Question

An inductor of inductance \(2\mu H\) is connected in series with a resistance, a variable capacitor and an AC source of frequency \(7\ kHz\). The value of capacitance for which maximum current is drawn into the circuit is \(\frac 1x\) F, where the value of \(𝑥 \) is ______. (Take \(\pi=\frac {22}{7}\) )

Answer 1

Correct Answer: 3872

For maximum current, the circuit must be in resonance, i.e., the inductive reactance equals the capacitive reactance: \[ \frac{1}{2 \pi f C} = 2 \pi f L. \] Rearrange to find \(C\): \[ C = \frac{1}{4 \pi^2 f^2 L}. \] Substitute values: \[ f = 7 \, \text{kHz} = 7 \times 10^3 \, \text{Hz}, \quad L = 2 \times 10^{-6} \, \text{H}. \] \[ C = \frac{1}{4 \pi^2 (7 \times 10^3)^2 (2 \times 10^{-6})}. \] Simplify: \[ C = \frac{1}{4 \times (3.14)^2 \times 49 \times 10^6 \times 2 \times 10^{-6}} = \frac{1}{3872}. \] Thus, \(x = 3872\).