8B-Soln07-v1.0.2

# 8B-Soln07-v1.0.2 - R I 2 R = 0 a-I 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 =

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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 a-I 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.

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