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. $\frac{2}{\ln 2}$
• B. $\ln 2$
• C. $\frac{1}{2}\ln 2$
• D. $2\ln 2$

Answer 1

The correct option is D $2\ln 2$

As both resistor and inductor are connected in series, so same current will be flowing through them
Energy dissipated by resistor = $V_{R}i$
and Energy stored in inductor = $V_{l}i$
where $V_i$ and $V_l$ are voltages across resistor and inductor.
As the energies are equal and current flowing is same, thus
$V_R=V_l=\frac{E}{2}$
We know that voltage across inductor at any time is an exponential function
$⟹V_0\left(e^{-\frac{t}{\tau }}\right)=\frac{E}{2}$
Here, $V_0=E$ when ($t=0$)
$⟹E\left(e^{-\frac{t}{\tau }}\right)=\frac{E}{2}$
$⟹\frac{t}{\tau }=\ln 2$
$⟹t=\tau \times \ln 2$
we know that $\tau =\frac{L}{R}$
So, $\tau =2$ sec
$\therefore$ $t=2\ln 2$ sec