ECE 201 Spring 2010Homework 37 SolutionsProblem 5(a)For Figure (a),I2eff=1TZT0i2(t) dt=19Z309 dt +Z6316 dt=253⇒ Ieff= 5/√3For Figure (b),I2eff=14Z109 dt +Z3216 dt= 25/4⇒ Ieff= 2.5(b)Current through RLis given by i(t) × 60/(30 + 60) = (2/3)i(t). Powerabsorbed by RLis then given by the square of the effective current throughit times the resistance. ThusP =49×253× 30=10009W1(c)P =49×254× 30=2503WProblem 7In all parts of the problem, the following identities will be used,Z2π/m0cos(mnx) dx = 0Z2π/m0sin(mnx) dx = 0Here m and n are integers.(a)V21eff=1TZT0v21(t) dt=1TZT0[102 + 40 cos(20t) + 2 cos(40t)] dt= 102⇒ V1eff= 10.0995(b)V22eff=1TZT0v22(t) dt=1TZT0[50{1 + cos(4t)} + 12.5{1 + cos ( 8 t)} + 50{cos(6t) + cos(2t)}] dt= 62.5⇒ V2eff= 7.9057(c)V23eff=1TZT0v23(t) dt=1TZT0[50{1 + cos(4t)} + 36.4277{1 + cos(8t)} + 6.2499{1 − cos ( 8 t)} + . . .] dt= 92.6776⇒ V3eff= 9.62692Problem 9(a)ZL=15+ j(30 × 5 × 10−3)−1= 0.8(4 − j3)⇒ VL= Iin× ZL=4√2(4 − j3)⇒ vL(t) = 20 cos(30t − 36.87◦) V(b)Pinst(t) = iin(t)vL(t)= 100 cos(30t) cos(30t − 36.87◦) WPaverage=VmIm2cos(θv− θi)= 50 cos 36.87◦= 40 WProblem 10(a)Is=506− 90◦6 + j12 − j4= 56− 143.13◦(magnitude = 5)(b)Paverage= VeffIeffcos(θv− θi)= 5 × 50 cos(−53.13◦)= 150 W3(c)Paverage= R|Is|2= 6 × 25= 150 W (same as part (b))(d)Is=506− 90◦30 + j50 − j10= 16− 143.13◦(magnitude = 1)Paverage= VeffIeffcos(θv− θi)= 1 × 50 cos(−53.13◦)= 30
View Full Document