we assume the current in R4 is to the right, we have
Vcd = I4R4 = (3.50 mA)(4.0 kΩ) = 14.0 V. We can now find the current in R8 :
I8 = Vcd/R8 = (14.0 V)/(8.0 kΩ) = 1.75 mA.
From conservation of current at the junction c, we have I = I4 + I8 = 3.50 mA + 1.75 mA = 5.25 mA.
If we go clockwise around the outer loop, starting at a, we have Vba – IR5 – I4R4 – å – Ir = 0, or
Vba = (5.25 mA)(5.0 kΩ) – (3.50 mA)(4.0 kΩ) – 12.0 V – (5.25 mA)(1.0 Ω) = 52 V.
If we assume the current in R4 is to the left, all currents are reversed, so we have
Vdc = 14.0 V; I8 = 1.75 mA, and I = 5.25 mA.
If we go counterclockwise around the outer loop, starting at a, we have – Ir + å – I4R4 – IR5 – Ir + Vab = 0, or Vba = – Vab = – Ir + å – I4R4 – IR5 – Ir;
We have two equations for two unknowns, with the solutions:
Rsh = 13 Ω, and Rser = 30 × 103 Ω = 30 kΩ.
r
I
IG
G
Ish
Rx
+
–
Rsh
Rser
V
a b
Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 19
Page 19 – 31
84. The resistance along the potentiometer is proportional to the length, so we find the equivalent resistance between points b and c: 1/Req = (1/xRpot) + (1/Rbulb), or
Req = xRpotRbulb/(xRpot + Rbulb).
We find the current in the loop from I = V/[(1 – x)Rpot + Req].
The potential difference across the bulb is Vbc = IReq , so the power expended in the bulb is
P = (27.4 V)2/(200 Ω) = 3.75 W. 85. (a) Normally there is no DC current in the circuit, so the voltage of the battery is across the capacitor. When there is an interruption, the capacitor voltage will decrease exponentially:
VC = V0 e – t/τ .
We find the time constant from the need to maintain 70% of the voltage for 0.20 s:
0.70V0 = V0 e – (0.20 s)/τ , or (0.20 s)/τ = ln(1.43) = 3.57,
which gives τ = 0.56 s. We find the required resistance from
τ = RC;
0.56 s = R(22 × 10–6 F), which gives R = 2.5 × 104 Ω = 25 kΩ. (b) In normal operation, there will be no voltage across the resistor, so the device should be connected between b and c. 86. (a) Because the capacitor is disconnected from the power supply, the charge is constant. We find the new voltage from Q = C1V1 = C2V2 ;
(10 pF)(10,000 V) = (1 pF)V2 , which gives V2 = 1.0 × 105 V = 0.10 MV.
(b) A major disadvantage is that, when the stored energy is used, the voltage will decrease exponentially, so it can be used for only short bursts.