An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω , is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ..... mH. $(Take π^{2} = 10)$

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Detailed Solution

C = 62.5 nF, R = 50Ω , f = 2 KHZ

i_max condition is z_min resonance condition

$\begin{array}{l} f = \frac{1}{2 π \sqrt{LC}} \\ L = \frac{1}{4 π^{2} f^{2} C} = \frac{1}{4 \times 10 \times 4 \times 10^{6} \times 62.5 \times 10^{- 9}} \\ = 100 mH \end{array}$

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