PHYS102
HOMEWORK05
SOLUTIONS
______________________________________________
Chapter21
Electric Current
P21.25
(a)
P
=
Δ
V
(
)
2
R
becomes
20.0 W
=
11.6 V
(
)
2
R
so
R
=
6.73
Ω
.
(b)
Δ
V
=
IR
so
11.6 V
=
I
6.73
Ω
(
)
and
I
=
1.72 A
e
=
IR
+
Ir
so
15.0 V
=
11.6 V
+
1.72 A
(
)
r
r
=
1.97
Ω
.
FIG. P21.25
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P21.29
If we turn the given diagram on its side, we find that it is the same as
figure (a). The
20.0
Ω
and
5.00
Ω
resistors are in series, so the
first reduction is shown in (b). In addition, since the
10.0
Ω
,
5.00
Ω
, and
25.0
Ω
resistors are then in parallel, we can solve
for their equivalent resistance as:
R
eq
=
1
1
10.0
Ω
+
1
5.00
Ω
+
1
25.0
Ω
(
)
=
2.94
Ω
.
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
Next, we work backwards through the diagrams applying
I
=
Δ
V
R
and
Δ
V
=
IR
alternately to every resistor, real and equivalent. The
12.94
Ω
resistor is connected across 25.0 V, so the current
through the battery in every diagram is
I
=
Δ
V
R
=
25.0 V
12.94
Ω
=
1.93 A
.
In figure (c), this 1.93 A goes through the
2.94
Ω
equivalent resistor to
give a potential difference of:
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 Spring '08
 N/A
 Physics, Current, Work, Resistor, Potential difference, PHYS102 HOMEWORK05 SOLUTIONS

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