San Jose State University Department of Electrical Engineering Exam Solution EE 98-MIT 6.x Fall 1 losed Book, losed Notes, and no electronic devices. Instructions: There are six problems. Interpretation of the problems will not be given during the exam. If you are unsure ab the meaning of a question, make an assumption, state what it is and continue. Problem Score Possible 1 15 1 3 1 4 5 15 6 3 Total 1 Problems are weighted as shown. Please write your name and Student Id (SID) at the bottom of this page and work all six problems. Show all your work. Work easier problem first. No credit will be given for work not shown. Use back of page if necessary but label clearly. Last Name: First Name: SID:
Problem 1: (15 points) A 5 F capacitor has a voltage as shown in the figure below. Find its current i 7ms t 1ms v t mv 1 i 5F t and draw it for v 7 7 1 t msec i t 1 Slope 7 t 3 3 V 1 1 11 Slope V sec 3 3 t 7 1 51 6 6 3 51 11 11 1mA i t Slope V sec 6 i t 51 ma t 7 3 1 1 3 1 1 Slope 3 3 7 1 1 4V sec i t 51 4 ma 6 7 t 1 3 3 1 1 11 Slope V sec 3 3 1 7 1 51 6 3 i t 51 11 1mA i t ma 7 1 7 1 t msec
Problem : (1 points) A voltage source is connected to a capacitor of F. Find the energy stored in the capacitor from 4t t to t 8sec, if v t e. (You don t have to express it as a numerical value. It can be expressed in terms of a power of e.) i t E v i v v 1 E v tv t 1 8t 8t E e e 64 e 1 E t t v t t t v t
Problem 3: (1 points) The circuit below is in D steady state. i 3H 3 4V 3F v 3A 1H a) Find it (5 points) In D steady-state inductors are replaced by shorts and capacitors by open circuits i 4V v 3A 3 4 i t 3 i t 3 4 5it9 15 5i t i t 3A b) Find v (5 points). v t 3 i t 3 v t 3 33 18V
Problem 4: ( points) The switch for a long period of time is to the left. Switch is moved to the right abruptly at t. 4 t 3F v t 3 3A a) Find the voltage vt for t 33 9 v t V (switch to the left position). (5 points) 4 v t v t 3 3A b) Find the voltage across the capacitor v t for t (switch to the left position). (5 points) 15 v t V c) Find the time constant for the resulting circuit for t (switch to the right position). (3 points) R 33 9sec
d) Find the expression for voltage across the capacitor right position). (7 points) v t and sketch it for t (switch to the 4 3F v t 3 3A 3F v 3 9V t 9 9 9 15 v t e 9 9 6 t v t e v V 15 9sec 9 t sec
Problem 5: (15 points) Derive the ordinary, second-order, constant coefficient differential equation in terms of circuit below with initial conditions i and vin 4V t vin v vl L 4H i 4F v t in the v 1) Write the ordinary second-order differential equation with constant coefficient describing the L network. i t d v d i t vin t v t L 1 t L KL 1 t L v in t v t i t d v v in v d v 1 vin v L d v t d v t L v vin 16 v 4 d v 16 v 4
Problem 6: (3 points) Given the ordinary, second-order, constant coefficient, differential equation below with initial condition v and d v 16 v 4 a) Use the 3 step method to solve the differential equation for v t Step 1: Find the particular solution (5points) For t Assume a particular solution Verify by substituting into DE P 16 vp 4 vp 4V vp t for example vp t Step ) Find the homogenous solution using 4-step method (1 points) Step a) Assume a homogenous solution vh t We assume a solution that its second derivative of which is proportional to itself. For example solution in the form of exponential H v t Ae st Step b) Find the characteristic equation Then, substitute in the DE. d vh 16 vh H d vh sae st st s Ae st st 16s Ae Ae 16s 1 Above equation is called characteristic equation Step c) Find the roots of characteristic equations 16s 1 1 s 16
s 1 1 j j.5 16 j 4 j j Angular frequency.5 rad sec H o 1 v t Ae A e j t jot Step d) General solution to the homogenous H v t Ae A e j.5t j.5t 1 3) Find the total solution v t v t v t P H v t 4 Ae A e j.5t j.5t 1 Zero state response. Initial condition v and Applying the first initial condition v t 1 v t 4 A A Applying the second initial condition t.5.5 j.5a e j.5a e A A j t j t 1 1 4 A1 A A1 A 4 A1 A Finally the total solution j.5t j.5t 4 v t e e Using Euler s identity e cos x jx e jx 4 4cos.5 v t t