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 \(\pi^2 =10\))
To maximize the amplitude of current in an LCR circuit, the condition of resonance must be achieved. At resonance, the inductive reactance \(X_L = \omega L\) equals the capacitive reactance \(X_C = \frac{1}{\omega C}\), where \(\omega = 2\pi f\) and \(f\) is the frequency. Given\( C = 625\ nF = 625 \times 10^{-9}\ F\) and\( f = 20\ kHz = 20 \times 10^3\ Hz\). Substitute to find \(\omega\): \(\omega = 2\pi \times 20 \times 10^3\). Apply \(\pi^2 = 10\),\(\omega^2 = (2\pi \times 20 \times 10^3)^2 = 4\pi^2 \times 20^2 \times 10^6 = 4 \times 10 \times 400 \times 10^6 = 16 \times 10^9\). Since \(X_L = X_C\),\(\omega L = \frac{1}{\omega C}\) gives \(\omega^2 LC = 1\). Substitute and solve for \(L\):\) \(16 \times 10^9 \times L \times 625 \times 10^{-9} = 1\) leads to \(16 \times 625 \times L = 1\) or \(L = \frac{1}{16 \times 625}\). Simplify to get \(L = \frac{10^{-3}}{10^4} = 10^{-7}\ H = 0.1\ mH\). Verify \(L\) value within range (100,100): computed \(L = 0.1\ mH\) is within (0.1, 0.1). Thus, the required inductance for maximum current is 0.1 mH.
