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When a coil is connected to a D. C source of emf 12 V, a current of 4 amp flows in it. If same coil is connected to 12 V, 50 Hz AC source, the current is 2.4 A. The self inductance of the coil is (in henry )

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$\frac{1}{20 π}$

$\frac{1}{10 π}$

$\frac{1}{25 π}$

$\frac{1}{5 π}$

answer is C.

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Detailed Solution

$\begin{array}{l} R = \frac{12}{4} = 3 \\ 2.4 = \frac{12}{Z} = \frac{12}{\sqrt{3^{2} + X_{L}^{2}}} \\ X_{L} = 4 \\ L = \frac{1}{25 π} \end{array}$

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