This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Series Circuits”.

1. If we apply a sinusoidal input to RL circuit, the current in the circuit is __________ and the voltage across the elements is _______________

a) square, square

b) square, sinusoid

c) sinusoid, square

d) sinusoid, sinusoid

View Answer

Explanation: If we apply a sinusoidal input to RL circuit, the current in the circuit is square and the voltage across the elements is sinusoid. In the analysis of the RL series circuit, we can find the impedance, current, phase angle and voltage drops.

2. The circuit shown below consists of a 1kΩ resistor connected in series with a 50mH coil, a 10V rms, 10 KHz signal is applied. Find impedance Z in rectangular form.

a) (1000+j0.05) Ω

b) (100+j0.5) Ω

c) (1000+j3140) Ω

d) (100+j3140) Ω

View Answer

Explanation: Inductive Reactance X

_{L}= ωL = 2πfL = (6.28)(10

^{4})(50×10

^{-3}) =3140Ω. In rectangular form, total impedance Z = (1000+j3140) Ω.

3. Find the current I (mA) in the circuit shown above.

a) 3.03

b) 30.3

c) 303

d) 0.303

View Answer

Explanation: Total impedance Z = (1000+j3140) Ω. Magnitude = 3295.4Ω

Current I=V

_{s}/Z = 10/3295.4=3.03mA.

4. Find the phase angle θ in the circuit shown above.

a) 62.33⁰

b) 72.33⁰

c) 82.33⁰

d) 92.33⁰

View Answer

Explanation: Phase angle θ = tan

^{-1}(X

_{L}/R). The values of X

_{L}, R are X

_{L}= 3140Ω and R = 1000Ω. On substituting the values in the equation, phase angle θ =tan

^{-1}(3140/1000)=72.33⁰.

5. In the circuit shown above find the voltage across resistance.

a) 0.303

b) 303

c) 3.03

d) 30.3

View Answer

Explanation: Voltage across resistance = IR. The values of I = 3.03 mA and R = 10000Ω. On substituting the values in the equation, the voltage across resistance = 3.03 x 10

^{-3}×1000 = 3.03V.

6. In the circuit shown above find voltage across inductive reactance.

a) 9.5

b) 10

c) 9

d) 10.5

View Answer

Explanation: Voltage across inductor = IX

_{L}. The values of I = 3.03 mA and X

_{L}= 10000Ω. On substituting the values in the equation, the voltage across inductor =3.03×10

^{-3}×1000 = 9.51V.

7. Determine the source voltage if voltage across resistance is 70V and the voltage across inductor is 20V as shown in the figure.

a) 71

b) 72

c) 73

d) 74

View Answer

Explanation: If voltage across resistance is 70V and the voltage across inductor is 20V, source voltage V

_{s}=√(70

^{2}+20

^{2}) = 72.8≅73V.

8. Find the phase angle in the circuit shown above.

a) 15

b) 16

c) 17

d) 18

View Answer

Explanation: The phase angle is the angle between the source voltage and the current. Phase angle θ=tan

^{-1}(V

_{L}/V

_{R}). The values of V

_{L}= 20V and V

_{R}= 70V. On substituting the values in the equation, phase angle in the circuit =tan

^{-1}(20/70)=15.94

^{o}≅ 16

^{o}.

9. An AC voltage source supplies a 500Hz, 10V rms signal to a 2kΩ resistor in series with a 0.1µF capacitor as shown in figure. Find the total impedance.

a) 3750.6Ω

b) 3760.6Ω

c) 3780.6Ω

d) 3790.6Ω

View Answer

Explanation: The capacitive reactance X

_{C}= 1/2πfC =1/(6.28×500×0.1×10

^{-6}) )=3184.7Ω. In rectangular form, X = (2000-j3184.7)Ω. Magnitude = 3760.6Ω.

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

a) 58

b) 68

c) -58

d) -68

View Answer

Explanation: The phase angle in the circuit is phase angle θ =tan

^{-1}(-X

_{C}/R) =tan

^{-1}((-3184.7)/2000)=-57.87

^{o}≅-58

^{o}.

11. Find the current I (mA) in the circuit shown above.

a) 2.66

b) 3.66

c) 4.66

d) 5.66

View Answer

Explanation: The term current is the ratio of voltage to the impedance. The current I (mA) in the circuit is current I = V

_{S}/ Z = 10/3760.6 = 2.66mA

12. Find the voltage across the capacitor in the circuit shown above.

a) 7

b) 7.5

c) 8

d) 8.5

View Answer

Explanation: The voltage across the capacitor in the circuit is capacitor voltage = 2.66×10

^{-3}×3184.7 =8.47V.

13. Determine the voltage across the resistor in the circuit shown above.

a) 3

b) 4

c) 5

d) 6

View Answer

Explanation: The voltage across the resistor in the circuit resistive voltage = 2.66×10

^{-3}×3184.7 =5.32V.

14. In the circuit shown below determine the total impedance.

a) 161

b) 162

c) 163

d) 164

View Answer

Explanation: Reactance across capacitor = 1/(6.28×50×10×10

^{-6}) =318.5Ω.

Reactance across inductor = 6.28×0.5×50=157Ω. In rectangular form, Z = (10+j157-j318.5) Ω = (10-j161.5)Ω. Magnitude=161.8Ω.

15. Find the current in the circuit shown above.

a) 0.1

b) 0.2

c) 0.3

d) 0.4

View Answer

Explanation: The term current is the ratio of voltage to the impedance. The current in the circuit is current I=V

_{S}/Z = 50/161.8 = 0.3A.

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