Electronic Devices and Circuits Multiple Choice Questions on “Sinusoidal Steady State Analysis”.
1. i(t) = ?
A. 20 cos (300t + 68.2) A
B. 20 cos(300t – 68.2) A
C. 2.48 cos(300t + 68.2) A
D. 2.48 cos(300t – 68.2) A
2. Vc(t) = ?
A. 0.89 cos (1000t – 63.43) V
B. 0.89 cos (1000t + 63.43) V
C. 0.45 cos (1000t + 26.57) V
D. 0.45 cos (1000t – 26.57) V
3. Vc(t) = ?
A. 2.25 cos (5t + 150) V
B. 2.25 cos (5t – 150) V
C. 2.25 cos (5t + 140.71) V
D. 2.25 cos (5t – 140.71) V
4. i(t) = ?
A. 2 sin (2t 5.77) A
B. cos (2t 84.23) A
C. 2 sin (2t 5.77) A
D. cos (2t 84.23) A
5. In the bridge shown, Z1 = 300 ohm, Z2 = 400 – j300 ohm, Z3 = 200 + j100 ohm. The Z4 at balance is
A. 400 + j300 ohm
B. 400 – j300 ohm
C. j100 ohm
D. -j900 ohm
Answer: B
6. i1(t) = ?
A. 2.36 cos (4t 41.07) A
B. 2.36 cos (4t 41.07) A
C. 1.37 cos (4t 41.07) A
D. 2.36 cos (4t 41.07) A
7. i2(t) = ?
A. 2.04 sin (4t 92.13) A
B. 2.04 sin (4t 2.13) A
C. 2.04 cos (4t 2.13) A
D. 2.04 cos (4t 92.13) A
8. In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be
A. R C
B. L C
C. R L
D. R R
Answer: A
Clarification: RC circuit causes a positive shift in the circuit.
9. P = 269 W, Q = 150 VAR (capacitive). The power in the complex form is
A. 150 – j269 VA
B. 150 + j269 VA
C. 269 – j150 VA
D. 269 + j150 VA
Answer: C
Clarification: S = P – jQ.
10. Q = 2000 VAR, pf = 0.9 (leading). The power in complex form is
A. 4129.8 j2000 VA
B. 2000 j4129.8 VA
C. 2000 j41.29.8 VA
D. 4129.8 j2000 VA
Answer: D
Clarification: Use cos T = 0.9 or T = 25.84 degrees.
Q = S sin T or S = 4588.6 VA
p = S cos T or P = 0.9 X 4588.6 4129.8 VA.