(c) The power dissipated in R3is ()()223330.263A5.000.346W.PiR==−Ω=(d) The power supplied by ε1is i31= (0.421 A)(3.00 V) = 1.26 W. (e) The power“supplied”by 2is i22= (–0.158 A)(1.00 V) = –0.158 W. The negative sign indicates that 2is actually absorbing energy from the circuit. 43. (a) We first find the currents. Let i1be the current in R1and take it to be positive if it is to the right. Let i2be the current in R2and take it to be positive if it is to the left. Let i3be the current in R3and take it to be positive if it is upward. The junction rule produces iii1230++=. The loop rule applied to the left-hand loop produces 111330iR iR−+=and applied to the right-hand loop produces 222330.−We substitute i3= –i2– i1, from the first equation, into the other two to obtain 123130iR iR iR−−−=and 3−Solving the above equations yield 3231(3.00 V)(2.005.00) (1.00 V)(5.00)0.421 A.(4.00)(2.00) (4.00)(5.00 ) (2.00)(5.00)RRRi+−
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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.