EE2111_A03_SOL4

EE2111_A03_SOL4 - EE2111 Homework #4 A03 Assigned 18...

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EE2111 Homework #4 A03 Assigned 18 September 03 Due 23 September 03 Problem 1 . Hambly P4.20 Solution: () () ( ) ( ) [] 0 3 4 4 4 1 0 0 25 2 4 25 100 0 , 1 25 75 100 5 . 12 5 . 12 = + = + = = = = = = + = t e e t i e i i t i R L A i t A t i t t t τ Problem 2 . Hambly P4.22 Solution: ( ) 0 0 = t A t i In DC steady state as t gets large, the inductor looks like a short circuit and so the 2 resistors are in parallel, Thus ( ) A i 1 = Furthermore 20 1 10 10 1 = + = = R L
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And so () ( ) ( ) [] () A e e t i e i i t i t t t 20 20 1 1 1 0 0 0 = + = + = τ Problem 3 . Hambly P4.25 Solution: If we apply KVL around the loop, we get the differential equation for capacitor voltage RC t t v RC dt t dv C C = + 1 This is a case of the general first order equation () () t f t ax dt t dx = + We found the solution to this equation in class + = t a at at d e f e e x t x 0 0 λ where λ is the variable of integration. In this problem () () ( ) 0 0 , , , 1 = = = = x t v t x RC t t f RC a C so that + = d e e RC t v t RC RC t C 0 1 0 We solve the integral using integration by parts ∫∫ = vdu uv udv
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where we identify ∫∫ = = = = 1 , 2 2 0 RC t t RC RC t RC e C R RCt d e RC RCte udv d e dv u λ and so () RC t C RCe t RC t v + + = Problem 4 . Hambly P4.26 Solution: (a) The capacitor voltages cannot change instantaneously, thus ( ) ( ) V v
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This note was uploaded on 10/13/2009 for the course ECE 2111 taught by Professor Labonte during the Spring '09 term at WPI.

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EE2111_A03_SOL4 - EE2111 Homework #4 A03 Assigned 18...

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