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Chapter 27
(# 1, 3, 5, 7, 14)
1. A standard flashlight battery can deliver about 2.0 W · h of energy before it runs
down. (a) If a battery costs US$0.80, what is the cost of operating a 100 W lamp for
8.0 h using batteries? (b) What is the cost if energy is provided at the rate of
US$0.06 per kilowatt-hour?
1. (a) The cost is (100 W · 8.0 h/2.0 W · h) ($0.80) = $3.2 02.
(b) The cost is (100 W · 8.0 h/103 W · h) ($0.06) = $0.048 = 4.8 cents.
3. A certain car battery with a 12.0 V emf has an initial charge of 120 A · h.
Assuming that the potential across the terminals stays constant until the battery is
completely discharged, for how many hours can it deliver energy at the rate of 100
W?
3. If P is the rate at which the battery delivers energy and t is the time, then E = P t is
the energy delivered in time t. If q is the charge that passes through the battery in time
t and  is the emf of the battery, then E = q. Equating the two expressions for E and
solving for t, we obtain
t 
q (120A  h)(12.0V)

 14.4h.
P
100W
5. In Fig. 27-25, the ideal batteries have emfs ξ1 = 12 V and ξ2 = 6.0 V and the
resistors have resistances R1 = 4.0 Ω and R2 = 8.0 Ω. What are (a) the current, the
dissipation rate in (b) resistor 1 and (c) resistor 2, and the energy transfer rate in (d)
battery 1 and (e) battery 2? Is energy being supplied or absorbed by (f) battery 1
and (g) battery 2?
5. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R1.
We use Kirchhoff’s loop rule: 1 – iR2 – iR1 – 2 = 0. We solve for i:
i
1   2
R1  R2

12 V  6.0 V
 0.50 A.
4.0   8.0 
A positive value is obtained, so the current is counterclockwise around the circuit.
If i is the current in a resistor R, then the power dissipated by that resistor is given by
P  i2 R .
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CHAPTER 27
(b) For R1, P1 = (0.50 A)2(4.0 ) = 1.0 W,
(c) and for R2, P2 = (0.50 A)2 (8.0 ) = 2.0 W.
If i is the current in a battery with emf , then the battery supplies energy at the rate P =i
provided the current and emf are in the same direction. The battery absorbs energy at the
rate P = i if the current and emf are in opposite directions.
(d) For 1, P1 = (0.50 A)(12 V) = 6.0 W
(e) and for 2, P2 = (0.50 A)(6.0 V) = 3.0 W.
(f) In battery 1 the current is in the same direction as the emf. Therefore, this battery
supplies energy to the circuit; the battery is discharging.
(g) The current in battery 2 is opposite the direction of the emf, so this battery absorbs
energy from the circuit. It is charging.
7. A car battery with a 12 V emf and an internal resistance of 0.040 Ω is being
charged with a current of 50 A. What are (a) the potential difference V across the
terminals, (b) the rate Pr of energy dissipation inside the battery, and (c) the rate
Pemf of energy conversion to chemical form? When the battery is used to supply 50 A
to the starter motor, what are (d) V and (e) Pr?
7. (a) The potential difference is V =  + ir = 12 V + (0.040 )(50 A) = 14 V.
(b) P = i2r = (50 A)2(0.040 ) = 1.0×102 W.
(c) P' = iV = (50 A)(12 V) = 6.0×102 W.
(d) In this case V =  – ir = 12 V – (0.040 )(50 A) = 10 V.
(e) Pr = i2r = 1.002 W.
14. A solar cell generates a potential difference of 0.10 V when a 500 Ω resistor is
connected across it, and a potential difference of 0.15 V when a 1000 Ω resistor is
substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c)
The area of the cell is 5.0 cm2, and the rate per unit area at which it receives energy
from light is 2.0 mW/cm2. What is the efficiency of the cell for converting light
energy to thermal energy in the 1000 Ω external resistor?
14. (a) Let the emf of the solar cell be  and the output voltage be V. Thus,
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V    ir   
VI
F
G
HR J
Kr
for both cases. Numerically, we get
0.10 V =  – (0.10 V/500 )r
0.15 V =  – (0.15 V/1000 )r.
We solve for  and r.
(a) r = 1.003 .
(b)  = 0.30 V.
(c) The efficiency is
V2 /R
0.15V

 2.3103  0.23%.
2
3
2
Preceived 1000   5.0cm   2.0 10 W/cm 