problem26_60

# problem26_60 - = = = = = = R I I I I I I So the other...

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26.60: a) Using the currents as defined on the circuit diagram below we obtain three equations to solve for the currents: . 0 4 3 0 ) ( 2 ) ( : loop Bottom . 0 3 2 0 ) ( 2 : loop Top . 14 2 3 0 ) ( 2 14 : loop Left 2 1 2 2 1 2 1 2 1 1 2 1 2 1 2 1 1 = - + - = - - + + - - = + + - = + + - - = - = - - - I I I I I I I I I I I I I I I I I I I I I Solving these equations for the currents we find: A. 0 . 2 A; 0 . 6 ; A 0 . 10 3 1 2 R 1 battery
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Unformatted text preview: = = = = = = R I I I I I I So the other currents are: A. . 6 A; . 4 A; . 4 2 1 2 1 1 5 4 2 = +-= =-= =-= I I I I I I I I I I R R R . 40 . 1 b) A . 10 V . 14 eq Ω = = = I V R...
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## This homework help was uploaded on 04/19/2008 for the course PHYS 4 taught by Professor Stuart during the Spring '08 term at UCSB.

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