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

Network Theory Questions & Answers for Exams 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

Answer: A
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{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

Answer: C
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(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

Answer: D
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(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

Answer: B
Clarification: f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(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

Answer: A
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 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

Answer: B
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 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

Answer: D
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{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

Answer: C
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(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

Answer: B
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(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

Answer: A
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(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

Answer: C
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(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_CI_T = (2.168 kΩ)(7.521 mA. = 16.306 V
∴ Voltage across the inductor, V_L = X_LI_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

Answer: D
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 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

Answer: D
Clarification: 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

Answer: A
Clarification: Dynamic resistance of the tank circuit, Z_DY = L/(R_LC.
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

Answer: C
Clarification: θ = 0° since X_L and X_C are cancelling, which means at resonance the circuit is purely resistive.