A 20 H inductor coil is connected to a 10 $$\Omega$$ 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:

For growth of current in an RL circuit,

$$I=\frac{E}{R}\left(1-e^{-Rt/L}\right)$$

Given,

$$L=20\ H,\qquad R=10\ \Omega$$

So,

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

Rate of Joule heating across resistance:

$$P_R=I^2R$$

Rate of storage of magnetic energy in inductor:

$$P_L=L I\frac{dI}{dt}$$

At the required instant,

$$LI\frac{dI}{dt}=I^2R$$

Now,

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

Substituting,

$$20\times \frac{E}{10}(1-e^{-t/2}) \times \frac{E}{20}e^{-t/2} = \left[\frac{E}{10}(1-e^{-t/2})\right]^2\times10$$

Simplifying,

$$e^{-t/2}=1-e^{-t/2}$$

$$2e^{-t/2}=1$$

$$e^{-t/2}=\frac{1}{2}$$

Taking logarithm,

$$-\frac{t}{2}=\ln\left(\frac{1}{2}\right)$$

$$\boxed{t=2\ln2\ \text{s}}$$