240_quiz6a

# 240_quiz6a - and the current i(0-) = i(0+) = 12/3 = 4A For...

This preview shows page 1. Sign up to view the full content.

Name ________________________ ECE 240 – Quiz #6 March 4, 2010 For the circuit shown below, assume that the switch is closed and the circuit is in steady state prior to t=0. The switch opens at t=0. Find: a. i(t), t > 0 b. v(t), t > 0 at t = 0-, the inductor is replaced by a short circuit. Thus the 2 + 6 ohm resistance is bypassed
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: and the current i(0-) = i(0+) = 12/3 = 4A For t > 0, the switch is open and only the inductor and 2 and 6 ohm resistors remain. Thus, L = 4 and R = 8 L/R = 1/2 Thus: i ( t ) = 4 e − t /(1/2) = 4 e − 2 t And v ( t ) = L di dt = 4( − 2)(4) e − 2 t = − 32 e − 2 t +-! ! " ! #\$ % & ’ ( ) ! *"+ ,-....
View Full Document

## This note was uploaded on 10/19/2011 for the course ECE 240 taught by Professor Drsharlenekatz during the Fall '10 term at CSU Northridge.

Ask a homework question - tutors are online