R
5
V
2
P
5
1
120 V
2
2
100 W
5
144
V
P
5
V
2
/
R
(
a
)
E
5
48.0 V
R
1
5
1.00
V
R
3
5
7.00
V
R
2
5
3.00
V
R
4
5
5.00
V
(
b
)
E
5
48.0 V
R
12
5
4.00
V
R
34
5
12.0
V
(
c
)
E
5
48.0 V
R
eq
5
3.00
V
I
3
5
I
4
5
4.00 A.
I
1
5
I
2
5
12.0 A
I
12
1
I
34
5
16.0 A.
I
34
5
48.0 V
12.0
V
5
4.00 A.
I
12
5
48.0 V
4.00
V
5
12.0 A
I
5
48.0 V
3.00
V
5
16.0 A.
R
eq
5
3.00
V
.
R
eq
5
R
1
1
R
2
.
R
eq
5
R
1
R
2
R
1
1
R
2
.
1
R
eq
5
1
R
1
1
1
R
2
9.00 A.
I
4
5
V
34
R
4
5
36.0 V
4.00
V
5
I
3
5
V
34
R
3
5
36.0 V
12.0
V
5
3.00 A
R
4
,
R
3
V
34
4.00 A.
I
2
5
V
12
R
2
5
24.0 V
6.00
V
5
I
1
5
V
12
R
1
5
24.0 V
3.00
V
5
8.00 A
R
2
,
R
1
V
12
60.0 V.
V
12
1
V
34
5
V
34
5
IR
34
5
1
12.0 A
21
3.00
V
2
5
36.0 V.
V
12
5
IR
12
5
1
12.0 A
21
2.00
V
2
5
24.0 V.
(
a
)
(
b
)
(
c
)
R
1
R
2
R
3
R
4
R
12
R
34
R
eq
E
E
E
I
5
60.0 V
5.00
V
5
12.0 A.
R
eq
5
5.00
V
.
Current, Resistance, and DirectCurrent Circuits
1911
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power.
Use the rated power to find the resistance of each bulb. In parallel, the full line voltage is
applied to each bulb.
Solve:
The resistance of each bulb is:
and
(a)
In series,
The current in each bulb is the current in the equivalent circuit,
The power consumed by each bulb is
and
(b)
In parallel,
for each bulb, so
and
(c)
P
is proportional to
and now
for each bulb, so they would each consume 4 times their rated power:
and
The power rating for each bulb would be exceeded and they would
quickly burn out.
Reflect:
In series, the bulb with the greatest resistance dissipates the greatest power. When the bulbs are connected
individually across the 120 V outlet, the bulb with the least resistance dissipates the greatest power.
19.56.
Set Up:
For resistors in series the equivalent resistance is
The current is the same in
each.
so
(b)
The 20.0
resistor would dissipate more than 2.00 W and would fail.
19.57.
Set Up:
Assume the unknown currents have the directions shown in Figure 19.57. We have used the
junction rule to write the current through the 10.0 V battery as
There are two unknowns,
and
so we will
need two equations. Three possible circuit loops are shown in the figure.
Figure 19.57
Solve: (a)
Apply the loop rule to loop (1), going around the loop in the direction shown:
and
(b)
Apply the loop rule to loop (3):
and
(c)
Reflect:
For loop (2) we get
so that with the currents we have calculated the loop rule is satisfied for this third loop.
5.00 V
1
5.00 V
2
10.0 V
5
0,
1
5.00 V
1
I
2
1
20.0
V
2
2
I
1
1
30.0
V
2
5
5.00 V
1
1
0.250 A
21
20.0
V
2
2
1
0.333 A
21
30.0
V
2
5
I
1
1
I
2
5
0.333 A
1
0.250 A
5
0.583 A
I
2
5
0.250 A.
1
10.0 V
2
1
20.0
V
2
I
2
2
5.00 V
5
0
I
1
5
0.333 A.
0
1
10.0 V
2
1
30.0
V
2
I
1
5
10.0 V
20.0
V
30.0
V
5.00 V
1
1
2
1
2
2
1
3
2
+
+
I
2
I
1
I
2
,
I
1
I
1
1
I
2
.
V
V
5
IR
eq
5
1
0.316 A
21
35.0
V
2
5
11.1 V.
R
eq
5
5.00
V 1
10.0
V 1
20.0
V 5
35.0
V
.
I
5
Å
P
R
20
5
Å
2.00 W
20.0
V
5
0.316 A.
P
5
I
2
R
R
eq
5
R
1
1
R
2
1
R
3
.
P
100
5
400 W.
P
75
5
300 W
P
60
5
240 W,
V
5
240 V
V
2
P
100
5
100 W.
P
75
5
75 W
P
60
5
60 W,
V
5
120 V
P
100
5
1
0.208 A
2
2
1
144
V
2
5
6.23 W.
P
75
5
1
0.208 A
2
2
1
192
V
2
5
8.31 W
P
60
5
I
2
R
60
5
1
0.208 A
2
2
1
240
V
2
5
10.4 W,
I
5
V
R
eq
5
120 V
576
V
5
0.208 A.
R
eq
5
R
60
1
R
75
1
R
100
5
576
V
.
R
100
5
V
2
P
100
5
1
120 V
2
2
100 W
5
144
V
.
R
75
5
V
2
P
75
5
1
120 V
2
2
75 W
5
192
V
R
60
5
V
2
P
60
5
1
120 V
2
2
60 W
5
240
V
,
P
5
V
2
R
5
I
2
R
.
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 Spring '09
 RODRIGUEZ
 Physics, Charge, Current, Resistance, Resistor, Electrical resistance, Series and parallel circuits, DirectCurrent Circuits

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