Question

The switch is closed at $t=0$. The time after which the rate of dissipation of energy in the resistor is equal to the rate at which energy is being stored in the inductor is:

• A. $\ln 2$
• B. $\frac{1}{2}\ln 2$
• C. $\frac{1}{4}\ln 2$
• D. $2\ln 2$

Answer 1

The correct option is A

$\ln 2$

Step 1: Given Data

The switch is closed at

$t=0$

$t$ is the instantaneous time.

Step 2: Formula Used

Powe dissipated= $i^{2}R$

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

$i$ is the current.

$R$ is the resistance.

$L$ is the inductance.

$E$ is the emf

$t$ is the instantaneous time.

$\tau =\frac{L}{R}$, $\tau$ is the time constant

Step 3: Calculation of time after which the rate of dissipation of energy,

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

Differentiating both sides with respect to $t$

$$\begin{gathered} \frac{di}{dt}=-\frac{E}{R}{e}^{\frac{-Rt}{L}}.(-\frac{R}{L}) \\ \frac{di}{dt}=\frac{E}{L}{e}^{{}^{\frac{-Rt}{L}}} \end{gathered}$$

Putting,$\tau =\frac{L}{R}$

$\frac{di}{dt}=\frac{E}{L}{e}^{\frac{-t}{\tau }}$

We know,

$i^{2}R=i\left(L\frac{di}{dt}\right)$

$$\begin{gathered} R\frac{E}{R}\left(1-e^{\frac{-t}{\tau }}\right)=L\frac{E}{L}{e}^{\frac{-t}{\tau }} \\ 1-e^{\frac{-t}{\tau }}=e^{\frac{-t}{\tau }} \end{gathered}$$

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

$\ln 2=\frac{t}{\tau }$

$t=\tau \ln 2=\ln 2$$seconds$

The time after which the rate of dissipation of energy in the resistor is equal to the rate at which energy is being stored in the inductor is $\ln 2$$seconds$

Hence, option A is the correct answer.