HW 2 ans

HW 2 ans - [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ icl.syr.edu ELE232 Electrical...

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- icl.syr.edu ELE232 – Electrical Fundamentals II – 23JAN2008 Homework 2 Answers Problem 1 : We know how to solve the circuit on the left for all its currents and voltages using differential equation methods. The circuit on the rights is equivalent to the one on the left and has been transformed from a time domain representation to the Laplace Transform domain . (a) For the circuit on the left, show that V C (t) = 1 e t RC u(t) using methods learned last semester. Ans : V C (t) = A + Be t t C , t C = RC . V C (t) = 1v when t > 5t C . A = 1v . V C (0) = 0v = A + B . B = -1 and V C (t) = 1 e t t C u(t) . The u(t) term is very important. It make it clear that V C (t) is zero when t < 0 is zero. (b) For the circuit on the right, show that V C (s) = 1 RC s s + 1 RC . Ans : Use the voltage divider rule : V C (t) = 1 s 1 sC R + 1 sC = 1 s RCs + 1 ( ) = 1 RC s s + 1 RC (c) Decompose V C (s) as the sum of two first order terms : V C (s) = A s + B s + 1 RC = 1 RC s s + 1 RC . Find values for A and B . Ans : V C (s) = A s + B s + 1 RC = 1 RC s s + 1 RC = A s + 1 RC + Bs s s + 1 RC From this equation, it follow that

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This note was uploaded on 05/02/2008 for the course ELE 232 taught by Professor Phelps during the Spring '08 term at Syracuse.

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HW 2 ans - [email protected] icl.syr.edu ELE232 Electrical...

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