Resonance Questions and Answers

Q: A broadcasting station transmits on a wavelength of 3000m. A capacitor of 0.0005 µF is connected in parallel to a variable inductor to form a closed oscillating circuit. The approximate value of inductor that must be adjusted in order to tune into the particular station is ________.

The inductor must be set to about 5 mH.The station's frequency follows from its wavelength, and the tuned circuit resonates when \(f = 1/(2\pi\sqrt{LC})\).\[ \begin{aligned} f &= \frac{c}{\lambda} = \frac{3 \times 10^{8}}{3000} = 10^{5}\ \text{Hz} \\ L &= \frac{1}{4\pi^{2} f^{2} C} \end{aligned} \]\[ \begin{aligned} L &= \frac{1}{39.48 \times 10^{10} \times 5 \times 10^{-10}} \\ &= \frac{1}{197.4} = 5.07 \times 10^{-3}\ \text{H} \end{aligned} \]That is about 5 mH. The other values would tune the circuit to different wavelengths: a larger inductance gives a lower frequency and a longer wavelength.

Q: A parallel circuit has R = 1kΩ, C= 50µF and L = 10mH. The quality factor at resonance is ________.

The quality factor is about 70.70.For a parallel resonant circuit with resistance R in parallel with L and C, the quality factor is \(Q = R\sqrt{C/L}\).\[ \begin{aligned} Q &= 1000\sqrt{\frac{50 \times 10^{-6}}{10 \times 10^{-3}}} \\ &= 1000 \times \sqrt{5 \times 10^{-3}} \\ &= 1000 \times 0.0707 = 70.7 \end{aligned} \]Using the series formula \(\frac{1}{R}\sqrt{L/C}\) would give a tiny value, and 100, 90.86 and 78.76 do not follow from these component values.

Q: In parallel resonance, the phase angle between applied voltage and line current is :

At parallel resonance the phase angle is zero.In a parallel resonant (tank) circuit the reactive current in the inductive branch is exactly cancelled by the reactive current in the capacitive branch. The line current is then only the small in-phase component that supplies the losses, so it is in phase with the applied voltage, the power factor is unity and the circuit's impedance is a maximum and purely resistive.A phase angle of 90° belongs to a purely inductive or capacitive circuit, 45° to a circuit whose resistance equals its net reactance, and 180° does not arise in a passive circuit.

Q: For an RLC series circuit, current at series resonance is __________.

At series resonance the current is maximum at unity power factor.At the resonant frequency the inductive reactance equals the capacitive reactance, so they cancel. The impedance of the series circuit falls to its minimum value, the resistance R alone, and the current V/R is the largest the circuit can draw. Since the impedance is purely resistive, the current is in phase with the voltage and the power factor is unity.Below resonance the circuit is capacitive (leading power factor) and above it inductive (lagging), and in both cases the current is less than at resonance. 'Minimum at unity power factor' describes the line current of a parallel resonant circuit, not a series one.

Q: What is the phase angle of a series RLC circuit at resonance ?

The phase angle at resonance is zero.A series RLC circuit is resonant when the inductive and capacitive reactances are equal, \(X_{L} = X_{C}\), at \(f_{0} = 1/(2\pi\sqrt{LC})\). The two reactances then cancel, the impedance reduces to R alone, its minimum value, and the current is in phase with the applied voltage.\[ \phi = \tan^{-1}\frac{X_{L} - X_{C}}{R} = 0 \]An angle of 90° would mean a purely reactive circuit, and 45° is reached at the half-power frequencies on either side of resonance, where \(|X_{L} - X_{C}| = R\).

A broadcasting station transmits on a wavelength of 3000m. A capacitor of 0.0005 µF is connected in parallel to a variable inductor to form a closed oscillating circuit. The approximate value of inductor that must be adjusted in order to tune into the particular station is ________.

A.3 mH
B.5 mH
C.7 mH
D.9 mH

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

The inductor must be set to about 5 mH.

The station's frequency follows from its wavelength, and the tuned circuit resonates when \(f = 1/(2\pi\sqrt{LC})\).

\[ \begin{aligned} f &= \frac{c}{\lambda} = \frac{3 \times 10^{8}}{3000} = 10^{5}\ \text{Hz} \\ L &= \frac{1}{4\pi^{2} f^{2} C} \end{aligned} \]

\[ \begin{aligned} L &= \frac{1}{39.48 \times 10^{10} \times 5 \times 10^{-10}} \\ &= \frac{1}{197.4} = 5.07 \times 10^{-3}\ \text{H} \end{aligned} \]

That is about 5 mH. The other values would tune the circuit to different wavelengths: a larger inductance gives a lower frequency and a longer wavelength.

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In a circuit, a resistance (R), a pure inductance (L), and a capacitance (C) are connected in parallel across a sinusoidal voltage source of V volt. The circuit current will lead the applied voltage if

A.\(I_{C} \lt I_{L}\)
B.\(I_{C} = I_{L}\)
C.\(I_{C} \gt I_{L}\)
D.\(I_{L} = 1.5 \times I_{C}\)

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The circuit current leads the voltage if \(I_{C} \gt I_{L}\).

In a parallel RLC circuit each branch has the full supply voltage. The resistor current is in phase with it, the capacitor current leads it by 90° and the inductor current lags it by 90°. The capacitor and inductor currents are in direct opposition, so the net reactive current is \(I_{C} - I_{L}\).

If \(I_{C}\) is larger, the net reactive current leads and so does the total current. If \(I_{C} \lt I_{L}\), including the case \(I_{L} = 1.5 I_{C}\), the current lags; if they are equal the circuit is at resonance and the current is in phase with the voltage.

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A parallel circuit has R = 1kΩ, C= 50µF and L = 10mH. The quality factor at resonance is ________.

A.100
B.90.86
C.70.70
D.78.76

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The quality factor is about 70.70.

For a parallel resonant circuit with resistance R in parallel with L and C, the quality factor is \(Q = R\sqrt{C/L}\).

\[ \begin{aligned} Q &= 1000\sqrt{\frac{50 \times 10^{-6}}{10 \times 10^{-3}}} \\ &= 1000 \times \sqrt{5 \times 10^{-3}} \\ &= 1000 \times 0.0707 = 70.7 \end{aligned} \]

Using the series formula \(\frac{1}{R}\sqrt{L/C}\) would give a tiny value, and 100, 90.86 and 78.76 do not follow from these component values.

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In parallel resonance, the phase angle between applied voltage and line current is :

A.45°
B.Zero
C.90°
D.180°

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

At parallel resonance the phase angle is zero.

In a parallel resonant (tank) circuit the reactive current in the inductive branch is exactly cancelled by the reactive current in the capacitive branch. The line current is then only the small in-phase component that supplies the losses, so it is in phase with the applied voltage, the power factor is unity and the circuit's impedance is a maximum and purely resistive.

A phase angle of 90° belongs to a purely inductive or capacitive circuit, 45° to a circuit whose resistance equals its net reactance, and 180° does not arise in a passive circuit.

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For an RLC series circuit, current at series resonance is __________.

A.Maximum at lagging power factor
B.Maximum at unity power factor
C.Minimum at unity power factor
D.Maximum at leading power factor

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

At series resonance the current is maximum at unity power factor.

At the resonant frequency the inductive reactance equals the capacitive reactance, so they cancel. The impedance of the series circuit falls to its minimum value, the resistance R alone, and the current V/R is the largest the circuit can draw. Since the impedance is purely resistive, the current is in phase with the voltage and the power factor is unity.

Below resonance the circuit is capacitive (leading power factor) and above it inductive (lagging), and in both cases the current is less than at resonance. 'Minimum at unity power factor' describes the line current of a parallel resonant circuit, not a series one.

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What is the phase angle of a series RLC circuit at resonance ?

A.Zero
B.90°
C.45°
D.30°

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The phase angle at resonance is zero.

A series RLC circuit is resonant when the inductive and capacitive reactances are equal, \(X_{L} = X_{C}\), at \(f_{0} = 1/(2\pi\sqrt{LC})\). The two reactances then cancel, the impedance reduces to R alone, its minimum value, and the current is in phase with the applied voltage.

\[ \phi = \tan^{-1}\frac{X_{L} - X_{C}}{R} = 0 \]

An angle of 90° would mean a purely reactive circuit, and 45° is reached at the half-power frequencies on either side of resonance, where \(|X_{L} - X_{C}| = R\).

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