EECE 210C Exam 2, Spring 2011

American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EECE 210 - Electric Circuits-Spring term 2011 Quiz Il - Solution Problem 1 Find V7, with respect to the terminals a-b in the circuit shown below when vs= 15 V, R; = R= 150 Q, and b= 1. Solution V. V, =R,i, +R,(1+b)i, Si, = 8 oars Shai, R, +R(1+b) Vn, =Re0 +b, = 2 R,; +R,(1+b) Problem 2 Find Rr, with respect to the terminals a-b for the circuit in problem 1. Solution ig =(1+b)i, =(14 bya 1 Vin __Ro(Gi+b)V, | Ry RAR, ig Ry +R,(1+b) (1+b)V, R,+R5(1+b) Ry, = Problem 3 The switch in the circuit shown below has been closed for a very long time before it opens at ¢ = 0. Find the energy stored in the inductor at = 0". 480 Solution As the switch has been closed for a long period of time, the inductor acts as a short circuit. Accordingly, we can remove the 48Q resistor and the current in the inductor is the same as the one in the 8 Q resistor. 8* 40 20 32 The total resistance is then given by: Ry =(8//40)+4= re +4= 5 +4= : Q : an. . 64 *3 The total current (the current in the 4Q resistor) is then given by: I, = 39 =6A . ne 40 The current in the 8Q resistor is then: Ig = 6 * a8 =5A 1 The total energy stored in the inductor is given by: W, =—L(I ) =4*25=100J L 2 8 Problem 4 In the circuit of problem 3, Find i(s) at f= 0.333 s. Solution Reg =(40 +8)//48 = 240 c= 4-8 933333 R 24 “ay fi, en 5 4 i(t)= me * => i(0.333) = Se = 0.924 Problem 5 Consider the figure shown below, determine the output voltage when I=] mA 2.2k0 Solution Yp =0 Volts, then v, =0 and no currents flow in the 150 Q resistor. Accordingly, _ Oven ~ 2.2103 Problem 6. Consider the Figure shown below, determine the output voltage given that v1=1 volts. i, =I => Vou, =-22001 n 6 Ohm 2 Ohm 12 Ohm WIN o——-VV- a iS yout = 4 0hm + NWA = 8 Ohm Solution . a 8 2 Using voltage divider rule, Yop —o a, I= al = “p 8+4 3 Determine V,,; as a function of v,,. 3 Vout’. Vout =v. Vv in = 0 => out n ae out n = n 6 12 ) Vv 3Vv out —s n => Vout = 3, 4 4 y) Vn =Vp > Vout =3Vn -(2}y => Vout =2V, Problem 7 For the circuit shown below, the switch has been in its position for a long period of time. oe 3s What is the value of the current in the inductor i, at t = 0°? Solution The inductor, and before the switch is open, acts as a short circuits. 20 pe 240 + (200//800) 200 °300+800— ing ip =ip Problem 8 The figure shown below is equivalent to a single inductor having an equivalent inductance of 15 Henries. Determine L Le Solution The 25 H inductor is in series with a parallel combination of 20H and 60 H inductors. The inductance of the equivalent inductor is: 25+((60*20)/(60+20))=40 H. The 30 H inductor is in parallel with a short circuit, which is equivalent to as short circuit. After making these simplifications, we have 40 11 Lg =L, #10, De Sy ee L,; 20 40 L 3L+40 Substituting, we obtain i. - 40k +105 3LL,, +40L.. = 40L+30L +400 eq «4 9 3L +40 400 - 40L gq ~ 3Lgg —70 Problem 9 Find the Thevenin equivalent voltage seen between terminals ab if V1 = 3V and R1 = 10Q. 50 409 R1 ¢\/\/\-9——$9—9 \V\V\9 \V\\0 a ance es Vv sov( * . 5A v e e ob Solution Let V2 be the node voltage to the left of the 5 Q resistor and V3 to the right. Using Nodal . V3; V3; -—50 analysis, we have: V2 = 50 Volts and — + +5=0 20 Rearranging the above equation, we obtain V3; +4V; —200+100=0=> 5V; =100 => V, =20 Volts V3 — Vay = V; => Vep = V3 -V, =20-V, Problem 10 The switch was open for a long time before it closes at t=0. Find the voltage across the capacitor after t= It (one time constant after closing the switch) given that R = 10 kQ. 100 pA 0.4 wF 20 kO 5k 10kQ ¢ AWN ? WA- WW, 9 wv LY R $ o- Solution First, let us determine the initial voltage across the capacitor when the switch is open. 1. Convert the current source in parallel with the 20 KQ resistir to a voltage source of 2 V in series with this resistor 2. Combine the two voltage sources into one source of 8 Volts 3. Use voltage divider rule to obtain the voltage across the 10 KQ resistor 8*10 80 10+5+20+R, 35+R, When the switch is open, the capacitor will discharge through the 10 KQ resistor. 0 — 6 4 € € 35+R, 35+R, V,(t) = Problem 11 The switch was closed for a long time. It opens at t=0. Find the total energy stored in the capacitors at t=0, given that C = 11F. 9) 1) 2 1SkO x 20 kO MA, AAA Ya - Cc = 40 kO é Solution When the switch is closed for a very long period of time, both capacitors acts as open circuits and the voltage across the 5uF and the capacitor C are the voltages across the 20 KQ and the 60KQ resistors, respectively. * * Vip =—15%40_ _15*40 _ gyi, 15+20+40 75 15*20 _ 15*20 15+20+40 75 =4 Volts Vo = Energy stored in the capacitors is given by E= 5 (5810-4?) +(C¥10-* V8?) Joules Problem 12 Assume in the following circuit L}=3 H, L2=2 H, L3=26H, and M=5H. Find the mesh equation around i2. on Solution . di, 3i5 +L, —+- 2 at Ip =i, —13 di di d 3i2 +L; + -M— +1(i, -i,) +L, <(i, -i, +i.) =0 2 - (iz —i,) 27 2 1 +1,) di di di Ly — +(L, +L.)—2-(M+L,)—!-i, +4, =0 oa wa | 2) 1, +415 Problem 13 In the circuit shown below, find the output voltage Vo if V=2V V=1V assuming that all operational amplifiers are ideal and operating in their linear region. vt 10K SK 5K 10K Solution Vv “UH 10K SK SK 10K v Voltage at node (a) is equal to V* = 2 Volts and the voltage at node b is equal to V=1 volts. Therefore, the current in the 5K resistor (which is the same in the 10K and 20K resistors) is given by: Va Vp =5*10° *i>i=0.2mA Voi — Va = 20*10° *0.2*107 = vo, =4+2=6 volts Vp — Von =10*10° *0.2*10 = vo, =1-2=-1 volts Vo = 2a —Vo,) =14 volts Problem 14 In the circuit shown below, calculate the maximum power that can be dissipated by R, if Vx=6V. b Vx © + S20 WAVAY w (2) 5 AVA, Lo 12A Solution Maximum power transfer is obtained when the load is equal to Ry. Accordingly, we need to determine the Thevenin equivalent Ry, =(2+1)/34+3=4.50 Vth = 12 _ V30 V3q is the voltage across the 3Q resistor which is obtained by converting the current source parallel to a resistor to a voltage source in series with the resistor, add the two voltage sources, and use voltage divider rule. _ (12—Vx)*3 _ 2-V,) Van = 32 6 2 Vy =12-02=Va) _ 12+, 2 2 Maximum power transfer is given by: Vin _ (12+V,)? _ (12+V,)? 4Ry, 4*4*4.5 4*18 P= (18)? For V, = 6 volts=> P= = 4.5 Watts = 4*18