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  • In the given figure, an inductor and a resistor are connected in series with a batter of emf E volt. <img alt="\frac{E^{\mathrm{a}}}{2 \mathrm{~b}} \mathrm{~J} / \mathrm{s}" src="https://entra

In the given figure, an inductor and a resistor are connected in series with a batter of emf E volt. \frac{E^{\mathrm{a}}}{2 \mathrm{~b}} \mathrm{~J} / \mathrm{s} represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of \frac{b}{a} will be ________

Option: 1

25


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\begin{aligned} & \mathrm{U}=\frac{1}{2} \mathrm{LI}^2 \\ & \mathrm{I}=\mathrm{I}_0\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \end{aligned}

\text{Rate of energy}, P=\frac{\mathrm{dU}}{\mathrm{dt}}

\begin{aligned} & \mathrm{P}=\mathrm{LI} \frac{\mathrm{dI}}{\mathrm{dt}} \\ & \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{L}\left(\mathrm{I} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}+\left(\frac{\mathrm{dI}}{\mathrm{dt}}\right)^2\right) \end{aligned}

For maximum rate , \frac{dp}{dt}=0

\begin{aligned} & \mathrm{I} \frac{\mathrm{d}^2 I}{\mathrm{dt}^2}=-\left(\frac{\mathrm{dI}}{\mathrm{dt}}\right)^2 \ldots . \\ & \mathrm{I}=\mathrm{I}_0\left(1-\mathrm{e}^{\mathrm{t} / \mathrm{t}}\right) \\ & \frac{\mathrm{dI}}{\mathrm{dt}}=\frac{\mathrm{I}_0}{\tau} \mathrm{e}^{-\mathrm{t} / \mathrm{t}} \\ & \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=-\frac{\mathrm{I}_0}{\tau^2} \mathrm{e}^{-\mathrm{t} / \tau} \end{aligned}

By equation (1),

\\I_0\left(1-e^{-t / \tau}\right) \times \frac{I_0}{\tau^2} e^{-t / \tau}=\frac{-I_0^2}{\tau^2} e^{-2 t / \tau}\\ \\Let\; , e^{-t / \tau}=x

\begin{aligned} & x-x^2=x^2 \\ & x=\frac{1}{2} \end{aligned}

Maximum power, 

\begin{aligned} & \mathrm{P}=\mathrm{LI} \frac{\mathrm{dI}}{\mathrm{dt}} \\ & \mathrm{P}=\mathrm{LI}_0\left(1-\frac{1}{2}\right)\left(\frac{\mathrm{I}_0}{\tau} \times \frac{1}{2}\right) \end{aligned}

\begin{aligned} & \mathrm{P}=\frac{\mathrm{LI} \mathrm{I}_0^2}{4 \times \frac{\mathrm{L}}{\mathrm{R}}}=\frac{\mathrm{I}_0^2 \mathrm{R}}{4} \\ & \mathrm{P}=\frac{\mathrm{E}^2}{4 \mathrm{R}} \\ & \mathrm{a}=2,2 \mathrm{~b}=4 \mathrm{R} \\ & \mathrm{b}=2 \mathrm{R}=50 \\ & \frac{\mathrm{b}}{\mathrm{a}}=25 \end{aligned}

Posted by

Sumit Saini

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