Solutions to Homework Problems CH4 and CH4.5: Circuit Theory, Exams of Electrical and Electronics Engineering

Download Solutions to Homework Problems CH4 and CH4.5: Circuit Theory and more Exams Electrical and Electronics Engineering in PDF only on Docsity! HomeWork #4 Due on Oct 11 (Tue) at 9:30am. (do not turn in) means that the grader does not check, thus no points are assigned. However these questions are also very useful for preparation to Midterm II. CH4.1 P4.4* The solution is of the form of Equation 4.17:      322 10exp100exp  tKRCtKVtv sC in which 2K is a constant to be determined. At 0t , we have   2100500 KvC  Solving, we find that 1502 K and the solution is    310exp150100  ttvC CH4.2 P4.15* In steady state, the equivalent circuit is: Thus, we have A 2 0 23 1   ii i CH4.3 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 CH4.3 (do not turn in) P4.30* (no points) In steady state with the switch closed, we have   0 for 0  tti because the closed switch shorts the source. In steady state with the switch open, the inductance acts as a short circuit and the current becomes   A 1i . The current is of the form     0 for exp21  tLRtKKti in which  20R , because that is the Thévenin resistance seen looking back from the terminals of the inductance with the switch open. Also, we have       1 21 1 000 Ki KKii   Thus, 12 K and the current (in amperes) is given by     0 for 20exp1 0 for 0   tt tti CH4.4 P4.37*Applying KVL, we obtain the differential equation:       0 for exp5  tttRi dt tdi L (1) We try a particular solution of the form:    tAtip  exp (2) in which A is a constant to be determined. Substituting Equation (2) into Equation (1), we have      ttRAtLA  exp5expexp which yields