What is the value of inductance $L$ for which the current is maximum in a series $LCR$ circuit with $C=10 \, \mu F $ and $ \omega=1000 \, S^{-1}$ ?

Question

In an LC circuit, the inductance $ L $ is 2 H and the capacitance $ C $ is 4 Î¼F. What is the frequency of oscillation of the circuit?

Answer 1

To find the value of inductance $L$ for which the current is maximum in a series $LCR$ circuit, we need to consider the condition of resonance. In a series $LCR$ circuit, resonance occurs when the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$.

The equations for inductive reactance and capacitive reactance are given by:

X_L = \omega L

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

At resonance:

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

Solving for L gives:

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

Substituting the given values, \omega = 1000 \, S^{-1} and C = 10 \, \mu F = 10 \times 10^{-6} \, F, we get:

L = \frac{1}{(1000)^2 \times 10 \times 10^{-6}}

L = \frac{1}{10^6 \times 10 \times 10^{-6}}

L = \frac{1}{10}

L = 0.1 \, H

Therefore, L = 100 \, mH.

This matches with option: 100 mH

Thus, the value of inductance for which the current is maximum is 100 mH.

Answer 2