Solutions to Electrical Engineering Problems: Midterm II, Exams of Electrical and Electronics Engineering

Download Solutions to Electrical Engineering Problems: Midterm II and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Midterm-II Solution Q1 (20 points) P4.15* In steady state, the equivalent circuit is: Thus, we have A 2 0 23 1   ii i Q2 (20 points) P4.24 The general form of the solution is    LtRKKtiL  exp21 At 0t , we have     2100(0 KKii LL  At t , the inductance behaves as a short circuit, and we have   11.0 KiL  Thus, the solution for the current is     0 for 10-0.1exp-0.1 0 for 0 6   tt ttiL The voltage is       0 for 10exp100 0 for 0 6    tt t dt tdi Ltv Q3. (20 points) E5.9 (a) The transformed network is: mA 13528.28 250250 9010       jZ sVI mA )135500cos(28.28)(  tti 13507.7  IV RR 4507.7  IV LjL  (b) The phasor diagram is shown in Figure 5.16b in the book. (c) i(t) lags vs(t) by .45 E5.11 The transformed network is: We write KVL equations for each of the meshes: 100)(100100 211  IIIj 0)(100100200 1222  IIII jj Simplifying, we have 100100)100100( 21  IIj 0)100100(100 21  II j Solving we find A. 01 and A 45414.1 21   II Thus we have ).1000cos()( and A )451000cos(414.1)( 21 ttitti   Q4. (20 points) P5.53 This is a capacitive load because the reactance is negative. kW 5.22100)15( 22  RIP rms kVAR 25.11)50()15( 22  XIQ rms 57.26)5.0(tantan 11         P Q θ power factor %44.89)cos(  θ