ECE 201 Spring 2010 Homework 21 Solutions Problem 31 a Using voltage division 20R V0 50R 0 4V0 vC 0 vC 0 b The Thevenin equivalent is given by R3 R3 R1 R2 0 8V0 Voc V0 Rth R1 R2 R3 4R c vC 0 8V0 vC 0 0 4V0 vC t vC vC t0 vC e t t0 th C R t 0 8V0 0 4V0 e 4RC 1 d vC T vC T 0 8V0 0 4V0 e 1 5 0 71075V0 e Rth C 4RC f Using the same relation as in part c with t0 T t T vC t 0 71075V0e 4RC g For simplicity let all quantities be normalized to 1 Then the plot for vC t would look as shown on the next page Problem 33 a 10 80 20 0 1 A iL 0 iL 0 2 Plot of v t for 0 t 4T C 0 8 0 7 0 6 C v t 0 5 0 4 0 3 0 2 0 1 0 0 5 10 15 20 25 t Figure 1 Plot of vC t for problem 31 b Here we have to consider two intervals 0 t 80 ms and 80 ms t 160 ms For 0 t 80 ms iL t iL iL 0 iL e tR L 20 0 1 0 2 e 25t 100 0 2 0 3e 25t iL 80 iL 80 0 2 0 3e 2 0 1594 A Again for the interval 80 ms t 160 ms iL t iL iL 80 iL e t 80 R L 0 1 0 1594 0 1 e 25 t 80 0 1 0 2594e 25 t 80 We can write vout t as the following vout t vin t R1 iL t 3 4 24e 25t 0 t 80 ms 2 20 752e 25 t 80ms 80 ms t 160 ms Plot of vout t for 0 t 160ms 30 20 vout t volts 10 0 10 20 30 0 0 02 0 04 0 06 0 08 t seconds 0 1 Figure 2 Plot of vout t for problem 33 4 0 12 0 14 0 16
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