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

This set of Network Theory Questions & Answers for Exams focuses on “Problems of Parallel Resonance Involving Quality Factor”.

1. For the parallel resonant circuit shown below, the value of the resonant frequency is _________

a) 1.25 MHz
b) 2.5 MHz
c) 5 MHz
d) 1.5 MHz

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.

2. For the parallel resonant circuit given below, the value of the inductive and capacitive reactance is _________

a) XL = 654.289 Ω; XC = 458.216 Ω
b) XL = 985.457 Ω; XC = 875.245 Ω
c) XL = 785.394 Ω; XC = 785.417 Ω
d) XL = 125.354 Ω; XC = 657.215 Ω

Explanation: Resonant Frequency, f = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.
Inductive Reactance, XL = 2πfL = (6.28) (1.25×106)(100×10-6)
= 785.394 Ω
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}$$
= $$\frac{1}{1.273×10^{-3}}$$ = 785.417 Ω.

3. For the parallel resonant circuit given below, the current through the capacitor and inductor are _________

a) IC = 10.892 mA; IL = 12.732 mA
b) IC = 12.732 mA; IL = 10.892 mA
c) IC = 10.892 mA; IL = 10.892 mA
d) IC = 12.732 mA; IL = 12.732 mA

Explanation: Resonant Frequency, f = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.
Inductive Reactance, XL = 2πfL = (6.28) (1.25×106)(100×10-6)
= 785.394 Ω
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}$$
= $$\frac{1}{1.273×10^{-3}}$$ = 785.417 Ω.
IC = $$\frac{V_A}{X_C}$$
= $$\frac{10}{785.417}$$ = 12.732 mA
IL = $$\frac{V_A}{X_L}$$
= $$\frac{10}{785.394}$$ = 12.732 mA.

4. For the parallel resonant circuit given below, the value of the equivalent impedance of the circuit is ________

a) 56.48 kΩ
b) 78.58 kΩ
c) 89.12 kΩ
d) 26.35 kΩ

Explanation: f = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.
Inductive Reactance, XL = 2πfL = (6.28) (1.25×106)(100×10-6)
= 785.394 Ω
Q = $$\frac{X_L}{R} = \frac{785.394}{7.85}$$ = 100.05
∴ ZEQ = QXL = (100.05)(785.394) = 78.58 kΩ.

5. For the parallel resonant circuit, the bandwidth of the circuit is ____________

a) ±6.25 kHz
b) ±8.56 kHz
c) ±10.35 kHz
d) ±6.37 kHz

Explanation: Resonant Frequency, f = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.
Inductive Reactance, XL = 2πfL = (6.28) (1.25×106)(100×10-6)
= 785.394 Ω
Q = $$\frac{X_L}{R} = \frac{785.394}{7.85}$$ = 100.05
∴ ∆F = $$\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}$$ = 12.5 kHz
Hence, bandwidth = $$\frac{∆F}{2}$$ = 6.25 kHz.

6. For the parallel resonant circuit given below, the cut-off frequencies are ____________

a) ∆f1 = 2.389 MHz; ∆f2 = 441.124 MHz
b) ∆f1 = 1.256 MHz; ∆f2 = 1.244 MHz
c) ∆f1 = 5.658 MHz; ∆f2 = 6.282 MHz
d) ∆f1 = 3.656 MHz; ∆f2 = 8.596 MHz

Explanation: Resonant Frequency, f = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}$$
= $$\frac{1}{6.28\sqrt{1.621×10^{-14}}}$$
= $$\frac{1}{799×10^{-9}}$$ = 1.25 MHz.
Inductive Reactance, XL = 2πfL = (6.28) (1.25×106)(100×10-6)
= 785.394 Ω
Q = $$\frac{X_L}{R} = \frac{785.394}{7.85}$$ = 100.05
∴ ∆F = $$\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}$$ = 12.5 kHz
Hence, bandwidth = $$\frac{∆F}{2}$$ = 6.25 kHz
∴ ∆f1 = f + $$\frac{∆F}{2}$$ = 1.25 MHz + 6.25 kHz = 1.256 MHz
∴ ∆f1 = f – $$\frac{∆F}{2}$$ = 1.25 MHz – 6.25 kHz = 1.244 MHz.

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

a) 10.262 kHz
b) 44.631 kHz
c) 50.288 kHz
d) 73.412 kHz

Explanation: Resonant Frequency, FR = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz.

8. For the series resonant circuit given below, the value of the inductive and capacitive reactance is _________

a) XL = 5.826 kΩ; XC = 5.826 kΩ
b) XL = 2.168 kΩ; XC = 2.168 kΩ
c) XL = 6.282 kΩ; XC = 6.282 kΩ
d) XL = 10.682 kΩ; XC = 10.682 kΩ

Explanation: Resonant Frequency, $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz
Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 103)(4.7 × 10-6)
= 2.168 kΩ
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}$$
= $$\frac{1}{4.613×10^{-4}}$$ = 2.168 kΩ.

9. For the series resonant circuit given below, the value of the equivalent impedance of the circuit is ________

a) 55 Ω
b) 47 Ω
c) 64 Ω
d) 10 Ω

Explanation: Resonant Frequency, $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz
Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 103)(4.7 × 10-6)
= 2.168 kΩ
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}$$
= $$\frac{1}{4.613×10^{-4}}$$ = 2.168 kΩ
We see that, XC = XL are equal, along with being 180° out of phase.
Hence the net reactance is zero and the total impedance equal to the resistor.
∴ ZEQ = R = 47 Ω.

10. For the series resonant circuit given below, the value of the total current flowing through the circuit is ____________

a) 7.521 mA
b) 6.327 mA
c) 2.168 mA
d) 9.136 mA

Explanation: Resonant Frequency, $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz
Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 103)(4.7 × 10-6)
= 2.168 kΩ
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}$$
= $$\frac{1}{4.613×10^{-4}}$$ = 2.168 kΩ
ZEQ = R = 47 Ω
IT = $$\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}$$ = 7.521 mA.

11. For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________

a) VC = 16.306 V; VL = 16.306 V
b) VC = 11.268 V; VL = 11.268 V
c) VC = 16.306 V; VL = 16.306 V
d) VC = 14.441 V; VL = 14.441 V

Explanation: Resonant Frequency, $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz
Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 103)(4.7 × 10-6)
= 2.168 kΩ
Capacitive Reactance, XC = $$\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}$$
= $$\frac{1}{4.613×10^{-4}}$$ = 2.168 kΩ
ZEQ = R = 47 Ω
IT = $$\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}$$ = 7.521 mA
∴ Voltage across the capacitor, VC = XCIT = (2.168 kΩ)(7.521 mA) = 16.306 V
∴ Voltage across the inductor, VL = XLIT = (2.168 kΩ)(7.521 mA) = 16.306 V.

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

a) 35.156
b) 56.118
c) 50.294
d) 46.128

Explanation: Resonant Frequency, $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}$$
= $$\frac{1}{6.28\sqrt{4.7×10^{-12}}}$$
= $$\frac{1}{1.362×10^{-5}}$$ = 73.412 kHz
Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 103)(4.7 × 10-6)
= 2.168 kΩ
Quality factor Q = $$\frac{X_L}{R} = \frac{2.168 kΩ}{47}$$ = 46.128.

13. For a parallel RLC circuit, the incorrect statement among the following is _____________
a) The bandwidth of the circuit decreases if R is increased
b) The bandwidth of the circuit remains same if L is increased
c) At resonance, input impedance is a real quantity
d) At resonance, the magnitude of input impedance attains its minimum value

Explanation: BW = 1/RC
It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.
At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.
In parallel RLC circuit, the admittance is minimum, at resonance. Hence the magnitude of input impedance attains its maximum value at resonance.

14. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________
a) 0
b) $$\frac{V}{2}$$
c) $$\frac{V}{3}$$
d) V

Explanation: Dynamic resistance of the tank circuit, ZDY = L/(RLC)
But given that RL = 0
So, ZDY = L/(0XC) = ∞
Therefore current through circuit, I = $$\frac{V}{∞}$$ = 0
∴ VD = 0.

15. For the circuit given below, the nature of the circuit is ____________

a) Inductive
b) Capacitive
c) Resistive
d) Both inductive as well as capacitive

Explanation: θ = 0° since XL and XC are cancelling, which means at resonance the circuit is purely resistive.

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