ch27-p043 - 43. (a) We first find the currents. Let i1 be...

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(c) The power dissipated in R 3 is () ( ) 2 2 33 3 0.263A 5.00 0.346W. Pi R == = (d) The power supplied by ε 1 is i 3 1 = (0.421 A)(3.00 V) = 1.26 W. (e) The power “supplied” by 2 is i 2 2 = (–0.158 A)(1.00 V) = –0.158 W. The negative sign indicates that 2 is actually absorbing energy from the circuit. 43. (a) We first find the currents. Let i 1 be the current in R 1 and take it to be positive if it is to the right. Let i 2 be the current in R 2 and take it to be positive if it is to the left. Let i 3 be the current in R 3 and take it to be positive if it is upward. The junction rule produces iii 123 0 + + = . The loop rule applied to the left-hand loop produces 11 13 3 0 iR iR += and applied to the right-hand loop produces 22 23 3 0. We substitute i 3 = – i 2 i 1 , from the first equation, into the other two to obtain 12 31 3 0 iR iR iR −−= and 3 Solving the above equations yield 3 2 3 1 (3.00 V)(2.00 5.00 ) (1.00 V)(5.00 ) 0.421 A. (4.00 )(2.00 ) (4.00 )(5.00 ) (2.00 )(5.00 ) RR R i +−
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