ch27-p031

# ch27-p031 - ) = 26.6 V. Therefore, by the loop rule, = V 1...

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By the junction rule, the current in R 2 is i 2 = i 4 + i 6 =1.05 A + 1.40 A = 2.45 A, so its voltage is V 2 = (2.00 )(2.45 A) = 4.90 V. The loop rule tells us the voltage across R 3 is V 3 = V 2 + V 4 = 21.7 V (implying that the current through it is i 3 = V 3 /(2.00 ) = 10.85 A). The junction rule now gives the current in R 1 as i 1 = i 2 + i 3 = 2.45 A + 10.85 A = 13.3 A, implying that the voltage across it is V 1 = (13.3 A)(2.00
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Unformatted text preview: ) = 26.6 V. Therefore, by the loop rule, = V 1 + V 3 = 26.6 V + 21.7 V = 48.3 V. 31. First, we note V 4 , that the voltage across R 4 is equal to the sum of the voltages across R 5 and R 6 : V 4 = i 6 ( R 5 + R 6 )= (1.40 A)(8.00 + 4.00 ) = 16.8 V. The current through R 4 is then equal to i 4 = V 4 / R 4 = 16.8 V/(16.0 ) = 1.05 A....
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