problem26_61

# problem26_61 - 5 4 4 10 12 1 2 2 1-= ⇒ =-I I I I and A...

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26.61: a) Going around the complete loop, we have: V. 22 . 0 ) 1 1 2 ( ) A 44 . 0 ( V 0 . 10 V 0 . 12 . A 44 . 0 0 ) 0 . 9 ( V 0 . 8 V 0 . 12 + = + + - - = - = = = - - = - IR ε V I I IR ε ab b) If now the points a and b are connected by a wire, the circuit becomes equivalent to the diagram shown below. The two loop equations for currents are (leaving out the units):
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Unformatted text preview: 5 . 4 4 10 12 1 2 2 1-= ⇒ = +--I I I I and A. 464 . 5 . 2 5 5 ) 2 4 ( 2 ) ( 5 4 2 5 4 8 10 1 1 1 1 2 1 2 3 2 = ⇒ = +----⇒ = +--=---I I I I I I I I I Thus the current through the 12-V battery is 0.464 A....
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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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