250+ TOP MCQs on Series Circuits and Answers

Network Theory Multiple Choice Questions 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

Answer: D
Clarification: 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

Answer: C
Clarification: Inductive Reactance X_L = ωL = 2πfL = (6.28)(10⁴)(50×10^-3) = 3140Ω. In rectangular form, total impedance Z = (1000+j3140) Ω.

3. Find the current I (mA. in the circuit shown below.

A. 3.03
B. 30.3
C. 303
D. 0.303

Answer: A
Clarification: 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 below.

A. 62.33⁰
B. 72.33⁰
C. 82.33⁰
D. 92.33⁰

Answer: B
Clarification: 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 below, find the voltage across resistance.

A. 0.303
B. 303
C. 3.03
D. 30.3

Answer: C
Clarification: 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 below, find voltage across inductive reactance.

A. 9.5
B. 10
C. 9
D. 10.5

Answer: A
Clarification: 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

Answer: C
Clarification: If voltage across resistance is 70V and the voltage across inductor is 20V, source voltage V_s=√(70²+20²) = 72.8≅73V.

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

A. 15
B. 16
C. 17
D. 18

Answer: B
Clarification: 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 the following figure. Find the total impedance.

A. 3750.6Ω
B. 3760.6Ω
C. 3780.6Ω
D. 3790.6Ω

Answer: B
Clarification: 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 below.

A. 58
B. 68
C. -58
D. -68

Answer: C
Clarification: 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 below.

A. 2.66
B. 3.66
C. 4.66
D. 5.66

Answer: A
Clarification: 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 below.

A. 7
B. 7.5
C. 8
D. 8.5

Answer: D
Clarification: 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 below.

A. 3
B. 4
C. 5
D. 6

Answer: C
Clarification: 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

Answer: B
Clarification: 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 below.

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer: C
Clarification: 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.