JEE Main 2023 — Physics Electromagnetism

The formula to calculate the current flowing through an L-R circuit is given by

$$i=\frac{E}{R}[1-{e}^{\frac{-Rt}{L}}]...(1)$$i=RE​[1−eL−Rt​]...(1)

The energy stored within the inductor can be calculated as follows:

$$\begin{matrix}{E}_{\text{induction }} & = & \frac{1}{2}L{i}^{2} \\ & = & \frac{1}{2}L\frac{{E}^{2}}{{R}^{2}}{[1-{e}^{\frac{-Rt}{L}}]}^{2}...(2)\end{matrix}$$Einduction ​​==​21​Li221​LR2E2​[1−eL−Rt​]2...(2)​

The rate of change of energy stored within the inductor is given by

$$\begin{matrix}\frac{d}{dt}({E}_{\text{induction }}) & = & 2[1-{e}^{\frac{-Rt}{L}}]\cdot {e}^{\frac{-Rt}{L}}\cdot \frac{R}{L}\cdot \frac{{E}^{2}L}{2{R}^{2}} \\ & = & \frac{{E}^{2}}{R}[{e}^{\frac{-Rt}{L}}-{e}^{\frac{-2Rt}{L}}]...(3)\end{matrix}$$dtd​(Einduction ​)​==​2[1−eL−Rt​]⋅eL−Rt​⋅LR​⋅2R2E2L​RE2​[eL−Rt​−eL−2Rt​]...(3)​

For maximum rate,

$$\frac{{d}^{2}E}{d{t}^{2}}=0...(4)$$dt2d2E​=0...(4)

From equation (3) and (4), it can be written that

$$\begin{matrix}\frac{d}{dt}[\frac{{E}^{2}}{R}[{e}^{\frac{-Rt}{L}}-{e}^{\frac{-2Rt}{L}}]] & = & 0 \\ & \Rightarrow & {e}^{\frac{-Rt}{L}}=2{e}^{\frac{-2Rt}{L}} \\ & \Rightarrow & {e}^{\frac{Rt}{L}}=2...(5)\end{matrix}$$dtd​[RE2​[eL−Rt​−eL−2Rt​]]​=⇒⇒​0eL−Rt​=2eL−2Rt​eLRt​=2...(5)​

From equation (3) and (5), it can be written, under maximum energy stored condition for the circuit, that

$$\begin{matrix}\frac{d}{dt}({E}_{\text{induction }}) & = & \frac{{E}^{2}}{R}[{2}^{-1}-{2}^{-2}] \\ & = & \frac{{E}^{2}}{4R}\end{matrix}$$dtd​(Einduction ​)​==​RE2​[2−1−2−2]4RE2​​

Substitute the values of the resistance and compare with the given expression to obtain the values of the parameters.

$$\begin{matrix}a & = & 2 \\ b & = & 2R \\ & = & 50\end{matrix}$$ab​===​22R50​

Hence, the required ratio is given by

$$\frac{b}{a}=25$$ab​=25