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hw 21

# hw 21 - A Again for the interval 80 ms ≤ t< 160 ms i L...

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ECE 201 Spring 2010 Homework 21 Solutions Problem 31 (a) Using voltage division, v C (0 - ) = 20 R 50 R V 0 = 0 . 4 V 0 = v C (0+) (b) The Thevenin equivalent is given by V oc = V 0 R 3 R 3 + ( R 1 || R 2 ) = 0 . 8 V 0 R th = R 1 || R 2 || R 3 = 4 R (c) v C ( ) = 0 . 8 V 0 v C (0+) = 0 . 4 V 0 v C ( t ) = v C ( ) + [ v C ( t 0 +) - v C ( )] e - t - t 0 R th C = 0 . 8 V 0 - 0 . 4 V 0 e - t 4 RC 1

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(d) v C ( T - ) = v C ( T +) = 0 . 8 V 0 - 0 . 4 V 0 e - 1 . 5 = 0 . 71075 V 0 (e) τ = R th C = 4 RC (f) Using the same relation as in part (c), with t 0 = T , v C ( t ) = 0 . 71075 V 0 e - t - T 4 RC (g) For simplicity, let all quantities be normalized to 1. Then the plot for v C ( t ) would look as shown on the next page. Problem 33 (a) i L (0+) = i L (0 - ) = - 10 80 + 20 = - 0 . 1 A 2
0 5 10 15 20 25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 t v C (t) Plot of v C (t) for 0<=t<=4T Figure 1: Plot of v C ( 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 , i L ( t ) = i L ( ) + [ i L (0+) - i L ( )] e - tR/L = 20 100 + [ - 0 . 1 - 0 . 2] e - 25 t = 0 . 2 - 0 . 3 e - 25 t i L (80 - ) = i L (80+) = 0

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Unformatted text preview: A Again, for the interval 80 ms ≤ t < 160 ms , i L ( t ) = i L ( ∞ ) + [ i L (80+)-i L ( ∞ )] e-( t-80) R/L =-. 1 + [0 . 1594 + 0 . 1] e-25( t-80) =-. 1 + 0 . 2594 e-25( t-80) We can write v out ( t ) as the following, v out ( t ) = v in ( t )-R 1 i L ( t ) 3 = 4 + 24 e-25 t , ≤ t < 80 ms =-2-20 . 752 e-25( t-80 ms ) , 80 ms ≤ t < 160 ms 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16-30-20-10 10 20 30 t(seconds) v out (t)(volts) Plot of v out (t) for 0<=t<=160ms Figure 2: Plot of v out ( t ) for problem 33 4...
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