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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Faculty of Engineering ELECTRICAL AND ELECTRONIC ENGINEERING DEPARTMENT EENG223 Circuit Theory I INFE221 – Electrical Circuits FINAL EXAM Spring 2009-10 10 June 2010 Duration: 120 minutes Instructor: O. Kukrer Solve all 6 Problems STUDENT’S NUMBER NAME SURNAME GROUP NO. Problem Points 1 2 3 4 5 6 TOTAL 15 15 15 20 15 20 100 EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q1. In the circuit of Figure 1, find the current io using mesh analysis (Other methods will not be accepted). (15 pts) io 8Ω 2Ω Figure 1 6Ω ix 4Ω + - 3ix 14 V SOLUTION: (a) io 8Ω i1 ix 4Ω 2Ω 6Ω i2 14 V + - 3ix i3 KVL eqn. for mesh 1: 16i1 2i2 6i3 0 (1) 6i2 2i1 14 0 (2) KVL eqn. for mesh 2: In meshes 2 and 3, i3 3ix ix i2 i3 3i2 5 (1) 16i1 2i2 18i2 0 i1 i2 4 5 (2) 6i2 i2 14 i2 4 A 2 i1 5 A i0 i1 5 A EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q2. In the circuit of Figure 2, use source transformation to find the voltage vx. (Other methods will not be accepted) (15 pts) 4Ω + vx Figure 2 15 V + - 3vx 2Ω SOLUTION: Transform the dependent current source to a voltage source 4Ω 2Ω + vx 15 V i + - + - 6vx KVL for the loop: 15 6i 6v x 0 vx 4i 30i 15 i 0.5 A vx 2 V OR a more difficult way: Transform the independent voltage source to a current source. Define the voltage across the dependent current source as v1. 4/3 Ω + 3.75 A 4Ω 2Ω v1 i1 3vx _ + _ v1 _ 4/3 Ω 4 v1 5 i1 5 4vx 3 Also, from the original circuit v1 15 vx 5V + i1 3vx vx 2 V 3vx EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q3. In the circuit in Figure 3, the op-amp is ideal. (a) Find the output voltage v0. (7 pts) (b) Find the current i0 (output current of the op-amp). (8 pts) 5 kΩ Figure 3 i0 v0 2 mA 2 kΩ SOLUTION: 5 kΩ if i=0 A i0 v0 v-=0 V iL 2 mA (a) i f 2 mA (b) KCL at the output node: i0 i f iL 2 kΩ v0 5 k i f 10 V iL v0 5 mA 2 k i0 7 mA EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q4. The switch in the circuit in Figure 4 has been closed for a long time. It is opened at t = 0. (a) Find the initial value of the capacitor voltage v(0+) just after the switch is opened. (5 pts) (b) Find the capacitor voltage v(t) for t > 0. (10 pts) (c) Determine the time instant t1 > 0 at which the capacitor voltage becomes equal to 10 V. (5 pts) t=0 Figure 4 + 10 Ω 9A 8Ω v(t) 0.25 F SOLUTION: (a) For t < 0 i By current division, + 10 Ω 9A 8Ω v(t) i 10 9 5 A 10 8 (b) v (0 ) 8i 40 V v(0 ) v(0 ) 40 V For t > 0 + v(t ) V0 e t / 8Ω v(t) 0.25 F (c) V0 40 V, RC 8 0.25 2 s v(t ) 40e0.5 t V v(t1 ) 40e 0.5t1 10 e 0.5t1 0.25 t1 2 ln(0.25) 2.7726 s EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q5. In the circuit in Figure 5, v(t ) 100 e 200 t V , i (t ) I 0 e 200 t A (a) Find L and I0. (5 pts) (b) Determine the time constant . (5 pts) (c) Find the energy dissipated in the resistor until t 5 ms (i.e. in the time interval 0 t 5 ms ). for t 0 . i Figure 5 + L (5 pts) v 5Ω - SOLUTION: (a) v(t ) R i(t ) 100 5I 0 I 0 20 A Also, v (t ) L di (t ) dt since L does not satisfy the passive sign convention 100 LI 0 ( 200) L 100 25 mH 200 20 (b) 1 5 ms 200 or L 25 mH 5 ms R 5 ( c) The energy dissipated in the resistor from t = 0 s to t = 5 ms is equal to the change in the energy stored in the inductor between those instants. 1 2 1 2 Li1 Li0 i0 i (0) 20 A, i1 i (5 ms) 20 e 2000.005 20 e 1 7.3576 A 2 2 1 1 WL L(202 7.35762 ) 0.025 345.87 4.3233 J 2 2 WL OR the energy dissipated may be found by integrating the power 5ms WR 5ms R i (t ) dt RI e 2 0 2 0 0 400 t 1 400t 5ms 5(1 e 4000.005 ) 4.3233 J dt 2000 e 0 400 EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam Q6. In the circuit in Figure 6, u(t ) is the unit step function. (a) Sketch graphs of the current sources i1 (t ) 2[1 u(t )] A and i2 (t ) 5u(t ) A. (4 pts) (b) Find and sketch the capacitor voltage v0 (t ) for all time. (16 pts) + v 0(t) Figure 6 0.1 F 10 Ω 2 [1 u(t) ] A 10 Ω 5 u(t) A SOLUTION: (a) i1 i2 5A 2A t t (b) For t < 0 under dc conditions: _ + v 0(0 ) 10 Ω 2A v0 (0 ) 2 10 20 V 10 Ω For t > 0 Under dc conditions (as t + v 0(t) Req + v 0() 0.1 F 10 Ω 10 Ω 5A 10 Ω 10 Ω 10 Ω 5A 10 Ω v0 (0 ) v0 (0 ) 20 V, v0 () 50 V, Req 20 ReqC 20 0.1 2 s v0 (t ) v0 ( ) v0 (0) v0 ( ) e t / 50 (20 50) e t /2 50 30 e t /2 v0 (V) 50 V 20 t