EE 202/Quiz 6 Section 1

Name: _______________________________________________

Feb 4, 2009

1. (4 pts) The inductor in the circuit below has an initial current of iL (0 − ) = −2 A. Then vC (t) = (in V): (1) 2 sin(4t)u(t) (2) 2 sin(2t)u(t) (3) 2 sin(0.5t)u(t) (6) 4 sin(0.25t)u(t) (4) 4 sin(0.5t)u(t) (5) −4 sin(0.5t)u(t) (8) None of above (7) −4 sin(0.25t)u(t)

−iL (0 − ) 1 2 0.5 VC (s) = Zin (s) = × =4 2 . Thus vC (t) = 4 sin(0.5t)u(t) 2 1 s s s + (0.5) s+ 4s Answer is (4).

2. (3 pts) The Thevenin equivalent impedance for the circuit below is Zth(s) =:

Vin (s) = (2s + 4)I in (s) + 2(I in (s) − 0.25VL (s)) = (2s + 6)I in (s) − sI in (s) = (s + 6)I in (s) Therefore Zin (s) = (s + 6) .

EE 202/Quiz 6 Section 2

Name: _______________________________________________

February 4, 2009

1. (4 pts) The inductor in the circuit below has an initial current of iL (0 − ) = 4 A and vC (0 − ) = 0 . Then vC (t) = (in V):

−iL (0 − ) 1 −4 0.5 VC (s) = Zin (s) = × = −8 2 . Thus vC (t) = −8 sin(0.5t)u(t) 2 1 s s s + (0.5) s+ 4s

2. (3 pts) The Thevenin equivalent impedance for the circuit below is Zth(s) =: (1) 1.5s + 6 (4) s + 4 (7) 2s + 6

(2) s + 2 (3) 3s + 6 (5) 1.5s + 2 (6) s + 6 (8) None of the above

Vin (s) = (2s + 4)I in (s) + 2(I in (s) − 0.25VL (s)) = (2s + 6)I in (s) − sI in (s) = (s + 6)I in (s) Therefore Zin (s) = (s + 6) .

qz6soln

1. (4 pts) The inductor in the circuit below has an initial current of i L (0 − )=−2 A. Then v C (t)= (in V): (1) 2sin(4t)u(t) (2) 2sin(2t)u...

qz6soln

Published on Jan 26, 2011

1. (4 pts) The inductor in the circuit below has an initial current of i L (0 − )=−2 A. Then v C (t)= (in V): (1) 2sin(4t)u(t) (2) 2sin(2t)u...

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