ECE 201 Section 4 Lecture 18 Prof Peter Bermel February 22 2012 Returning Homeworks HW 13 N 91 83 20 HW 14 N 97 89 15 2 22 2012 ECE 201 4 Prof Bermel Exams Posted Exam 1 solution on Blackboard is it useful as is Exam 2 review session tentatively scheduled for Wed March 7 at normal class time 2 22 2012 ECE 201 4 Prof Bermel Solving Differential Equations Using KVL RL circuits obey RC circuits can be written as 1 General form decomposes into homogeneous and inhomogeneous equations 0 2 22 2012 ECE 201 4 Prof Bermel Solving Differential Equations Can solve each part separately Homogeneous part Inhomogeneous part constant 2 22 2012 ECE 201 4 Prof Bermel Solving Differential Equations Use separation of variables and the following rules 1 ln ln ln ln ln ln ln 2 22 2012 ECE 201 4 Prof Bermel Solving Differential Equations We can integrate our homogeneous ODE To obtain 0 0 1 ln 2 2 0 6 1 2 22 2012 1 ECE 201 4 Prof Bermel Solving Differential Equations Overall solution is given by For RC or RL circuits we obtain 2 2 0 6 constant Where the time constant is given by RL circuits L R RC circuits RC 2 22 2012 ECE 201 4 Prof Bermel Example 1 What is the voltage response of this circuit as a function of time if V 0 1 mV Vx 2 Vx 5 2 22 2012 ECE 201 4 Prof Bermel 10 mF Solution Vx 2 Vx 10 mF 5 Calculate Th venin equivalent for left hand side No independent sources so Voc 0 Th venin resistance obtained from KCL 7 2 5 7 3 10 10 3 2 22 2012 ECE 201 4 Prof Bermel Solution 10 mF 10 3 Applying homogeneous solution from before 2 6 Since RC 1 30 s Vo 1 mV Qo 10 C 10 Charge exceeds that of known universe in 2 22 2012 ln A B AC ECE 201 4 Prof Bermel 5 06 s Example 2 What is the current flow in this circuit before and after the switch is flipped at time t 0 I 3 I 6 9H 9V t 0 2 22 2012 ECE 201 4 Prof Bermel 3 6 9V 9H Solution 3 I 6 9H 9V I 3 6 9V t 0 For t 0 I 9 V 9 1 A For t 0 KVL yields 9 3 I 9 H dI dt Since L R 3 assume solution takes the form I 3 Be t 3 2 22 2012 ECE 201 4 Prof Bermel 9H Solution 3 I 6 9H 9V I 3 6 9V t 0 For t 0 9 3 3 G H I 2 J 2H 9 I B is set by continuity condition for current 0 0 3 G 2 Thus B 3 and 3 1 2 0 2 22 2012 ECE 201 4 Prof Bermel 9H Homework HW 16 solution posted today HW 17 due today by 4 pm in EE 325B HW 18 due Fri DeCarlo Lin Chapter 8 Problem 4 Problem 5 Problem 9 b 2 22 2012 ECE 201 4 Prof Bermel
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