problem26_23

# problem26_23 - = Ω-Ω ⇒ I I I I Along with 3 2 1 I I I...

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26.23: After reversing the polarity of the 10-V battery in the circuit of Fig. 26.42, the only change in the equations from Problem 26.22 is the upper loop where the 10 V battery is: Upper loop: ( 29 ( 29 0 V 00 . 5 00 . 4 00 . 1 00 . 3 00 . 2 V 0 . 10 2 1 = - + - + - - I I ( 29 ( 29 . A 00 . 3 0 00 . 5 00 . 5 V 0 . 15 2 1 2 1 - = + = - - - I I I I Lower loop: ( 29 ( 29 0 0 . 10 00 . 4 00 . 1 V 00 . 5 3 2 = - + + I I ( 29 ( 29 . A 00 . 1 2 0 0 . 10 00 . 5 V 00 . 5 3 2 3 2 - = -
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Unformatted text preview: = Ω-Ω + ⇒ I I I I Along with , 3 2 1 I I I + = we can solve for the three currents and find: . A 200 . , A 40 . 1 , A 60 . 1 3 2 1-=-=-= I I I b) ( 29 ( 29 ( 29 ( 29 . V 4 . 10 00 . 3 A 60 . 1 00 . 4 A 40 . 1 = Ω + Ω + = ab V...
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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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