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 :

Question

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) (2/ℓ n2) (B) ℓ n 2  (C) 2 ℓ n 2  (D) (1/2) ℓ n 2

Tardigrade Admin · Accepted Answer

LIdI = I2R L×(E/10) (-e-t/2) ×(-1/2) = (E/10) (1-e-t/2) ×10 e-t/2 = 1- e-t/2 t = 2ℓ n 2

Q. 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 :

$\frac{2}{ℓ n 2}$

$ℓ n 2$

$2 ℓ n 2$

$\frac{1}{2} ℓ n 2$

$L I d I = I^{2} R$
$L \times \frac{E}{10} (- e^{- t /2}) \times \frac{- 1}{2} = \frac{E}{10} (1 - e^{- t /2}) \times 10$
$e^{- t /2} = 1 - e^{- t /2}$
$t = 2 ℓ n 2$

Q. 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 :

ℓn22​

ℓn2

2ℓn2

21​ℓn2