ch27-p007

# ch27-p007 - 7(a Let i be the current in the circuit and...

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7. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 . We use Kirchhoff’s loop rule: ε 1 iR 2 iR 1 2 = 0. We solve for i : i RR = + = + = 12 12 6 0 80 050 VV 4.0 A . . .. ΩΩ A positive value is obtained, so the current is counterclockwise around the circuit. If i is the current in a resistor R , then the power dissipated by that resistor is given by 2 PiR = . (b) For R 1 , P 1 = 2 1 iR = (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = 2 2 = (0.50 A) 2 (8.0 ) = 2.0 W. If i is the current in a battery with emf , then the battery supplies energy at the rate P =i provided the current and emf are in the same direction. The battery absorbs energy at the
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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