# TOP RLC Circuit Questions and Answers with explanation – 2022

Ques.21. It is advisable to have _____  power factor.

1. Low
2. Large
3. Zero
4. No

Explanation:-

The power factor of a circuit is a measure of its effectiveness in utilizing the apparent power. Thus 0.5 pf. of a circuit means that it will utilize only 50% of apparent power whereas OR p.f. would mean 80% utilization of apparent power. For this reason, we wish that the power factor of the circuit should be as large as possible (maximum value is 1).

22. At zero frequency the capacitive reactance is ______

1. Zero
2. Maximum
3. Infinity
4. Minimum

Explanation:-

The capacitive reactance of RLC series circuit is

XC = 1/2πfC

Now f = 0

XC =1/2π0C

XC = 1/0 = ∞

At zero frequency the capacitive reactance is infinity and the capacitor behaves like an open at 0 Hz frequency.

Ques.23. At zero frequency the Inductive reactance is ________

1. Zero
2. Maximum
3. Infinity
4. Minimum

Explanation:-

The Inductive reactance of RLC series circuit is

XL = 2πfL

Now f = 0

XL = 2π0L

XL = 0

At zero frequency the Inductive reactance is zero and the capacitor behaves like short at 0 Hz frequency.

Ques.24. At frequencies below resonance, XC > XL  the current _______ the source voltage

1. Lags
2. Become twice
3. Become half

Explanation:-

At a frequency below the resonant frequency, current leads the source voltage because the capacitive reactance is greater than the inductive reactance. The phase angle decreases as the frequency approach the resonant value and is 0° at resonance.

Ques.25. At frequencies above resonance, XL > XC and the current _______ the source voltage.

1. Lags
2. Become twice
3. Become half

Explanation:-

At frequencies above resonance, the current lags behind the source voltage, because the inductive reactance is greater than capacitive reactance. As the frequency goes higher, the phase angle approaches 90°.

Ques.26. What will be the phase difference of an RLC circuit illustrated by the impedance triangle?

1. 37°
2. 73°
3. 3.7°
4. 7.3°

Explanation:-

The impedance triangle when XC > XL

tanφ = (XC − XL)/R

tanφ = 300/400

φ = 37°

Ques.27. Find the current in polar form of series RLC circuit when the resistance is 75 kΩ, capacitive reactance is 60 kΩ , inductive reactance is 25 kΩ and the source voltage is 10∠0°.

1. 1.21∠25.0° µs
2. 0.121∠25.0° µs
3. 121∠25.0° µs
4. 121∠2.50° µs

Explanation:-

Given

R = 75 kΩ

XC = 60 kΩ

XL = 25 kΩ

VS = 10∠0°

Total impedance

Z = R + jXC − jXL

= 75 + 60j − 25j

Z = 75 − j35 kΩ

Converting it to polar form

$Z = \sqrt {{R^2} + {{({X_{tot}})}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_{tot}}}}{R}} \right)$

The negative sign for the angle is used to indicate that the circuit is capacitive.

$Z = \sqrt {{{(75)}^2} + {{(35)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{35}}{{75}}} \right)$

Z = 82.8∠−25° kΩ

Apply Ohm’s law to find the current

I = VS/Z

= 10∠0°/82.8∠−25°

= 121∠25° µA

Ques.28. In series RLC circuit if R = 100 ohm, XL = 500 Ω and XC = ? then find the impedance at resonance condition.

1. 50∠0°
2. 100∠0°
3. 10∠0°
4. 1∠0°

Explanation:-

For RLC resonance condition XC = XL

Given

R = 100Ω

XL = 500 Ω

∴ XC = XL = 500 Ω

The impedance at resonance is

Zr = R + jXL − jXC

= 100 + 500j − 500j

Z = 100∠0°

Ques.29. In series RLC circuit if R = 100 ohm, L = 5 mH and C = 47 pF, then find the resonant frequency.

1. 3.28 kHz
2. 32.8 kHz
3. 328 kHz
4. 0.328 kHz

Explanation:-

The resonant frequency of series RLC circuit is given by

$\begin{array}{l} {f_r} = \frac{1}{{2\pi \sqrt {LC} }}\\ \\ = \frac{1}{{2\pi \sqrt {5 \times 47} }} \end{array}$

fr = 328 kHz

Ques.30. Find I, VR, VL and VC at resonance for the circuit in Figure

1. I = 3.33 mA, VR = 7.33 V, VL = 3.33V,  VC = 3.33 V
2. I = 7.33 mA, VR = 7.33 V, VL = 3.33V,  VC = 3.33 V
3. I = 333 mA, VR = 7.33 V, VL = 3.33V,  VC = 3.33 V
4. I = 7.33 mA, VR = 7.33 V, VL = 7.33V,  VC = 7.33 V

Answer.1. I = 3.33 mA, VR = 7.33 V, VL = 3.33V,  VC = 3.33 V

Explanation:-

Given

R = 1.5 kΩ

XC = 1 kΩ

XL = 1 kΩ

At resonance, I is maximum and equal to VS/R

I = VS/R = 5/1.5

I = 3.33 mA

Apply Ohm’s law to obtain the voltage magnitudes.

VR = IR = 3.33 × 2.2 = 7.33 V

VL = IXL = 3.33 × 1 = 3.33 V

VC = IXC = 3.33 × 1 = 3.33 V

All of the source voltage is dropped across the resistor. Also, VL and VC are equal in magnitude but opposite in phase. This causes these voltages to cancel, making the total reactive voltage zero.

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