Physics 21
Fall, 2004
Solution to HW8
26
P25
Determine the magnitudes and directions of the
currents through
R
1
and
R
2
in Fig. 2639.
9.0 V
6.0 V
15
Ω
22
Ω
I
1
I
3
I
2
1
2
We can choose loop 1 so it involves only one unknown
current. The equations are
current:
I
1
=
I
2
+
I
3
loop 1:
6 + 9 = 22
I
1
loop 2:
9 = 15
I
2
+22
I
1
Solve loop 1 to get
I
1
=15
/
22 = 0
.
682 A. Substitute
I
1
into
loop 2 and ±nd
I
2
=
−
6
/
15 =
−
0
.
4 A. Rounding o², we get
I
1
=0
.
68 A
leftward through 22 Ω
I
2
=0
.
40 A
leftward through 15 Ω
26
P27
Determine the magnitudes and directions of the
currents through each resistor shown in ±g. 2640. The bat
teries have emfs of
E
1
=9
.
0V and
E
2
=12
.
0 V, and the
resistors have values of
R
1
=15Ω
,
R
2
=20Ω,
R
3
=40Ω
.
15
Ω
20
Ω
40
Ω
9V
12 V
1
2
I
3
I
1
I
2
The equations are
current:
I
1
+
I
3
=
I
2
loop 1:
9 = 15
I
1
+20
I
2
loop 2:
12 = 40
I
3
+20
I
2
We can solve these by eliminating
I
3
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 Fall '07
 Hickman
 Physics, Current, Resistor, loop, Electrical resistance, Series and parallel circuits

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