250+ TOP MCQs on Problems of Series Resonance Involving Quality Factor and Answers

Network Theory Questions & Answers for Exams on “Problems of Series Resonance Involving Quality Factor”.

1. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?
A. 50
B. 100
C. 150
D. 200

Answer: C
Clarification: Q = (frac{ω}{ω1 – ω2} = frac{f}{f2-f1})
Here, f = 1.5 × 10⁶ Hz
f1 = (1.5 × 10⁶ – 5 × 10³)
f2 = (1.5 × 10⁶ + 5 × 10³)
So, f2 –f1 = 10 × 10³ Hz
∴ Q = (frac{1.5 × 10^6}{10 × 10^3}) = 150.

2. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Effective resistance of the circuit is?
A. 2 Ω
B. 3 Ω
C. 5.5 Ω
D. 4.7 Ω

Answer: D
Clarification: R = (frac{f2-f1}{2πf^2 L})
Here, f = 1.5 × 10⁶ Hz
f1 = (1.5 × 10⁶ – 5 × 10³)
f2 = (1.5 × 10⁶ + 5 × 10³)
So, f2 – f1 = 10 × 10³ Hz
∴ R = 4.7 Ω.

3. Consider a circuit consisting of two capacitors C₁ and C₂. Let R be the resistance and L be the inductance which are connected in series. Let Q₁ and Q₂ be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?
A. Q = (frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2})
B. Q = (frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2})
C. Q = (frac{(C_1 – C_2) Q_1 Q_2}{Q_2 C_2-Q_1 C_1})
D. Q = (frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2})

Answer: A
Clarification: ωL = (frac{1}{ωC}) and Q₁ = (frac{ωL}{R} = frac{1}{ωC_1 R})
X_S = (frac{C_1 – C_2}{ωC_1 C_2}), R_S = (frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2})
Q_X = (frac{X_S}{R_S} = frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}).

4. Consider a circuit consisting of two capacitors C₁ and C₂. Let R be the resistance and L be the inductance which are connected in series. Let Q₁ and Q₂ be the quality factor for the two capacitors. While measuring the Q value by the Parallel Connection method, the value of the Q factor is?
A. Q = (frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2})
B. Q = (frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2})
C. Q = (frac{(C_1 – C_2) Q_1 Q_2}{Q_2 C_2-Q_1 C_1})
D. Q = (frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2})

Answer: B
Clarification: (frac{1}{R_P} = frac{ωC_1}{Q_2} – frac{1}{RQ_1^2}), X_P = (frac{1}{ω(C_2-C_1)})
Q = (frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}).

5. A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type AC voltmeter connected in parallel. If the meter readings are V_a and V_b respectively. Then the form factor may be estimated as?
A. (frac{V_a}{V_b} )
B. (frac{1.11V_a}{V_b})
C. (frac{sqrt{2} V_a}{V_b})
D. (frac{πV_a}{V_b})

Answer: B
Clarification: Form factor of the wave = (frac{RMS ,value}{Mean ,value})
Moving iron instrument will show rms value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage that is, with form factor of 1.11.
∴ Mean value of the applied voltage = (frac{V_b}{1.11})
∴ Form factor = (frac{V_a}{V_b/1.11} = frac{1.11V_a}{V_b}).

6. For the resonant circuit given below, the value of the quality factor of the circuit is __________

A. 5.6
B. 7.1
C. 8.912
D. 12.6

Answer: B
Clarification: f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(10 × 10^{-6})(200 × 10^{-12})}})
= (frac{1}{6.28sqrt{2 × 10^{-15}}})
= (frac{1}{888 × 10^{-9}}) = 1.13 MHz
Inductive Reactance, X_L = 2πfL = (6.28) (1.13 × 10⁶)(10 × 10^-6)
= 70.96 Ω
∴ Q = (frac{X_L}{R} = frac{70.96}{10}) = 7.1.

7. For the series resonant circuit shown below, the value of the resonant frequency is _________

A. 36.84 kHz
B. 40.19 kHz
C. 25.28 kHz
D. 15.9 kHz

Answer: D
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}})
= (frac{1}{6.28sqrt{10^{-10}}})
= (frac{1}{6.28 × 10^{-5}}) = 15.9 kHz.

8. For the series resonant circuit given below, the value of the quality factor is ___________

A. 15
B. 36
C. 25
D. 10

Answer: D
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}})
= (frac{1}{6.28sqrt{10^{-10}}})
= (frac{1}{6.28 × 10^{-5}}) = 15.9 kHz.
Inductive Reactance, X_L = 2πfL = (6.28) (15.92 × 10³)(5 × 10^-6)
= 0.5 kΩ
∴ Quality factor Q = (frac{X_L}{R} = frac{0.5 kΩ}{50}) = 10.

9. For the series resonant circuit given below, the bandwidth is ____________

A. ±351 Hz
B. ±796 Hz
C. ±531 Hz
D. ±225 Hz

Answer: B
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}})
= (frac{1}{6.28sqrt{10^{-10}}})
= (frac{1}{6.28 × 10^{-5}}) = 15.9 kHz.
Inductive Reactance, X_L = 2πfL = (6.28) (15.92 × 10³)(5 × 10^-6)
= 0.5 kΩ
∴ Quality factor Q = (frac{X_L}{R} = frac{0.5 kΩ}{50}) = 10
∴ ∆F = (frac{f}{Q} = frac{15.9 kHz}{10}) = 1.592 kHz
∴ Bandwidth = (frac{∆F}{2}) = ±796 Hz.

10. For the series circuit given below, the value of the cut-off frequencies are ____________

A. 78.235 kHz; 16.215 kHz
B. 13.135 kHz; 81.531 kHz
C. 16.716 kHz; 15.124 kHz
D. 50.561 kHz; 18.686 kHz

Answer: C
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}})
= (frac{1}{6.28sqrt{10^{-10}}})
= (frac{1}{6.28 × 10^{-5}}) = 15.9 kHz.
Inductive Reactance, X_L = 2πfL = (6.28) (15.92 × 10³)(5 × 10^-6)
= 0.5 kΩ
∴ Quality factor Q = (frac{X_L}{R} = frac{0.5 kΩ}{50}) = 10
∴ ∆F = (frac{f}{Q} = frac{15.9 kHz}{10}) = 1.592 kHz
∴ Bandwidth = (frac{∆F}{2}) = ±796 Hz.
Therefore, f₂ = f + (frac{∆F}{2}) = 16.716 kHz and f₁ = f – (frac{∆F}{2}) = 15.124 kHz.

11. In a series resonance type BPF, C = 1.8 pf, L = 25 mH, R_F = 52 Ω and R_L = 9 kΩ. The Resonant frequency f is __________

A. 75.1 kHz
B. 751 kHz
C. 575 kHz
D. 57.5 kHz

Answer: B
Clarification: f = (frac{1}{2πsqrt{LC}} = frac{1}{2πsqrt{0.025×1.8×10^{-12}}})
= (frac{1}{2π} × frac{10^6}{sqrt{0.45}})
= (frac{1}{2π} × frac{10^6}{0.67})
= 751000
∴ f = 751 kHz.

12. In a series resonance type BPF, C = 1.8 pf, L = 25 mH, R_F = 52 Ω and R_L = 9 kΩ. The Bandwidth is ___________

A. 331 kHz
B. 575 kHz
C. 331 Hz
D. 575 Hz

Answer: C
Clarification: R = (frac{R_f×R_L}{R_f+R_L}) ≈ 52 Ω
∴ Q_factor = (frac{ω_r L}{R} = frac{2π×751×25×10^{-3}}{52} )
= (frac{π×37580}{52}) = 2270
Now, B_W = (frac{f_r}{Q} = frac{751 × 10^3}{2270})
∴ B_W = 331 Hz.

13. A 50 μF capacitor, when connected in series with a coil having a resistance of 40Ω, resonates at 1000 Hz. The inductance of the coil for the resonant circuit is ____________
A. 2.5 mH
B. 1.2 mH
C. 0.5 mH
D. 0.1 mH

Answer: C
Clarification: At resonance, X_L = X_C
Or, 2πfL = (frac{1}{2πfC} )
∴ L = (frac{1}{4π^2 f^2 C} )
= (frac{1}{4π^2 × (1000)^2 × 50 × 10^{-6}} )
= (frac{1}{39.46×50} )
= 0.0005
So, L = 0.5 mH.

14. A coil (which can be modelled as a series RL circuit) has been designed for a high Quality Factor (Q) performance. The voltage is rated at and a specified frequency. If the frequency of operation is increased 10 times and the coil is operated at the same rated voltage. The new value of Q factor and the active power P will be?
A. P is doubled and Q is halved
B. P is halved and Q is doubled
C. P remains constant and Q is doubled
D. P decreases 100 times and Q is increased 10 times

Answer: D
Clarification: ω₂ L = 10 ω₁ LR will remain constant
∴ Q₂ = (frac{10 ω_1 L}{R}) = 10 Q₁
That is Q is increased 10 times.
Now, I₁ = (frac{V}{ω_1 L} )
For a high Q coil, ωL >> R,
I₂ = (frac{V}{10 ω_1 L} = frac{I_1}{10})
∴ P₂ = R ((frac{I_1}{10})^2 = frac{P_1}{100})
Thus, P decreases 100 times and Q is increased 10 times.

15. A 50 Hz, bar primary CT has a secondary with 800 turns. The secondary supplies 7 A current into a purely resistive burden of 2 Ω. The magnetizing ampere-turns are 300. The phase angle is?
A. 3.1°
B. 85.4°
C. 94.6°
D. 175.4°

Answer: A
Clarification: Secondary burden is purely resistive and the resistance of burden is equal to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The loss component of no-load current is to be neglected i.e. I_e = 0. I_M = 300 A.
Secondary winding current I_S = 7 A
Reflected secondary winding current = n I_S = 5600 A
∴ tan θ = (frac{I_M}{nI_S} ). So, θ = 3.1°.