# Network Theory Questions and Answers – Problems of Series Resonance Involving Quality Factor

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

Explanation: 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

Explanation: 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 = $$\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}$$

Explanation: ωL = $$\frac{1}{ωC}$$ and Q1 = $$\frac{ωL}{R} = \frac{1}{ωC_1 R}$$
XS = $$\frac{C_1 – C_2}{ωC_1 C_2}$$, RS = $$\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}$$
QX = $$\frac{X_S}{R_S} = \frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}$$.

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 = $$\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}$$

Explanation: $$\frac{1}{R_P} = \frac{ωC_1}{Q_2} – \frac{1}{RQ_1^2}$$, XP = $$\frac{1}{ω(C_2-C_1)}$$
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 Va and Vb 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) $$\frac{πV_a}{V_b}$$

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 = $$\frac{V_a}{V_b/1.11} = \frac{1.11V_a}{V_b}$$.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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

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, XL = 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

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{10^{-10}}}$$
= $$\frac{1}{6.28 × 10^{-5}}$$ = 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

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{10^{-10}}}$$
= $$\frac{1}{6.28 × 10^{-5}}$$ = 15.9 kHz.
Inductive Reactance, XL = 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

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{10^{-10}}}$$
= $$\frac{1}{6.28 × 10^{-5}}$$ = 15.9 kHz.
Inductive Reactance, XL = 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

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{10^{-10}}}$$
= $$\frac{1}{6.28 × 10^{-5}}$$ = 15.9 kHz.
Inductive Reactance, XL = 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, RF = 52 Ω and RL = 9 kΩ. The Resonant frequency f is __________

a) 75.1 kHz
b) 751 kHz
c) 575 kHz
d) 57.5 kHz

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, RF = 52 Ω and RL = 9 kΩ. The Bandwidth is ___________

a) 331 kHz
b) 575 kHz
c) 331 Hz
d) 575 Hz

Explanation: R = $$\frac{R_f×R_L}{R_f+R_L}$$ ≈ 52 Ω
∴ Qfactor = $$\frac{ω_r L}{R} = \frac{2π×751×25×10^{-3}}{52}$$
= $$\frac{π×37580}{52}$$ = 2270
Now, BW = $$\frac{f_r}{Q} = \frac{751 × 10^3}{2270}$$
∴ BW = 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

Explanation: At resonance, XL = XC
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

Explanation: 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°

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. Ie = 0. IM = 300 A.
Secondary winding current IS = 7 A
Reflected secondary winding current = n IS = 5600 A
∴ tan θ = $$\frac{I_M}{nI_S}$$. So, θ = 3.1°.

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]