Question

A series LCR circuit consists of a variable capacitor connected to an inductor of inductance 50 mH,resistor of resistance 100 Ω and an AC source of angular frequency 500 rad/s. The value of capacitance so that maximum current may be drawn into the circuit is:

• A. 60 μF
• B. 50 μF
• C. 100 μF
• D. 80 μF
• E. 25 μF

Answer 1

The Correct Option is D

Given parameters:

• Inductance \( L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H} \)
• Resistance \( R = 100 \, \Omega \)
• Angular frequency \( \omega = 500 \, \text{rad/s} \)

 

Condition for maximum current: \[ \text{Resonance occurs when } \omega = \frac{1}{\sqrt{LC}} \]

Capacitance calculation: \[ C = \frac{1}{\omega^2 L} = \frac{1}{(500)^2 \times 50 \times 10^{-3}} \] \[ C = \frac{1}{250000 \times 0.05} = \frac{1}{12500} = 80 \times 10^{-6} \, \text{F} \] \[ C = 80 \, \mu\text{F} \]

Thus, the correct option is (D): \( 80 \, \mu\text{F} \).

Answer 2

1. Condition for maximum current:

Maximum current is drawn in a series LCR circuit when the circuit is at resonance. This occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C).

2. Reactance formulas:

Inductive reactance:

\[X_L = \omega L\]

Capacitive reactance:

\[X_C = \frac{1}{\omega C}\]

where:

• ω is the angular frequency
• L is the inductance
• C is the capacitance

3. Resonance condition:

At resonance, \(X_L = X_C\):

\[\omega L = \frac{1}{\omega C}\]

4. Solve for C:

\[C = \frac{1}{\omega^2 L}\]

5. Substitute the given values:

\[\omega = 500 \, rad/s\]

\[L = 50 \, mH = 50 \times 10^{-3} \, H\]

\[C = \frac{1}{(500^2)(50 \times 10^{-3})} = \frac{1}{250000 \times 0.05} = \frac{1}{12500} = 80 \times 10^{-6} \, F = 80 \, \mu F\]