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PHYS 241
Ch. 26 Homework Solutions
26.7.I
DENTIFY
:
First do as much seriesparallel reduction as possible.
S
ET
U
P
:
The 45.0Ω and 15.0Ω resistors are in parallel, so first reduce them to a single equivalent
resistance. Then find the equivalent series resistance of the circuit.
E
XECUTE
:
1/
R
p
= 1/(45.0 Ω) + 1/(15.0 Ω) and
R
p
= 11.25 Ω. The total equivalent resistance is
18.0 Ω + 11.25 Ω + 3.26 Ω = 32.5 Ω. Ohm’s law gives
I
= (25.0 V)/(32.5 Ω) = 0.769 A.
E
VALUATE
:
The circuit appears complicated until we realize that the 45.0Ω and 15.0Ω resistors are in
parallel.
26.11.
I
DENTIFY
:
For resistors in parallel, the voltages are the same and the currents add.
eq
1
2
1
1
1
R
R
R
=
+
so
1
2
eq
1
2
,
R R
R
R
R
=
+
For resistors in series, the currents are the same and the voltages add.
eq
1
2
R
R
R
=
+
.
S
ET
U
P
:
The rules for combining resistors in series and parallel lead to the sequences of equivalent
circuits shown in Figure 26.11.
E
XECUTE
:
eq
5.00
R
=
Ω
. In Figure 26.11c,
60.0 V
12.0 A
5.00
I
=
=
Ω
. This is the current through each of
the resistors in Figure 26.11b.
12
12
(12.0 A)(2.00
)
24.0 V
V
IR
=
=
Ω =
.
34
34
(12.0 A)(3.00
)
36.0 V
V
IR
=
=
Ω =
. Note that
12
34
60.0 V
V
V
+
=
.
12
V
is the voltage across
1
R
and
across
2
R
, so
12
1
1
24.0 V
8.00 A
3.00
V
I
R
=
=
=
Ω
and
12
2
2
24.0 V
4.00 A
6.00
V
I
R
=
=
=
Ω
.
34
V
is the voltage across
3
R
and across
4
R
, so
34
3
3
36.0 V
3.00 A
12.0
V
I
R
=
=
=
Ω
and
34
4
4
36.0 V
9.00 A
4.00
V
I
R
=
=
=
Ω
.
E
VALUATE
:
Note that
1
2
3
4
I
I
I
I
+ = +
.
Figure 26.11
26.13.
I
DENTIFY
:
In both circuits, with and without
R
4
, replace series and parallel combinations of resistors
by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the
currents and voltages in the original circuit. Use
2
P
I R
=
to calculate the power dissipated in each
bulb.
(a) S
ET
U
P
:
The circuit is sketched in Figure 26.13a.
E
XECUTE
:
2
3
4
,
, and
R
R
R
are in
parallel, so their equivalent resistance
eq
R
is given by
eq
2
3
4
1
1
1
1
R
R
R
R
=
+
+
Figure 26.13a
1
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eq
1
3
and
1.50
.
4.50
R
R
=
=
Ω
Ω
The equivalent circuit is drawn in Figure 26.13b.
(
29
1
eq
0
I R
R

+
=
E
1
eq
I
R
R
=
+
E
Figure 26.13b
1
9.00 V
1.50 A and
1.50 A
4.50
1.50
I
I
=
=
=
Ω +
Ω
Then
(
29
(
29
1
1
1
1.50 A 4.50
6.75 V
V
I R
=
=
Ω =
(
29
(
29
eq
eq
eq
eq
1.50 A,
1.50 A 1.50
2.25 V
I
V
I R
=
=
=
Ω =
For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent
resistor, so
2
3
4
2.25 V.
V
V
V
=
=
=
2
3
4
2
3
4
2
3
4
2.25 V
0.500 A,
0.500 A,
0.500 A
4.50
V
V
V
I
I
I
R
R
R
=
=
=
=
=
=
=
Ω
E
VALUATE
:
Note that
2
3
4
1.50 A,
I
I
I
+
+
=
which is
eq
.
I
For resistors in parallel the currents add and
their sum is the current through the equivalent resistor.
(b) S
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 Fall '08
 Milsom
 Magnetism, Resistance, Work

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