7. (a) Let ibe the current in the circuit and take it to be positive if it is to the left in R1. We use Kirchhoff’s loop rule: ε1– iR2– iR1– 2= 0. We solve for i: iRR=−+=−+=12126 080050VV4.0A....ΩΩA positive value is obtained, so the current is counterclockwise around the circuit. If iis the current in a resistor R, then the power dissipated by that resistor is given by 2PiR=. (b) For R1, P1= 21iR=(0.50 A)2(4.0 Ω) = 1.0 W, (c) and for R2, P2= 22=(0.50 A)2(8.0 Ω) = 2.0 W. If iis the current in a battery with emf , then the battery supplies energy at the rate P =iprovided the current and emf are in the same direction. The battery absorbs energy at the
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