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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

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

a) X_{L} = 654.289 Ω; X_{C} = 458.216 Ω

b) X_{L} = 985.457 Ω; X_{C} = 875.245 Ω

c) X_{L} = 785.394 Ω; X_{C} = 785.417 Ω

d) X_{L} = 125.354 Ω; X_{C} = 657.215 Ω

View Answer

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, X

_{L}= 2πfL = (6.28) (1.25×10

^{6})(100×10

^{-6})

= 785.394 Ω

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}\)

= \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω.

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

a) I_{C} = 10.892 mA; I_{L} = 12.732 mA

b) I_{C} = 12.732 mA; I_{L} = 10.892 mA

c) I_{C} = 10.892 mA; I_{L} = 10.892 mA

d) I_{C} = 12.732 mA; I_{L} = 12.732 mA

View Answer

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, X

_{L}= 2πfL = (6.28) (1.25×10

^{6})(100×10

^{-6})

= 785.394 Ω

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}\)

= \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω.

I

_{C}= \(\frac{V_A}{X_C} \)

= \(\frac{10}{785.417} \) = 12.732 mA

I

_{L}= \(\frac{V_A}{X_L} \)

= \(\frac{10}{785.394} \) = 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

View Answer

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, X

_{L}= 2πfL = (6.28) (1.25×10

^{6})(100×10

^{-6})

= 785.394 Ω

Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05

∴ Z

_{EQ}= QX

_{L}= (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

View Answer

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, X

_{L}= 2πfL = (6.28) (1.25×10

^{6})(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) ∆f_{1} = 2.389 MHz; ∆f_{2} = 441.124 MHz

b) ∆f_{1} = 1.256 MHz; ∆f_{2} = 1.244 MHz

c) ∆f_{1} = 5.658 MHz; ∆f_{2} = 6.282 MHz

d) ∆f_{1} = 3.656 MHz; ∆f_{2} = 8.596 MHz

View Answer

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, X

_{L}= 2πfL = (6.28) (1.25×10

^{6})(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

∴ ∆f

_{1}= f + \(\frac{∆F}{2}\) = 1.25 MHz + 6.25 kHz = 1.256 MHz

∴ ∆f

_{1}= 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

View Answer

Explanation: Resonant Frequency, F

_{R}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz.

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

a) X_{L} = 5.826 kΩ; X_{C} = 5.826 kΩ

b) X_{L} = 2.168 kΩ; X_{C} = 2.168 kΩ

c) X_{L} = 6.282 kΩ; X_{C} = 6.282 kΩ

d) X_{L} = 10.682 kΩ; X_{C} = 10.682 kΩ

View Answer

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, X

_{L}= 2πfL = (6.28) (73.142 × 10

^{3})(4.7 × 10

^{-6})

= 2.168 kΩ

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\)

= \(\frac{1}{4.613×10^{-4}}\) = 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

View Answer

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, X

_{L}= 2πfL = (6.28) (73.142 × 10

^{3})(4.7 × 10

^{-6})

= 2.168 kΩ

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

We see that, X

_{C}= X

_{L}are equal, along with being 180° out of phase.

Hence the net reactance is zero and the total impedance equal to the resistor.

∴ Z

_{EQ}= 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

View Answer

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, X

_{L}= 2πfL = (6.28) (73.142 × 10

^{3})(4.7 × 10

^{-6})

= 2.168 kΩ

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

Z

_{EQ}= R = 47 Ω

I

_{T}= \(\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA.

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

a) V_{C} = 16.306 V; V_{L} = 16.306 V

b) V_{C} = 11.268 V; V_{L} = 11.268 V

c) V_{C} = 16.306 V; V_{L} = 16.306 V

d) V_{C} = 14.441 V; V_{L} = 14.441 V

View Answer

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, X

_{L}= 2πfL = (6.28) (73.142 × 10

^{3})(4.7 × 10

^{-6})

= 2.168 kΩ

Capacitive Reactance, X

_{C}= \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

Z

_{EQ}= R = 47 Ω

I

_{T}= \(\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA

∴ Voltage across the capacitor, V

_{C}= X

_{C}I

_{T}= (2.168 kΩ)(7.521 mA) = 16.306 V

∴ Voltage across the inductor, V

_{L}= X

_{L}I

_{T}= (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

View Answer

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, X

_{L}= 2πfL = (6.28) (73.142 × 10

^{3})(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

View Answer

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

View Answer

Explanation: Dynamic resistance of the tank circuit, Z

_{DY}= L/(R

_{L}C)

But given that R

_{L}= 0

So, Z

_{DY}= L/(0X

_{C}) = ∞

Therefore current through circuit, I = \(\frac{V}{∞}\) = 0

∴ V

_{D}= 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

View Answer

Explanation: θ = 0° since X

_{L}and X

_{C}are cancelling, which means at resonance the circuit is purely resistive.

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