A series LCR circuit has L = 0.01 H, R = 10 $Ω$ and C = 1 $μ$ F and it is connected to ac voltage of amplitude (V_m) 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately :

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196 mA

312 mA

466 mA

238 mA

answer is C.

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Detailed Solution

Resonant frequency, $ω_{0} = \frac{1}{\sqrt{LC}} = 10^{4} rad / \sec$

$\begin{array}{l} ω^{'} = 0.4 \times 10^{4} = 4000 rad / \sec \\ X_{C}^{'} = 250 Ω \\ X_{L}^{'} = 40 Ω \\ i_{0} = \frac{V_{0}}{\sqrt{R^{2} + {(X_{C}^{'} - X_{L}^{'})}^{2}}} = 238 mA \end{array}$

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