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Chapter 27: Circuits
Problem 1
A wire of resistance 5.0
Ω
is connected to a battery whose emf
E
is 2.0 V and whose
internal resistance is 1.0
Ω
. In 2.0 min, how much energy is (a) transferred from chemical to
electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated
as thermal energy in the battery?
Answer
(a) 80 J; (b) 66.7 J; (c) 13.3 J
Solution
(a)
The energy transferred from chemical to electrical in the battery is
U
=
Pt
=
ε
it
=
2
t
R
tot
Where the total resistance for the wire and internal resistance in series is
R
tot
= 5.0
Ω
+ 1.0
Ω
= 6.0
Ω
Therefore
U
=
2.0
V
( )
2
∗
2min
6
Ω
=
80
J
(b)
The thermal energy in the 5.0
Ω
wire is
U
=
Pt
=
i
2
R
wire
t
=
R
tot
⎛
⎝
⎜
⎞
⎠
⎟
2
R
wire
t
=
2.0
V
6.0
Ω
⎛
⎝
⎜
⎞
⎠
⎟
2
∗
5.0
Ω ∗
2.0min
=
66.7
J
(c)
The “missing” 13.3 J = 80 J – 66.7 J is dissipated as thermal energy because of
the internal resistance.
Problem 3
A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf.
By how much is the chemical energy of the battery reduced?
Answer
10.8 kJ
The energy transferred is the product of the power and the time and the power is the product
of the current and the potential difference.
U = i
E
t = (5.0 A)*(6.0 V)*(6.0 min) = 10.8 kJ
272
Chapter 27: Circuits
Problem 4
A standard flashlight battery can deliver about 2.0 Wh of energy before it runs down. (a) If
a battery costs 80¢, what is the cost of operating a 100W lamp for 8.0 h using batteries?
(b) What is the cost if power provided by an electric utility company, at 12 cents per kWh,
is used?
Answer
(a) $320. (b) 4.8¢.
Solution
(a)
The total amount of energy required to operate a 200W lamp for 8.0 h is
U = P*t = (100 W)*(8 h) = 800 Wh.
To obtain 800 Wh from flashlight batteries would require 400 batteries (n = 800 Wh/2.0
Wh). If each battery cost 80 cents then the total cost would be
cost = 400($0.80) = $320. This is not a cheap way to light your home!
(b)
The cost from the power company at the rate of 6 cents per kwh is
cost = ($ 0.06/kwh)*(800 Wh) = 4.8¢.
But it is difficult to keep the power company in the glove compartment of your car.
Problem 12
(a) In Fig. 2729 what value must R have if
the current in the circuit is to be 1.0 mA?
Take
E
1
= 2.0 V,
E
2
= 3.0V, and r
1
= r
2
= 3.0
Ω
. (b) What is the rate at which thermal
energy appears in R?
Answer
(a) 994
Ω
. (b) 9.94
×
10
4
W.
E
1
E
2
r
1
r
2
R
i
Figure 2729
(a)
We assume that the current is going counterclockwise in the circuit and apply the
loop rule. Traversing the loop in a clockwise sense beginning at the upper left corner
 ir
2
+
E
2

E
1
 ir
1
 iR = 0
or
R
=
ε
2
−
1
i
−
r
2
−
r
1
=
3.0
V
−
2.0
V
1.0
mA
−
3.0
Ω −
3.0
Ω
=
994
Ω
.
(b)
The power dissipated in R is
P = i
2
R = (1.0 mA)
2
(994
Ω
) = 9.94
×
10
4
W
Chapter 27: Circuits
273
Problem 22
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 (a) the internal resistance and (b) the emf of the solar cell? (c) The area of the cell is 5.0
cm
2
and the rate per unit area at which it receives energy from light is 2.0 mW/cm
2
. What is
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