This set of Network Theory Questions & Answers for Exams focuses 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

View Answer

Explanation: Q = \(\frac{ω}{ω1 – ω2} = \frac{f}{f2-f1}\)

Here, f = 1.5 × 10

^{6}Hz

f1 = (1.5 × 10

^{6}– 5 × 10

^{3})

f2 = (1.5 × 10

^{6}+ 5 × 10

^{3})

So, f2 –f1 = 10 × 10

^{3}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

View Answer

Explanation: R = \(\frac{f2-f1}{2πf^2 L}\)

Here, f = 1.5 × 10

^{6}Hz

f1 = (1.5 × 10

^{6}– 5 × 10

^{3})

f2 = (1.5 × 10

^{6}+ 5 × 10

^{3})

So, f2 – f1 = 10 × 10

^{3}Hz

∴ R = 4.7 Ω.

3. Consider a circuit consisting of two capacitors C_{1} and C_{2}. Let R be the resistance and L be the inductance which are connected in series. Let Q_{1} and Q_{2} 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 = \(\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

View Answer

Explanation: ωL = \(\frac{1}{ωC}\) and Q

_{1}= \(\frac{ωL}{R} = \frac{1}{ωC_1 R}\)

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 C_{1} and C_{2}. Let R be the resistance and L be the inductance which are connected in series. Let Q_{1} and Q_{2} 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 = \(\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

View Answer

Explanation: \(\frac{1}{R_P} = \frac{ωC_1}{Q_2} – \frac{1}{RQ_1^2}\), 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) \(\frac{πV_a}{V_b}\)

View Answer

Explanation: 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

View Answer

Explanation: f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(10 × 10^{-6})(200 × 10^{-12})}}\)

= \(\frac{1}{6.28\sqrt{2 × 10^{-15}}}\)

= \(\frac{1}{888 × 10^{-9}}\) = 1.13 MHz

Inductive Reactance, X

_{L}= 2πfL = (6.28) (1.13 × 10

^{6})(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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X

_{L}= 2πfL = (6.28) (15.92 × 10

^{3})(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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X

_{L}= 2πfL = (6.28) (15.92 × 10

^{3})(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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X

_{L}= 2πfL = (6.28) (15.92 × 10

^{3})(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, f

_{2}= f + \(\frac{∆F}{2}\) = 16.716 kHz and f

_{1}= 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

View Answer

Explanation: f = \(\frac{1}{2π\sqrt{LC}} = \frac{1}{2π\sqrt{0.025×1.8×10^{-12}}}\)

= \(\frac{1}{2π} × \frac{10^6}{\sqrt{0.45}}\)

= \(\frac{1}{2π} × \frac{10^6}{0.67}\)

= 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

View Answer

Explanation: 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 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

View Answer

Explanation: 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

View Answer

Explanation: ω

_{2}L = 10 ω

_{1}LR will remain constant

∴ Q

_{2}= \(\frac{10 ω_1 L}{R}\) = 10 Q

_{1}

That is Q is increased 10 times.

Now, I

_{1}= \(\frac{V}{ω_1 L} \)

For a high Q coil, ωL >> R,

I

_{2}= \(\frac{V}{10 ω_1 L} = \frac{I_1}{10}\)

∴ P

_{2}= 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°

View Answer

Explanation: 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°.

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