HW2S - EE202L Homework #2 Solutions 7.13 10 K + 12 V - t=0...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EE202L Homework #2 Solutions 7.13 Find Vc(t) for t 0. Identify the forced & natural response. Vc(0 - ) = V 3 12 ) 10 20 10 ( 10 = ! + + Vc(0 + ) = V(0 - ) = 3 V Vc( ) = V 6 12 10 10 10 = ! + R th =10k || 10k = 5k τ = R th · C = 5k (1 × 10 -6 ) = 5 × 10 -3 Vc(t) = Vc( ) + [Vc(0) – Vc( )]· ! t e " = 6 - 3· 200t e ! 7.18 Find i L (t) for t > 0 and sketch the waveform i L (0 - ) = 5/5k = 1mA i L (0 + ) = i L (0 - ) = 1mA + 12 V - A B t = 0 20 K Ω 10 K Ω 10 K Ω 1 μ F + Vc(t) - + 12 V - 20 K Ω 10 K Ω 10 K Ω + Vc(t) - t < 0 + 12 V - 10 K Ω 10 K Ω + Vc(t) - t = Forced response Natural response + 5 V - t = 0 100 mH 1 K Ω i L (t) 5 K Ω 5 V i L (t) 5 K Ω t < 0 + -
Background image of page 1

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

View Full DocumentRight Arrow Icon
i L ( ) = 5/1k = 5mA R th = 1 k Ω τ = L/R th = 100 mH/1 k Ω = 10 -4 = 0.0001 S i L (t) = i L ( ) + [i L (0)-i L ( )] e -t/ τ = 5 + (1-5) e -10000t = 5 – 4e -10000t mA 7.21 Find Vc(t) for t 2ms t < 0. Vc(0 - ) = 0 V = Vc(0 + ) If we assume that the switch is closed at t = 0, and leave it there for a long time. t
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/24/2008 for the course EE 202L taught by Professor Katsouleas during the Spring '08 term at USC.

Page1 / 4

HW2S - EE202L Homework #2 Solutions 7.13 10 K + 12 V - t=0...

This preview shows document pages 1 - 3. Sign up to view the full document.

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