# Sinusoidal Steady State Analysis MCQ’s

This set of Electronic Devices & Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Sinusoidal Steady State Analysis”.

1. 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

2. 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

3. 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

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

Circuit for Q.6 and Q.7

5. 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

6. 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

7. 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

8. 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

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

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