PHYS102
Homework 5 Solutions
Spring
08
____________________________________________________________
Chapter 21 Current and Direct Current Circuits 25, 29, 31
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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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
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 Spring '08
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 Physics, Current, Work, Resistor, Potential difference, direct current circuits, Solutions Spring08 ____________________________________________________________

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