250+ TOP MCQs on Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit and Answers

Network Theory Questions & Answers for Exams on “Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit”.

1. A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10^-6 F. System voltage is 100 V. The frequency, at which the maximum voltage appears across the capacitor, is?
A. 5937.81 Hz
B. 7370 Hz
C. 7937.81 Hz
D. 981 Hz

Answer: C
Clarification: Frequency f_c at which V_C is maximum f_c = (frac{1}{2π}Big[frac{1}{0.01×0.01×10^{-6}} – frac{50×50}{2×0.01×0.01}Big]^{0.5})
So, f_c = 7937.81 Hz.

2. A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10^-6 F. System voltage is 100 V. At this frequency, the circuit impedance is?
A. 50 – j2.5 Ω
B. 50 – j2.8 Ω
C. 50 + j2.5 Ω
D. 50 + j2.8 Ω

Answer: A
Clarification: Z = R + j (ωL + (frac{1}{ωC})) = 50 + j (left(2π × 7937.81 × 0.01 – frac{1}{2π×7937.81×0.04×10^{-6}}right)) = 50 – j2.5 Ω.

3. Given, R = 10 Ω, L = 100 mH and C = 10 μF. The values of ω₀, ω₁ and ω₂ respectively are?

A. ω₀ = 1000 rad/s, ω₁ = 951.25 rad/s, ω₂ = 1075.54 rad/s
B. ω₀ = 1000 rad/s, ω₁ = 975.25 rad/s, ω₂ = 1051.25 rad/s
C. ω₀ = 1000 rad/s, ω₁ = 951.25 rad/s, ω₂ = 1051.25 rad/s
D. ω₀ = 1000 rad/s, ω₁ = 825.30 rad/s, ω₂ = 1075.54 rad/s

Answer: C
Clarification: ω₀ = (frac{1}{sqrt{100×10^{-3}×10×10^{-6}}}) = 1000 rad/s
Now, ω₁ = (frac{-R + sqrt{R^2 + frac{4L}{C}}}{2L} = frac{-10 + frac{sqrt{100 + 4 × 100 × 10^{-3}}}{10×10^{-6}}}{2×100×10^{-3}}) = 951.25 rad/s
And, ω₂ = (frac{+R+ sqrt{R^2+ 4L/C}}{2L}) = 1051.25 rad/s.

4. Given, R = 10 Ω, L = 100 mH and C = 10 μF. Selectivity is?

A. 10
B. 1.2
C. 0.15
D. 0.1

Answer: D
Clarification: Selectivity = (frac{1}{Q})
Q = (frac{ωL}{R} = frac{1000×100×10^{-3}}{10})
∴ Q = 10
So, Selectivity = 0.1.

5. Two series resonant filters are shown below. Let the 3 dB bandwidth of filter 1 be B₁ and that of filter 2 be B₂. The value of (frac{B_1}{B_2}) is ____________

A. 0.25
B. 1
C. 0.5
D. 0.75

Answer: A
Clarification: For series resonant circuit, 3dB bandwidth is (frac{R}{L})
B₁ = (frac{R}{L_1})
B₂ = (frac{R}{L_2} = frac{4R}{L_1})
Hence, (frac{B_1}{B_2}) = 0.25.

6. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the half power frequencies of the circuit are?
A. 50.53 Hz, 49.57 Hz
B. 52.12 HZ, 49.8 Hz
C. 55.02 Hz, 48.95 Hz
D. 50 HZ, 49 Hz

Answer: A
Clarification: Bandwidth, BW = (frac{f_o}{Q} = frac{50}{47.115}) = 1.061 Hz
f₂, higher half power frequency = f₀ + (frac{BW}{2})
∴ f₂ = 50 + (frac{1.061}{2}) =50.53 Hz
f₁, lower half power frequency = f₀ – (frac{BW}{2})
∴ f₁ = 100 – (frac{1.59}{2}) = 49.47 Hz.

7. At what frequency will the current lead the voltage by 30° in a series circuit with R = 10Ω and C = 25 μF?
A. 11.4 kHz
B. 4.5 kHz
C. 1.1 kHz
D. 24.74 kHz

Answer: A
Clarification: From the impedance diagram, 10 – jX_C = Z∠-30°
∴ – X_C = 10 tan (-30°) = -5.77 Ω
∴ X_C = 5.77 Ω
Then, X_C = (frac{1}{2πfC}) or, f = (frac{1}{2πX_C C}) = 1103 HZ = 1.1 kHz.

8. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, what is the value of the capacitor?
A. 676 μF
B. 575 μF
C. 591 μF
D. 610 μF

Answer: A
Clarification: We know that f_o = (frac{1}{2πsqrt{LC}})
Or, 50 = (frac{1}{2πsqrt{5×10^{-3}×C}})
Or, C = (frac{1}{4π^2×50^2×15×10^{-3}}) ≈ 676 μF.

9. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the Q factor of the circuit is?
A. 82.63
B. 47.115
C. 27.38
D. 92.38

Answer: B
Clarification: We know that, Q = (frac{ωL}{R})
Or, Q = (frac{2π×50×15×10^{-3}}{100×10^{-3}}) = 47.115.

10. For the circuit given below, if the frequency of the source is 50 Hz, then a value of t_o which results in a transient free response is?

A. 0
B. 1.78 ms
C. 7.23 ms
D. 9.21 ms

Answer: B
Clarification: For the ideal case, transient response will die out with time constant.
T = (frac{L}{R} = frac{0.01}{5}) = 0.002 s = 2 ms
Practically, T will be less than 2 ms.

11. In the circuit given below, the magnitudes of V_L and V_C are twice that of V_K. Calculate the inductance of the coil, given that f = 50.50 Hz.

A. 6.41 mH
B. 5.30 mH
C. 3.18 mH
D. 2.31 mH

Answer: C
Clarification: V_L = V_C = 2 VR
∴ Q = (frac{V_L}{V_R}) = 2
But we know, Q = (frac{ωL}{R} = frac{1}{ωCR})
∴ 2 = (frac{2πf × L}{5})
Or, L = 3.18 mH.

12. A circuit with a resistor, inductor and capacitor in series is resonant with frequency f Hz. If all the component values are now doubled, then the new resonant frequency will be?
A. 2 f
B. f
C. (frac{f}{4})
D. (frac{f}{2})

Answer: D
Clarification: Resonance frequency = (frac{1}{2πsqrt{LC}} = frac{1}{2π2sqrt{LC}})
Hence, (frac{f}{f_o} = frac{frac{1}{2πsqrt{LC}}}{frac{1}{2π2sqrt{LC}}}) = 2
∴ f = 2f_o.

13. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is?
A. 25.52
B. 35.35
C. 45.45
D. 20.02

Answer: B
Clarification: Quality factor Q of the series RLC circuit is given by, Q = (frac{1}{R} sqrt{frac{L}{C}})
Q_new = (frac{1}{2R} sqrt{frac{2L}{C}} = frac{1}{2} × frac{1}{R} sqrt{frac{2L}{C}} = frac{1}{2} × sqrt{2} × Q) = 35.35.

14. In a series RLC circuit R = 10 kΩ, L = 0.5 H and C = (frac{1}{250}) μF. Calculate the frequency when the circuit is in a state of resonance.
A. 4.089 × 10⁴ Hz
B. (frac{11111.1}{π})Hz
C. 4.089π × 10⁴ Hz
D. (frac{1}{π}) × 10⁴ Hz

Answer: B
Clarification: Resonance frequency = (frac{1}{2πsqrt{LC}})
= (frac{1}{2πsqrt{frac{1}{250} × 10^{-6} × 0.5}} = frac{11111.1}{π})Hz.

15. In a series RLC circuit for lower frequency and for higher frequency, power factors are respectively ______________
A. Leading, Lagging
B. Lagging, Leading
C. Independent of Frequency
D. Same in both cases

Answer: A
Clarification: A Leading power factor means that the current in the circuit leads the applied voltage. This condition occurs in capacitive circuits. On the other hand, a lagging power factor indicates that the current lags the voltage and this condition happens in an inductive circuit.