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  • An inductor of inductance  and a resistor

An inductor of inductance \mathrm{L=\frac{\tau R}{2}} and a resistor of resistance R is connected to a battery of emf V as shown in the figure. The potential difference across the resistance at a time, \mathrm{t=\tau \ln 2} after the switch S is closed is ( \mathrm{\tau } is constant)

Option: 1

\mathrm{\frac{V}{4}}


Option: 2

\mathrm{\frac{3 V}{4}}


Option: 3

\mathrm{\frac{V}{2}}


Option: 4

\mathrm{\frac{2 \mathrm{~V}}{3}}


Answers (1)

\mathrm{\begin{aligned} & \frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{i}_0}{\mathrm{z}} \mathrm{e}^{-\mathrm{t} / \tau} \\ & \mathrm{V}_{\mathrm{R}}=\mathrm{V}-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=\mathrm{V}-\frac{\tau \mathrm{R}}{2} \times \frac{\mathrm{i}_0}{\tau} \mathrm{e}^{-\mathrm{t} / \tau} \\ & =\mathrm{V}-\frac{\mathrm{V}}{2} \mathrm{e}^{-\mathrm{t} / \tau}=\mathrm{V}-\frac{\mathrm{V}}{2} \mathrm{e}^{-\frac{1}{\mathrm{\tau}} \times \operatorname{\tau ln} 2} \\ & =\mathrm{V}-\frac{\mathrm{V}}{4}=\frac{3 \mathrm{~V}}{4} \end{aligned}}

Posted by

Kshitij

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