A 20, Henry inductor coil is connected to a 10, ohm resistance in series as shown in figure. time at

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7.Alternating Current

hard

A $20\, Henry$ inductor coil is connected to a $10\, ohm$ resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is

A

$\frac{2}{{\ln \,2}}$

B

$ln\,2$

C

$\frac{1}{2}\,\ln\,2$

D

$2\,ln\,2$

(JEE MAIN-2019)

Solution

$L I D I=I^{2} R$

$L \times \frac{E}{10}\left(-e^{-t / 2}\right) \times \frac{-1}{2}=\frac{E}{10}\left(1-e^{-t / 2}\right) \times 10$

$e^{-1 / 2}=1-e^{-1 / 2} \quad ; \quad t=2 \ell n 2$

Standard 12

Physics

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