KING FAHD UNIVERSITY OF PETROLEUM & MINERALS ELECTRICAL ENGINEERING DEPARTMENT EE 202-122-FINAL EXAM DATE: Dec 30, 2013 TIME: 8:00 AM-11:00 AM SER# ID# KEY

Name Section#

Max Q1 MCQ (1-12) Q2

36 18

Q3

18

Q4

10

Q5

18

Total

100

First Order Circuit Solution x (t ) =X f + ( X o − X f )e − (t −t o ) / τ where x(t) is a voltage or current Second Order Circuit Solution x (t ) = X f + A1e s1t + A 2e s 2t x (t ) = X f + B1e −αt cos(ωd t ) + B2e −αt sin(ωd t ) x (t ) = X f + D1te −αt + D 2e −αt where x(t) is a voltage or current

Q1 (Multiple Choice Parts from 1 to 12 )

(Part 1)

Consider the 1F capacitor shown below

i (t ) i (t )

1F 1

+ v C (t ) − If

v C (0) = 1 V

0

and if the current

2

t

i (t ) passing through the capacitor

as shown above , then, the value of the capacitor voltage at 𝑡 = +∞,

v C ( ∞ ) , is

(a) v C ( ∞ ) =1 V (b) v C ( ∞ ) = +∞ V (c) v C ( ∞ ) =2 V (d) v C ( ∞ ) =−1 V

o = For the following circuit, the voltage source is v s 60 cos(5000t + 30 ) V

(Part 2)

The Thevenin voltage between the terminals ab in (Volts) is:

(a) 202.8∠114.3o

(b) 202.8∠ − 114.3o

(c) 68.7∠17.9o

(d) 68.7∠ − 17.9o

(e) 53.7∠56.6o

(f) 53.7∠ − 56.6o

(g) 26.8∠33.4o

(h) 26.8∠ − 33.4o

(Part 3)

The Thevenin Impedance between the terminals ab in (Ohms) is:

(a) 21.3∠96.5o

(b) 21.3∠ − 96.5o

(c) 2.68∠26.6o

(d) 2.68∠ − 26.6o

(e) 31.3∠94.4o

(f) 31.3∠ − 94.4o

(g) 11.63∠153.4o

(h) 11.63∠ − 153.4o

If the following circuit is operating at a frequency ω = 5000 rad/s .

(Part 4)

The Impedance Zˆ ab (seen between the terminals ab) is:

(a) 81.1∠46.4o

(b) 81.1∠ − 46.4o

(c) 51.9∠24.4o

(d) 51.9∠ − 24.4o

(e) 66.4∠38.9o

(f) 66.4∠ − 38.9o

(g) 86.5∠53.5o

(h) 86.5∠ − 53.5o

(Part 5)

The Impedance Zˆcd (seen between the terminals cd) is:

(a) 51.7∠30.7 o (e) 58.3∠51.4o

(b) 51.7∠ − 30.7 o (f) 58.3∠ − 51.4o ,

(c) 30.4∠26.2o

(d) 30.4∠ − 26.2o

(g) 32.05∠2.11o

(h) 32.05∠ − 2.11o

Consider the RLC circuit with two current sources as shown below.

+

2Ω

cos(1000t + 45o ) A

200 µ F

v L (t )

1 mH

sin(1000t + 45o ) A

−

(Part 6)

The voltage

v L (t ) across the inductor is

(a) 1.5 cos (1000 t + 58o )

(V)

(b) 1.5 sin (1000 t + 58o )

(V)

(c) 1.5 cos (1000 t + 58o ) + 1.5 sin (1000 t + 58o ) (d) 1.5 cos (1000 t + 58o ) − 1.5 sin (1000 t + 58o )

(V) (V)

For the RC circuit shown below the capacitor was initially charged to 1 volt. After 1 second from closing the switch, the voltage across the capacitor becomes 2 volts (Vc(1) = 2 ) .

(Part 7) (a)

2 −e

−1

1−e

−1

(Part 8)

(a)

2 3

s

Vs

The value of the voltage source

V

(b)

1−e

−1

2 −e

−1

V

2

(c) e

−1

V

(d)

is e

−1

V

(e) 2e

−1

V

(f) e

−1

V

2

The time constant of the circuit shown below after the switch closes is:

(b)

3 2

s

(c) −

2 3

s

(d) −

3 2

s

(e)

63 3

s

(f) −

63 3

s

Consider the circuit shown below

(Part 9) (a) − 8 V

The voltage Vx in the circuit:

(b) − 5 V

(Part 10)

The current

(a) 4 A

(b) 2 A

(d) − 35 V

(c) 35 V

Ix

(e) None of the above

in the circuit:

(c) − 2 A

(d) − 4 A

(e) None of the above

For the circuit shown below , the switch was in position a for a long time . At t = 0 , the switch moves to position b

1H

t =0

1Ω

b

a

4Ω

10 A

1F 8

+ v (t ) −

1H 2 i (t )

6Ω 3Ω

+ −

18 V

+ (Part 11) di (0 ) is dt

(a) 68 A/s

(b) 17 A/s

(c) 34 A/s

(d) 0 A/s

(Part 12) v (t ) is (a) overdamped

(b) underdamped

(c) critical damped

(d) None of the above

Q2 In the circuit shown below, the switch has three positions: a, b & c. The switch was initially at a for a long time. At t=0, the switch moves to position b and stays at position b for one second. At t=1, the switch moves to position c and stays there indefinitely. Find i L (t ) and v L (t ) for t > 0. a

1Ω

b c

4V

+ −

5Ω

1H i L (t )

Solution 1 : Finding i (t) then v L (t ) = di (t)/dt

+ v L (t ) −

Q2 -Continue

Solution 2 : Finding v L (t ) then i (t) = ∫ v L (τ )d τ

Solution 2 : continue

Q3 For the circuit shown below , if v a 100cos(200t + 30o ) V and v b = 50cos(200t ) V . = a) Find and draw the Phasor Circuit. b) In the Phasor domain, find the node voltage equations (without simplification) and put them in the box shown. Do not simplify the node voltage equations. Note that: v1 and v2 and the reference node are assigned as shown in the circuit. c) Find v1(t) and v2(t).

(a)

(3)

(6)

(9)

(b) The node voltage equations (without simplifications) are:

(10)

Q4 Find 𝑣(𝑡) in the following circuit

0.5 mH

+ 5 volts

5Ω

v(t)

-

20 μF 100cos(10000t+30º) V

Q5 In the circuit shown below, the switch opens at t = 0 . Find the current i (t ) for all times .

2H

+ −

100 V

10 Ω

t =0

15 Ω

1 F 24

1 F 8

3 H 2

i (t )

Solution Method - 1 finding v C (t ) then solving for i (t ) = C (1) or (0)

(2) or (0)

R 15 t>0 α = = = 5 rad / s 2L 2 3 2

(1) or (0)

ω = o

1= LC

α > ωo ⇒ overdamped v (t )= A1e s t + A 2e s t 1

s 1,2 =−α ± α 2 − ωo2 ⇒ s 1 =−2 s 2 =−8 v C (t ) = A1e −2t + A 2e −8t

2

dv C (t ) dt (2) or (0)

1 = 4 rad / s 3 1 2 24

t