quiz7 - 1. The inductor in the circuit below has an initial...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 202/Quiz 7 Morning Name: _______________________________________________ Feb 1, 2010 1. The Thevenin equivalent impedance for the circuit below is Z th (s) =: Solution: V in ( s ) = (2 s + 4) I in ( s ) + 2( I in ( s ) 0.25 V L ( s )) = (2 s + 6) I in ( s ) sI in ( s ) = ( s + 6) I in ( s ) Therefore Z in ( s ) = ( s + 6).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
EE 202/Quiz 7 Afternoon Name: _______________________________________________ February 1, 2010
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. The inductor in the circuit below has an initial current of i L (0 ) = 4 A and v C (0 ) = 0 . Then v C ( t ) = (in V): Solution. V C ( s ) = Z in ( s ) i L (0 ) s = 1 s + 1 4 s 4 s = 8 0.5 s 2 + (0.5) 2 . Thus v C ( t ) = 8sin(0.5 t ) u ( t )...
View Full Document

This document was uploaded on 02/03/2012.

Page1 / 2

quiz7 - 1. The inductor in the circuit below has an initial...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online