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EE 202/Quiz 7 Morning

Name: _______________________________________________

Feb 1, 2010

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

Solution:

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 7 Afternoon

Name: _______________________________________________

February 1, 2010

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

Solution.

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

quiz7  

EE 202/Quiz 7 Morning Name: _______________________________________________ Feb 1, 2010 1. The Thevenin equivalent impedance for the circuit...

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