Question

An LCR series circuit of capacitance \(625\ nF\) and resistance of \(50 \ \Omega\) is connected to an AC source of frequency \(20\ kHz\) For maximum value of amplitude of current in circuit, the value of inductance is ___ \(mH\). (Take \(\pi^2 =10\))

Answer 1

Correct Answer: 100

\(f=\frac {1}{2π\sqrt {LC​​}}\)

\(2000\ Hz=\frac {1}{2\pi \sqrt {L×62.5×10^{−9​}​}}\)

\(L=\frac {1}{4\pi^2×2000^2×62.5×10^{−9}}\)
\(L=0.1\ H\)
\(L=100\ mH\)

So, the answer is \(100\).

Answer 2

1. The condition for maximum current amplitude in an LCR circuit is resonance, where: \[ f = \frac{1}{2 \pi \sqrt{LC}}. \]
2. Rearrange to solve for \(L\): \[ L = \frac{1}{(2 \pi f)^2 C}. \]
3. Substituting the values: - \(f = 2000 \, \text{Hz}\), - \(C = 62.5 \, \text{nF} = 62.5 \times 10^{-9} \, \text{F}\),
\[ L = \frac{1}{(2 \pi \cdot 2000)^2 \cdot 62.5 \times 10^{-9}}. \]
4. Simplify: \[ L = \frac{1}{4 \pi^2 \cdot 4 \cdot 10^6 \cdot 62.5 \times 10^{-9}}. \]
Using \(\pi^2 = 10\): \[ L = \frac{1}{4 \cdot 10 \cdot 10^6 \cdot 62.5 \times 10^{-9}}. \]
\[ L = \frac{1}{2.5 \times 10^{-2}} = 100 \, \text{mH}. \]
Thus, the inductance is 100 mH. At resonance, the inductive reactance cancels out the capacitive reactance. Use \(f = \frac{1}{2 \pi \sqrt{LC}}\) to find the inductance.