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2. Left Circuit
a)
R
eq
= 2
R
I
batt
=
V
batt
/ 2
R
b)
Junction Rules:
I
1
=
I
2
+
I
3
A
Loop Rules:
V
batt

I
2
(4
R
) = 0
a

I
3
R

I
3
(2
R
) 
I
3
R
+
I
2
(4
R
) = 0
b
Solving Equation
b
gives
I
2
=
I
3
b
’
Plugging
b
’ into
A
gives
I
1
= 2
I
2
A
’
Solving
a
gives
I
2
=
V
batt
/ 4
R
a
’
Plugging
a
’ into
A
’, finally, gives
I
1
=
V
batt
/ 2
R
, (which is what we found in part (a)), so
I
1
=
V
batt
/ 2
R
I
2
=
V
batt
/ 4
R
I
3
=
V
batt
/ 4
R
Right Circuit
a)
R
eq
= 15
R
/ 8
I
batt
= 8
V
batt
/ 15
R
b)
Junction Rules:
I
1
=
I
2
+
I
3
A
Loop Rules:
V
batt

I
2
(4
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Unformatted text preview: R )  I 2 R = 0 aI 3 R I 3 (2 R ) + I 2 R + I 2 (4 R ) = 0 b Solving Equation b gives I 3 = 5/3 I 2 b ’ Plugging b ’ into A gives I 1 = 8/3 I 2 A ’ Solving a gives I 2 = V batt / 5 R a ’ Plugging a ’ into A ’, finally, gives I 1 = 8 V batt / 15 R , (which is what we found in part (a)), so I 1 = 8 V batt / 15 R I 2 = V batt / 5 R I 3 = V batt / 3 R...
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This note was uploaded on 03/10/2012 for the course PHYSICS 8B taught by Professor Shapiro during the Spring '07 term at University of California, Berkeley.
 Spring '07
 SHAPIRO

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