sp2004_exam1soln - SOLUTIONS EXAM#1 ECE 210 SPRING 2004...

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S OLUTIONS — E XAM #1 ECE 210, S PRING 2004 1. (a) (i) e π /2 (ii) 4e π /3 (iii) 1 2 e 3 π /2 1. (b) t = 0 50 V + 40 k 60 k v c (0 ) = 30 V v c (0 ) = 50 V · 60 100 = 30 V t = v c ( ) = –100 V 20 k 20 k + 5 mA τ = R eq C = 40 k · 25 · 10 –7 F = 10 msec
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2. 66 k –i 1 i 0 25 k + 68 k 12 k + 34 k 0.4 mA V 0 + V a 4 V + 0.1 mA = –i 1 The 12 k and 68 k resistors constitute a voltage divider. (a) V 1 V + = 68 80 · 4 = 3.4 V (b) –i 1 = 3.4 V 34 k = 0.1 mA V a (voltage drop across 66 k resistor) has the polarity shown and V a = 0.1 mA × 66 k = 6.6 V (c) i 25 = 0.4 mA & i 0 = –0.5 mA (d) V 0 = 3.4 + 6.6 = 10.0 V 3. (a) t < 0 5 S.C. + 5 V i(t = 0 ) = 1 A
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(b) i(t = 0 + ) = 1 A KVL: V s – v(t) – i(t) · 1 – 2 di(t) dt = 0 di(t) dt + i(t) 2 = V s - v(t) 2 (c) If v(t) = 0, then i(t) = 5 - 4e - t 2 A where i(t = ) = 5 A, τ = L R = 2 sec, and i(t) = i( ) + i(0 + ) - i( ) { }
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This note was uploaded on 06/30/2010 for the course ECE 210 taught by Professor Whoever during the Spring '07 term at University of Illinois at Urbana–Champaign.

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sp2004_exam1soln - SOLUTIONS EXAM#1 ECE 210 SPRING 2004...

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