ECE 212H1F Circuit Analysis October 30, :10-19: Reza Iravani 02 Reza Iravani 03 Piero Triverio. (Non-programmable Calculators Allowed)

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Please Print Clearly Last Name: First Name: Student Number: Your Tutorial Section (CIRCLE ONE): 01 Thu. 9-11 RS211 02 Thu. 9-11 GB119 03 Tue. 10-12 SF2202 04 Tue. 10-12 SF3201 05 Tue. 13-15 GB304 06 Tue. 13-15 GB412 (YOU LOSE FOUR POINTS FOR INCORRECT TUTORIAL SECTION INFORMATION) Your Lecture Section (CIRCLE ONE): 01 Reza Iravani 02 Reza Iravani 03 Piero Triverio (YOU LOSE ONE POINT FOR INCORRECT LECTURE SECTION INFORMATION) Mid-term Answer All Questions (Non-programmable Calculators Allowed) 100 minutes [25 Marks] Instructions: Do not unstaple this exam book. Answer each question neatly and concisely. Write the final answer in the box where provided. The exam has 5 questions and 10 pages, including the cover page, the 2 nd order-circuit aidsheet and an extra sheet. Solutions steps and final answers must not be in pencil. Q1 /8 Q2 /3 Q3 /6 Q4 /5 Q5 /3 Total /25 University of Toronto 1 of 10

Question 1 a) Find the expressions for V AB and V O in terms of V S in the circuit in Figure Q1.a. (The op amps are ideal). [4 points] x Figure Q1.a For ideal op amps: v + =v - and i + =i - =0 KCL at node x: Voltage division of and : From (1) and (2), And V AB = V O = University of Toronto 2 of 10

b) For the circuit in Figure Q1.b, (The op amps are ideal): Figure Q1.b. 1) Find v A (t) in terms of v s1 (t) and v s2 (t). [2 points] Using the inverting circuit equation: v A (t) = 2) Find v B (t) in terms of v s1 (t) and v s2 (t). [1 point] Using the non-inverting circuit equation: v B (t) = 3) Find the expression for i o (t). [1 point] Current flow through a capacitor: i o (t) = University of Toronto 3 of 10

Question 2 The switch in the circuit in Figure Q2 opens at t = 0. (It has been closed for a long time). i o Figure Q2 1) Calculate the voltage v o (t) for t 0. [2 points] In the given network:, and Circuit time constant, Using 1 st order circuit equation: For v o (t) = or v o (t) = for 2) Sketch v o (t) waveform for t 0. [1 point] v o (t)[v] 8 6 4 2 t [msec] -2 1 2 3 4 5 University of Toronto 4 of 10

Question 3 The switch in Figure Q3 has been in position 1 for a long time and is moved to position 2 at t = 0. For L = 4 mh, C =10 µf and I s = 2 A. i L (t)= i C (t) Figure Q3 a) Determine the values of R that make the circuit response (i)-overdamped, (ii)-critically damped and (iii)-underdamped. [3 points] Using the 2 nd order equation [a series RLC circuit] page 9: Laplace Transform of this equation: Compare it with the 2 nd order standard form: Then, & R Overdamped response For the overdamped response For the critically damped response For the underdamped response R= Critically damped response Underdamped response R University of Toronto 5 of 10

b) If R = 100 Ω, find v o (t) for t 0. [2 points] R = 100 Ω 40 Ω Overdamped response V for From the given circuit: V,, V & V Using the initial conditions, we can calculate & as follows: A for Solving (1) & (2) & A for V for v o (t)= V for c) Plot v o (t) (from part b above) for t - 0.5 msec. [1 point] v o (t)[v] 200 100 t [msec] -0.5 0.5 1 1.5 2 2.5-100 -200 University of Toronto 6 of 10

Question 4 For the network given in Figure Q4: Figure Q4 1) Write the mesh-current equations using the three loop currents identified on the figure. [3 points] KVL at loop A : Using the dependent current source equation (loop B ), we have: and from (loop A ) From loop C : The mesh-current equations 2) Use the mesh-current analysis to find the phasor voltage V o. [2 points] KVL at Loop C : Solving equations (1) & (2) for, V O = University of Toronto 7 of 10

Question 5 In the circuit shown in Figure Q4, determine the value of the inductance such that the current is in phase with the source voltage, i.e., they have the same phase angle. [3 points] Figure Q4 The voltage source in time domain: The voltage source in phasor domain: It is required to have the current in phase with the voltage source; this means both, the voltage source and the current, have the same phase angle ( ). Or in other words, In the given circuit: Solving for L L= University of Toronto 8 of 10

University of Toronto 9 of 10

Extra sheet University of Toronto 10 of 10