A 20 H inductor coil is connected to a 10 Ω 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

Question

A 20 H inductor coil is connected to a 10 Ω 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:

TestHub · Accepted Answer

Rate of dissipation of energy in resistor $= i^{2} R$
Rate of energy stored in inductor $= \frac{d}{d t} (\frac{1}{2} L i^{2}) = L i \frac{d i}{d t}$
$i^{2} R = L i \frac{d i}{d t}$
$\frac{d i}{d t} = i \frac{R}{L} \dots (i)$
In $L - R$ circuit:

$i = i_{0} (1 - e^{- \frac{t}{τ}}) (∴ τ = \frac{L}{R} = 2)$
$\frac{d i}{d t} = \frac{i _{0}}{τ} e^{- t / τ}$
From equation (i),

$\frac{i _{0}}{τ} e^{- t / τ} = i_{0} (1 - e^{- \frac{t}{τ}}) \frac{R}{L}$
$e^{- t / τ} = 1 - e^{- t / τ}$
$e^{- \frac{t}{τ}} = \frac{1}{2} \Rightarrow t = τ ln 2$
$= 2 ln 2$

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