RECITATION 7
9.
(a) and (b)
Assume
1
i
travels to the right,
2
i
up and
3
i
down.
We use the loop rule twice,
clockwise around the outer loop and clockwise around the right loop:
(
29
1
1
3
3
10 V
0 V;
i R
i R


=
(
29
2
2
3
3
5.0 V
0 V.
i R
i R


=
We use the junction rule
(
29
3
1
2
i
i
i
=
+
to eliminate
3
i
in the loop rule equations.
Substituting the
given data into the loop rule equations yields
(
29
(
29
1
2
10 V
8.0
4.0
;
i
i
=
Ω +
Ω
(
29
(
29
2
1
5.0 V
8.0
4.0
.
i
i
=
Ω +
Ω
Solving for
1
2
and
,
i
i
we find
1
1.25 A.
i
=
and
2
0 A
i
=
.
Therefore
3
1
2
1.25 A.
i
i
i
=
+
=
As assumed
above,
3
i
is traveling down in the picture.
11.
(a) The resistors
R
2
,
R
3
and
R
4
are in parallel with each other, and that combination is in
series with
R
1
.
The equivalent resistance for this network is
1
1
eq
1
2
3
4
1
1
1
1
1
1
100
118.75
120
.
50
50
75
R
R
R
R
R


=
+
+
+
=
Ω +
+
+
=
Ω ≈
Ω
Ω
Ω
Ω
(b)
Based on our result from part (a), the current through the battery must be
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 Spring '08
 Goksu
 Physics, Resistor, Electrical resistance, Series and parallel circuits

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