This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 1 of 6 EECS 40/42/100, Spring 2007 Prof. ChangHasnain Homework #5 Due at 6 pm in 240 Cory on Wednesday, 2/28/07 Total Points: 100 • Put (1) your name and (2) discussion section number on your homework. • You need to put down all the derivation steps to obtain full credits of the problems. Numerical answers alone will at best receive low percentage partial credits. • No late submission will be accepted expect those with prior approval from Prof. ChangHasnain. 1. Consider the following circuit: [11 points] The switch in the circuit shown has been in position a for a long time, At t = 0 the switch is moved to position b. Calculate: (a) The initial voltage on the capacitor Just prior to the switch flipping at time t=0, the circuit is in steadystate, so the capacitor acts like an open circuit. Thus the voltage across the capacitor is given by a voltage divider: V V k k k v c 80 ) 120 ( 8 4 8 ) ( = + = ! Because capacitor voltage is continuous: V v v c c 80 ) ( ) ( = ! = + [2 points] (b) The final voltage on the capacitor Again, the circuit has reached steady state, so the capacitor acts like an open circuit. Thus all the current goes through the 40k resistor. UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 2 of 6 By KVL: V k A k mA v v v k k c 60 ) 10 ( ) 40 ( 5 . 1 ) ( 10 40 ! = ! ! = ! = " [2 points] (c) The time constant for t > 0 To find the Thevenin equivalent resistance, we zero the current source and get k k k R eq 50 40 10 = + = So the time constant is " = R eq C = (50 k )(0.02 μ F ) = 1 ms [3 points] (d) The length of time required for capacitor voltage to reach zero after the switch is moved to position b. Solving the differential equation we get v c ( t ) = K 1 + K 2 exp{ " t /1 ms } V K V K V v V K v c c 140 80 60 ) ( 60 ) ( 2 2 1 = ! = + " = + " = = # [2 points] v c ( t ) = " 60 V + 140 V exp{ " t /1 ms } So to find the time it takes to reach zero, we set this equal to zero and solve for t. = " 60 V + 140 V exp{ " t /1 ms } # t = 1 ms $ ln(7/3) = 0.8473 ms [2 points] 2. Hambley, P4.33 [10 points] The differential equation obtained using KCL: ) ( ) ( ) ( = + !...
View
Full Document
 Spring '07
 ChangHasnain
 Derivative, Constant of integration, EECS Department

Click to edit the document details