EE411_HW6_Soln - v 4 ) ( = + At t=0 + , switch is switched...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Ans: Assume the voltage across the 1/25 F capacitor is v c (t). It is obviously that before switch is closed, v c (0 - )=0. Since the voltage across the capacitor cannot change instantaneously, we have v c (0 + )=v c (0 - )=0. In order to calculate dt dv c ) 0 ( + , we apply KCL at the right plate of the 1/25F capacitor 0 ) 0 ( 0 ) 0 ( ) 0 ( 4 ) 0 ( ) 0 ( 25 1 = = = = Ω + + + + + dt dv i i v dt dv c L L c c Apply KVL, we have j s t v dt t dv dt t v d t v dt t dv dt t v d t v dt t dv dt d t v t v dt t dv dt t di L t v t i c c c c c c c c c c c L c L 608 . 4 125 . 15 5000 ) ( 250 ) ( 4 121 ) ( 50 ) ( 5 . 2 ) ( 400 121 ) ( 100 1 50 4 ) ( ) ( 25 1 4 1 ) ( 6 4 ) ( ) ( 25 1 50 ) ( ) ( ) ( 2 , 1 2 2 2 2 ± = = + + = + + = Ω + + + × Ω + = + + So we can write )) 608 . 4 sin( ) 608 . 4 cos( ( 20 ) ( 125 . 15 t B t A e t v t c + + = 6467 . 65 0 608 . 4 125 . 15 ) 0 ( 20 0 20 ) 0 ( = = + = = = + = + + B B A dt dv A A v c c
Background image of page 1

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

View Full DocumentRight Arrow Icon
So () 0 ) 608 . 4 sin( 4117 . 16 ) 608 . 4 cos( 4 5 4 ) ( ) ( )) 608 . 4 sin( 6467 . 65 ) 608 . 4 cos( 20 ( 20 ) ( 125 . 15 125 . 15 > + + = Ω = + = t t t e t v t i t t e t v t c t c
Background image of page 2
Ans: (a)At t=0 - , we have V v 4 ) 0 ( = Since the voltage across a capacitor cannot change instantaneously, we have V
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: v 4 ) ( = + At t=0 + , switch is switched to the left, because the inductor current cannot change instantaneously, we have ) ( = + L i And hence 8 ) ( 5 . 4 ) ( − = ⇒ = Ω + + + dt dv V dt dv C (b)This is a parallel RLC circuit. We have 3 1 2 1 1 2 1 , 1 2 1 2 2 , 1 j s LC RC ± − = − ± − = = = = = ω α So )} 3 sin( ) 3 cos( { ) ( t B t A e t v t + = − Using the initial conditions, we have 309 . 2 3 4 8 3 ) 1 ( ) ( 4 ) ( − = − = ⇒ − = + − = = = + + B B A dt dv A v )} 3 sin( 309 . 2 ) 3 cos( 4 { ) ( t t e t v t − = −...
View Full Document

This note was uploaded on 02/24/2011 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas.

Page1 / 4

EE411_HW6_Soln - v 4 ) ( = + At t=0 + , switch is switched...

This preview shows document pages 1 - 4. Sign up to view the full document.

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