TOP RLC Circuit Questions and Answers with explanation – 2022

Answer.2. Large

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).

Answer.3. Infinity

Explanation:-

The capacitive reactance of RLC series circuit is

X_C = 1/2πfC

Now f = 0

X_C =1/2π0C

X_C = 1/0 = ∞

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

Answer.1. Zero

Explanation:-

The Inductive reactance of RLC series circuit is

X_L = 2πfL

Now f = 0

X_L = 2π0L

X_L = 0

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

Answer.4. Lead

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.

Answer.1. Lags

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°.

Answer.1. 37°

Explanation:-

The impedance triangle when X_C > X_L

tanφ = (X_C − X_L)/R

tanφ = 300/400

φ = 37°

Answer.3. 121∠25.0° µs

Explanation:-

Given

R = 75 kΩ

X_C = 60 kΩ

X_L = 25 kΩ

V_S = 10∠0°

Total impedance

Z = R + jX_C − jX_L

= 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 = V_S/Z

= 10∠0°/82.8∠−25°

= 121∠25° µA

Answer.2. 100∠0°

Explanation:-

For RLC resonance condition X_C = X_L

Given

R = 100Ω

X_L = 500 Ω

∴ X_C = X_L = 500 Ω

The impedance at resonance is

Z_r = R + jX_L − jX_C

= 100 + 500j − 500j

Z = 100∠0°

Answer.3. 328 kHz

Explanation:-

The resonant frequency of series RLC circuit is given by

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

f_r = 328 kHz

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

Explanation:-

Given

R = 1.5 kΩ

X_C = 1 kΩ

X_L = 1 kΩ

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

I = V_S/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

V_L = IX_L = 3.33 × 1 = 3.33 V

V_C = IX_C = 3.33 × 1 = 3.33 V

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