142
Chapter 18
PROBLEM SOLUTIONS
18.1
From
Δ
V
I R
r
=
+
(
)
, the internal resistance is
r
V
I
R
=
−
=
−
=
Δ
Ω
Ω
9 00
72 0
4 92
.
.
.
V
0.117 A
18.2
(a)
When the three resistors are in series, the equivalent resistance of the circuit is
R
R
R
R
eq
=
+
+
=
(
)
=
1
2
3
3 9 0
27
.
Ω
Ω
(b)
The terminal potential difference of the battery is applied across the series combination of
the three
9 0
.
Ω
resistors, so the current supplied by the battery and the current through
each resistor in the series combination is
I
V
R
=
=
=
Δ
Ω
eq
V
27
A
12
0 44
.
(c)
If the three
9 0
.
Ω
are now connected in parallel with each other, the equivalent
resistance is
1
1
9 0
1
9 0
1
9 0
3
9 0
R
eq
=
+
+
=
.
.
.
.
Ω
Ω
Ω
Ω
or
R
eq
=
=
9 0
3
3 0
.
.
Ω
Ω
When this parallel combination is connected to the battery, the potential difference across
each resistor in the combination is
Δ
V
=
12 V
, so the current through each of the
resistors is
I
V
R
=
=
=
Δ
Ω
12
1 3
V
9.0
A
.
18.3
For the bulb in use as intended,
R
V
bulb
V
W
=
(
)
=
(
)
=
Δ
Ω
2
2
120
75 0
192
P
.
Now, assuming the bulb resistance is unchanged,
the current in the circuit shown is
I
V
R
=
=
+
+
=
Δ
Ω
Ω
Ω
eq
V
0.800
A
120
192
0 800
0 620
.
.
and the actual power dissipated in the bulb is
P
=
=
(
)
(
)
=
I R
2
2
0 620
192
73 8
bulb
A
W
.
.
Ω
18.4
(a)
When the
8 00
.

Ω
resistor is connected across the 9.00V terminal potential difference of
the battery, the current through both the resistor and the battery is
I
V
R
=
=
=
Δ
Ω
9 00
1 13
.
.
V
8.00
A
(b)
The relation between the emf and the terminal potential difference of a battery supplying
current
I
is
Δ
V
Ir
=
−
ε
, where
r
is the internal resistance of the battery. Thus, if the battery
has
r
=
0 15
.
Ω
and maintains a terminal potential difference of
Δ
V
=
9 00
.
V
while sup
plying the current found above, the emf of this battery must be
ε
=
+
=
+
(
)
(
)
=
+
(
)
Δ
Ω
V
Ir
9 00
1 13
0 15
9 00
0 17
.
.
.
.
.
V
A
Ω
Ω
=
9 17
.
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DirectCurrent Circuits
143
18.5
(a)
The equivalent resistance of the two parallel
resistors is
R
p
=
+
⎛
⎝
⎜
⎞
⎠
⎟
=
−
1
7 00
1
10 0
4 12
1
.
.
.
Ω
Ω
Ω
Thus,
R
R
R
R
ab
p
=
+
+
=
+
+
(
)
=
4
9
4 00
4 12
9 00
17 1
.
.
.
.
Ω
Ω
(b)
I
V
R
ab
ab
ab
=
(
)
=
=
Δ
Ω
34 0
17 1
1 99
.
.
.
V
A
, so
I
I
4
9
1 99
=
=
.
A
Also,
Δ
Ω
V
I
R
p
ab
p
(
)
=
=
(
)
(
)
=
1 99
4 12
8 18
.
.
.
A
V
Then,
I
V
R
p
7
7
8 18
1 17
=
(
)
=
=
Δ
Ω
.
.
V
7.00
A
and
I
V
R
p
10
10
8 18
0 818
=
(
)
=
=
Δ
Ω
.
.
V
10.0
A
18.6
(a)
The parallel combination of the
6 0
.
and 12
Ω
Ω
resistors has an equivalent resistance of
1
1
6 0
1
12
2
1
12
1
R
p
=
+
=
+
.
Ω
Ω
Ω
or
R
p
1
12
3
4 0
=
=
Ω
Ω
.
Similarly, the equivalent resistance of the
4 0
.
and 8.0
Ω
Ω
parallel combination is
1
1
4 0
1
8 0
2
1
8 0
2
R
p
=
+
=
+
.
.
.
Ω
Ω
Ω
or
R
p
2
8
3
=
Ω
The total resistance of the series combination between points
a
and
b
is then
R
R
R
ab
p
p
=
+
+
=
+
+
=
1
2
5 0
4 0
5 0
8 0
3
35
3
.
.
.
.
Ω
Ω
Ω
Ω
Ω
(b)
If
Δ
V
ab
=
35 V
, the total current from
a
to
b
is
I
V
R
ab
ab
ab
=
=
(
)
=
Δ
Ω
35
3 0
V
35
3
A
.
and the potential differences across the two parallel combinations are
Δ
Ω
V
I
R
p
ab
p
1
1
3 0
4 0
12
=
=
(
)
(
)
=
.
.
A
V
and
Δ
Ω
V
I
R
p
ab
p
2
2
3 0
8 0
3
8 0
=
=
(
)
⎛
⎝
⎜
⎞
⎠
⎟
=
.
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 Resistor, Electrical resistance, Ω, Series and parallel circuits, equivalent resistance, Δv

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