# Network Theory Questions and Answers – Parallel Circuits

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: Capacitive reactance = 1/(6.28×5×103×0.2×10-6)=159.2Ω.Current in the resistance branch IR=VS/R=20/100 = 0.2A.Current in capacitive branch IC=VS/XC = 20/159.2 = 0.126A.Total current IT = (IR+j IC) A = (0.2+j0.13) A. In polar form, IT = 0.24∠33⁰ A.

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

a) 31⁰
b) 32⁰
c) 33⁰
d) 34⁰

Explanation: We obtained IT = 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

Explanation: Z = VS/IT = 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

Explanation: Since the voltage across each element is the same as the applied voltage, the current in resistive branch is IR = VS/R = 20∠0⁰/50∠0⁰=0.4A. Current in the inductive branch is IL = VS/XL = 20∠0⁰/30∠90⁰= 0.66∠-90⁰. Total current is IT = 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.

Explanation: The current lags behind the voltage by 58.8⁰. Total impedance Z = VS/IT = 20∠0⁰ / 0.77∠-58.8⁰ = 25.97∠-58.8⁰Ω. So the total impedance in the circuit shown is 25.97∠-58.8⁰Ω.
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6. Determine Z in the figure shown below.

a) 26∠-20.5⁰
b) 26∠20.5⁰
c) 25∠-20.5⁰
d) 25∠20.5⁰

Explanation: First the inductive reactance is calculated. XL=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. ZT = 10 + (20)(j31.42)/(20+j31.42)=24.23+j9.06=25.87∠20.5⁰.

7. Find IT 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

Explanation: The current lags behind the applied voltage by 20.5⁰. Total current IT = VS/ZT = 20/25.87∠20.5o=0.77∠-20.5o.

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

a) 20.5⁰
b) 20⁰
c) 19.5⁰
d) 19⁰

Explanation: As IT = 0.77∠-20.5o, 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

Explanation: Capacitive reactance XC = 1/2πfC = 1/(6.28×50×100×10-6)=31.83Ω. Capacitive susceptance BC = 1/XC = 1/31.83 = 0.031S. Conductance G=1/R = 1/50 = 0.02S. Total admittance Y=√(G2+Bc2)=√(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

Explanation: Phase angle θ=tan-1(R/Xc). Resistance R = 50Ω and capacitive reactance Xc = 31.83Ω. So the phase angle in the circuit = tan-1(50/31.83)=57.52⁰.

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