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

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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

Resistance offered by the coils is R = V/I = 12/4 = 3 Ohm

Impedance offered by the coil is $Z = \sqrt{R^{2} + {X_{L}}^{2}} = 12 / 2.4 = 5 Ω$

$\sqrt{3^{2} + {X_{L}}^{2}} = 5 \Rightarrow X_{L} = ω L = 4 Ω$

$\Rightarrow L = \frac{4}{2 πf} = \frac{4}{100 π} = \frac{1}{25 π} H$

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