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

View Answer

Explanation: Capacitive reactance = 1/(6.28×5×10

^{3}×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 above.

a) 31⁰

b) 32⁰

c) 33⁰

d) 34⁰

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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 above.

a) 25∠-58.8⁰

b) 25∠-58.8⁰.

c) 26∠-58.8⁰.

d) 26∠58.8⁰.

View Answer

Explanation: 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

View Answer

Explanation: 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 above.

a) 0.66∠-20.5⁰

b) 0.66∠20.5⁰

c) 0.77∠20.5⁰

d) 0.77∠-20.5⁰

View Answer

Explanation: 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 above.

a) 20.5⁰

b) 20⁰

c) 19.5⁰

d) 19⁰

View Answer

Explanation: 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

View Answer

Explanation: 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=√(G

^{2}+B

_{c}

^{2})=√(0.02

^{2}+0.031

^{2})=0.037S. Total impedance Z = 1/Y = 1/0.037 = 27.02Ω.

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

a) 56⁰

b) 56.5⁰

c) 57.5⁰

d) 57⁰

View Answer

Explanation: 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.

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