HW-8Solutions-01-26-08

HW-8Solutions-01-26-08 - Physics 21, Spring 2008 January...

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Unformatted text preview: Physics 21, Spring 2008 January 26, 2008 HW-8 Solutions 8-1 . (HRW 27-6) In Fig. 27-25 , the ideal batteries have emfs and and the resistances are and . If the potential at P is defined to be 100 V, what is the potential at Q? Solution: 6. The current in the circuit is i = (150 V 50 V)/(3.0 + 2.0 ) = 20 A. So from V Q + 150 V (2.0 ) i = V P , we get V Q = 100 V + (2.0 )(20 A) 150 V = 10 V. 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 Kirchhoffs loop rule: 1 iR 2 iR 1 2 = 0. We solve for i : i R R = + = + = 1 2 1 2 12 6 0 8 0 050 V V 4.0 A . . . . 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 P i R = . (b) For R 1 , P 1 = 2 1 i R = (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = 2 2 i R = (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 rate P = i...
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This note was uploaded on 02/20/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-8Solutions-01-26-08 - Physics 21, Spring 2008 January...

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