Physics 21, Spring 2008
February 8, 2008
HW9 Solutions
91.
(HRW 2718)
Figure
2735
shows a resistor of resistance
connected to
an ideal battery of emf
by means of two copper wires.
Each wire has length 20.0 cm and radius 1.00 mm. In dealing
with such circuits in this chapter, we generally neglect the
potential differences along the wires and the transfer of energy
to thermal energy in them. Check the validity of this neglect for
the circuit of Fig.
2735
: What is the potential difference
across (a) the resistor and (b) each of the two sections of wire?
At what rate is energy lost to thermal energy in (c) the resistor
and (d) each section of wire?
Solution:
18. (a) For each wire,
R
wire
=
ρ
L/A
where
A
=
π
r
2
.
Consequently, we have
R
wire
=
(1.69
×
10
−
8
m
Ω⋅
)(0.200 m)/
π
(0.00100 m)
2
= 0.0011
Ω
.
The total resistive load on the battery is therefore
tot
R
= 2
R
wire
+
R
=
2(
0.0011
Ω) +
6.00
Ω = 6.0022
Ω
.
Dividing this into the battery emf gives the current
tot
12.0 V
1.9993 A
6.0022
i
R
ε
=
=
=
Ω
.
The voltage across the
R
= 6.00
Ω
resistor is therefore
V
iR
=
=
(1.9993 A)(6.00
Ω
) = 11.996 V
≈
12.0 V.
(b) Similarly, we find the voltagedrop across each wire to be
wire
wire
V
iR
=
=
(1.9993 A)(0.0011
Ω
) = 2.15 mV.
(c)
P
=
i
2
R
= (1.9993 A)(6.00
Ω
)
2
= 23.98 W
≈
24.0 W.
(d) Similarly, we find the power dissipated in each wire to be 4.30 mW.
92.
(HRW 2722) 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?
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 Spring '08
 Kim
 Resistance, Energy, Resistor, Electrical resistance, Voltmeter reading

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