Physics 115A Autumn 2000 Chapter 18, Homework Problem Solutions
2. The rate at which electrons pass any point in the wire is the current:
I = 1.00 A = (1.00 C/s)/(1.60 × 10–19 C/electron) = 6.25 × 1018 electron/s.
8. (a)We find the resistance from
V = IR;
120 V = (9.0 A)R, which gives R = 13 Ω.
(b)The charge that passes through the hair dryer is
∆Q = I ∆t = (9.0 A)(15 min)(60 s/min) = 8.1 × 103 C.
9. We find the resistance from
V = IR;
12 V = (0.50 A)R, which gives R = 24 Ω.
The energy taken out of the battery is
E = Pt = IVt = (0.50 A)(12 V)(1 min)(60 s/min) = 3.6 × 102 J.
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 Ω · m)/(1.68 × 10–8 Ω · m)][(10.0 m)/(15.0 m)][(2.5 mm)/(2.0 mm)]2
= 1.6, or RAl = 1.6RCu.
21. (a)For the resistance of each wire we have
RCu = CuLCu/A = (1.68 × 10–8 Ω · m)(5.0 m)/π(0.50 × 10–3 m)2 = 0.107 Ω.
RAl = AlLAl/A = (2.65 × 10–8 Ω · m)(5.0 m)/π(0.50 × 10–3 m)2 = 0.169 Ω.
Thus the total resistance is
R = RCu + RAl = 0.107 Ω + 0.169 Ω = 0.28 Ω.
(b) We find the current from
V = IR;
80 V = I(0.28 Ω), which gives I = 2.9 × 102 A.
(c) The current must be the same for the two wires, so we have
VCu = IRCu = (2.9 × 102 A)(0.107 Ω) = 31 V.
VAl = IRAl = (2.9 × 102 A)(0.169 Ω) = 49 V.
23. For an ohmic resistor, we have
P = IV = V2/R, or R = V2/P = (240 V)2/(3.3 × 103 W) = 17 Ω.
28. We find the operating resistance from
P = V 2/R;
60 W = (240 V) 2/R, which gives R = 9.6 × 102 Ω.
If we assume that the resistance stays the same, for the lower voltage we have
P = V2/R = (120 V)2/(9.6 × 102 Ω) = 15 W.
At one-quarter the power, the bulb will be much dimmer.
30. The cost for a year would be
Cost = E(rate) = Pt(rate)
