250+ TOP MCQs on Parallel Circuits and Answers

Network Theory Multiple Choice Questions on “Parallel Circuits”.

1. A signal generator supplies a sine wave of 20V, 5 kHz to the circuit as shown. Determine total current IT.

A. 0.21∠33⁰
B. 0.22∠33⁰
C. 0.23∠33⁰
D. 0.24∠33⁰

Answer: D
Clarification: Capacitive reactance = 1/(6.28×5×10³×0.2×10^-6)=159.2Ω.Current in the resistance branch I_R=V_S/R=20/100 = 0.2A.Current in capacitive branch I_C=V_S/X_C = 20/159.2 = 0.126A.Total current I_T = (I_R+j I_C) A = (0.2+j0.13) A. In polar form, I_T = 0.24∠33⁰ A.

2. Find the phase angle in the circuit shown below.

A. 31⁰
B. 32⁰
C. 33⁰
D. 34⁰

Answer: C
Clarification: We obtained I_T = 0.24∠33⁰ A, the phase angle between applied voltage and current is 33⁰. The phase angle between the applied voltage and total current is 33⁰.

3. Determine the total impedance in the circuit.

A. 73.3∠33⁰
B. 83.3∠-33⁰
C. 83.3∠33⁰
D. 73.3∠-33⁰

Answer: B
Clarification: Z = V_S/I_T = 20∠0⁰ / 0.24∠33⁰ = 83.3∠-33⁰Ω. The phase angle indicates that the total line current is 0.24 A and leads the voltage by 33⁰.

4. A 50Ω resistor is connected in parallel with an inductive reactance of 30Ω. A 20V signal is applied to the circuit. Find the line current in the circuit.

A. 0.77∠-58.8⁰
B. 0.77∠58.8⁰
C. 0.88∠-58.8⁰
D. 0.88∠58.8⁰

Answer: A
Clarification: Since the voltage across each element is the same as the applied voltage, the current in resistive branch is I_R = V_S/R = 20∠0⁰/50∠0⁰=0.4A. Current in the inductive branch is I_L = V_S/X_L = 20∠0⁰/30∠90⁰= 0.66∠-90⁰. Total current is I_T = 0.4-j0.66 = 0.77∠-58.8⁰.

5. Determine total impedance in the circuit shown below.

A. 25∠-58.8⁰
B. 25∠-58.8⁰.
C. 26∠-58.8⁰.
D. 26∠58.8⁰.

Answer: C
Clarification: The current lags behind the voltage by 58.8⁰. Total impedance Z = V_S/I_T = 20∠0⁰ / 0.77∠-58.8⁰ = 25.97∠-58.8⁰Ω. So the total impedance in the circuit shown is 25.97∠-58.8⁰Ω.

6. Determine Z in the figure shown below.

A. 26∠-20.5⁰
B. 26∠20.5⁰
C. 25∠-20.5⁰
D. 25∠20.5⁰

Answer: B
Clarification: First the inductive reactance is calculated. X_L=6.28 x 50 x 0.1 = 31.42Ω. In figure the 10Ω resistance is in series with the parallel combination of 20Ω and j31.42Ω. Z_T = 10 + (20)(j31.42)/(20+j31.42)=24.23+j9.06=25.87∠20.5⁰.

7. Find I_T in the figure shown below.

A. 0.66∠-20.5⁰
B. 0.66∠20.5⁰
C. 0.77∠20.5⁰
D. 0.77∠-20.5⁰

Answer: D
Clarification: The current lags behind the applied voltage by 20.5⁰. Total current I_T = V_S/Z_T = 20/25.87∠20.5^o=0.77∠-20.5^o.

8. Find the phase angle θ in the circuit shown below.

A. 20.5⁰
B. 20⁰
C. 19.5⁰
D. 19⁰

Answer: A
Clarification: As I_T = 0.77∠-20.5^o, the phase angle θ between the voltage and the current in the circuit is 20.5⁰.

9. Find the impedance in the circuit shown below.

A. 25
B. 26
C. 27
D. 28

Answer: C
Clarification: Capacitive reactance X_C = 1/2πfC = 1/(6.28×50×100×10^-6)=31.83Ω. Capacitive susceptance B_C = 1/X_C = 1/31.83 = 0.031S. Conductance G=1/R = 1/50 = 0.02S. Total admittance Y=√(G²+B_c²)=√(0.02²+0.031²)=0.037S. Total impedance Z = 1/Y = 1/0.037 = 27.02Ω.

10. Determine the phase angle in the circuit shown below.

A. 56⁰
B. 56.5⁰
C. 57.5⁰
D. 57⁰

Answer: C
Clarification: Phase angle θ=tan^-1(R/X_c). Resistance R = 50Ω and capacitive reactance X_c = 31.83Ω. So the phase angle in the circuit = tan^-1(50/31.83)=57.52⁰.