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Chapter 28
135
P28.8
120
1234
V
eq
==
+
+
+
F
H
G
I
K
J
IR
I
AAAA
ρρρρ
, or
I
ρ
A
=
+++
120
1111
V
af
ej
∆
V
I
A
A
2
2
2
120
29 5
=
A
V
V
.
P28.9
If we turn the given diagram on its side, we find that it is the same as figure
(a). The 20 0
.
Ω
and 500
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
294
1
10 0
1
500
1
25 0
..
.
.
ΩΩΩ
Ω
ch
.
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
V
R
=
∆
and
∆
VI
R
=
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
V
R
=
∆
Ω
25 0
193
.
.
V
12.94
A.
In figure (c), this 1.93 A goes through the 2 94
Ω
equivalent resistor to give a
potential difference of:
∆Ω
R
=
568
.
A
V
.
From figure (b), we see that this potential difference is the same across
∆
V
ab
,
the 10
Ω
resistor, and the 500
Ω
resistor.
(b)
Therefore,
∆
V
ab
=
V
.
(a)
Since the current through the 20 0
Ω
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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