Download Solutions

Chapter 25 p. 1
Homework #6
P24-13 The uniform electric field between the plates is related to the potential difference across the plates:
E = V/d.
For a parallel-plate capacitor, we have
Qmax = CVmax = (e0A/d)(Emaxd) = e0AEmax
= (8.8510–12 C2/N · m2)(8.510–4 m2)(3.0106 V/m) = 2.310–8 C =
23 nC
C2
C1
d
P24-29 (a) From the circuit, we see that C1 and C2 are in series
and find their equivalent capacitance from
C3
1/C5 = (1/C1) + (1/C2) = (1/C) + (1/C), which gives C5 = C/2.
c
From the new circuit, we see that C3 and C5 are in parallel,
with an equivalent capacitance
C4
C6 = C3 + C5 = C + C/2 = 3C/2.
a
b
From the new circuit, we see that C4 and C6 are in series
C5
and find their equivalent capacitance from
1/Ceq = (1/C4) + (1/C6)
= (1/C) + [1/(3C/2)], which gives Ceq =
3C/5.
c
(b) The charge on the equivalent capacitor is also the
C3
charge on C4 and C6 :
Qeq = Q4 = Q6 = CeqVab = (3C/5)V = 3CV/5.
We find the potential difference between c and b from
Vcb = Q6/C6 = (3CV/5)/(3C/2) = 2V/5.
The charge on C5 is also the charge on C1 and C2 :
Q5 = Q1 = Q2 = C5Vcb = (C/2)(2V/5) = CV/5.
We find the potential differences from
Vcd = Q1/C1 = (CV/5)/(C) = V/5;
Vdb = Q2/C2 = (CV/5)/(C) = V/5;
Vac = Q4/C4 = (3CV/5)/(C) = 3V/5.
The charge on C3 is
Q3 = C3Vcb = (C)(2V/5) = 2CV/5.
Thus we have
Q1 = Q2 = CV/5, Q3 = 2CV/5, Q4 = 3CV/5;
V1 = V2 = V/5, V3 = 2V/5, V4 = 3V/5.
a
C4
C6
c
a
b
C4
P25-1. The rate at which electrons pass any point in the wire is the current:
I = 1.50 A = (1.50 C/s)/(1.6010–19 C/electron) =
9.381018 electron/s.
P25-9. (a) We find the resistance from
V = IR;
120 V = (7.5 A)R, which gives R =
16 .
(b) The charge that passes through the hair dryer is
Q = I t = (7.5 A)(15 min)(60 s/min) =
6.8103 C.
P25-10 We find the potential difference across the bird’s feet from
V = IR = (2500 A)(2.510–5 /m)(4.010–2 m) =
2.510–3 V.
P25-13. From the expression for the resistance, R = L/A, we form the ratio
RAl/RCu = (Al/Cu)(LAl/LCu)(ACu/AAl) = (Al/Cu)(LAl/LCu)(dCu/dAl)2
= [(2.6510–8  · m)/(1.6810–8  · m)][(10.0 m)/(20.0 m)][(2.5 mm)/(2.0 mm)] 2
= 1.2, or
RAl = 1.2RCu.
b
Chapter 25 p. 2
P26-8. Because resistance increases when resistors are connected in series, the maximum resistance is
Rseries = R1 + R2 + R3 = 500  + 900  + 1400  = 2800  =
2.80 k.
Because resistance decreases when resistors are connected in parallel, we find the minimum
resistance from
1/Rparallel = (1/R1) + (1/R2) + (1/R3) = [1/(500 )] + [1/(900 )] + [1/(1400 )],
which gives Rparallel =
261 .
P26-12. (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)/(0.60 A) =
23 .
The power dissipated in each bulb is
Pbulb = IVbulb = (0.60 A)(13.8 V) =
8.3 W.
a
R
b
R
c
I
V
P26-13. For the parallel combination, the total current from the source is
I = NIbulb = 8(0.340 A) = 2.72 A.
The voltage across the leads is
Vleads = IRleads = (2.72 A)(1.5 ) = 4.1 V.
The voltage across each of the bulbs is
Vbulb = V – Vleads = 110 V – 4.1 V = 106 V.
We find the resistance of a bulb from
Rbulb = Vbulb/Ibulb = (106 V)/(0.340 A) =
310 .
The power dissipated in the leads is IVleads and the total power used
is IV, so the fraction wasted is
IVleads/IV = Vleads/V = (4.1 V)/(110 V) = 0.037 =
3.7%.
R bul b
R leads
I
V
P26-66. We find the current through the patient (and nurse) from the series circuit:
I = V/(Rmotor + Rbed + Rnurse + Rpatient)
= (220 V)/(104  + 0 + 104  + 104 ) = 7.310–3 A =
7.3 mA.
R
d