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  • An inductor of inductance L = 400 mH and resistors of resistance  and

An inductor of inductance L = 400 mH and resistors of resistance \mathrm{R}_1=2 \Omega and \mathrm{R}_2=2 \Omega are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is:

Option: 1

\mathrm{6 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}}


Option: 2

\mathrm{\frac{12}{\mathrm{t}} \mathrm{e}^{-3 \mathrm{t}} \mathrm{V}}


Option: 3

\mathrm{6\left(1-\mathrm{e}^{\frac{-1}{02}}\right) \mathrm{V}}


Option: 4

\mathrm{12 \mathrm{e}^{-5 t} \mathrm{~V}}


Answers (1)

best_answer

\mathrm{E}(\text { across } \mathrm{BC})=\mathrm{L} \frac{\mathrm{dI}_2}{\mathrm{dt}}+\mathrm{R}_2 \mathrm{I}_2          (using Kirchhoff’s law) --------(i)

\mathrm{I}_2=\mathrm{I}_0\left(\mathrm{I}-\mathrm{e}^{-\mathrm{u} / \mathrm{t}_0}\right)

\mathrm{\begin{aligned} & \text { At } \mathrm{t}=0, \mathrm{I}_0=\frac{\mathrm{E}}{\mathrm{R}_2}=\frac{12}{2}=6 \mathrm{~A} \\ & \tau_{\mathrm{L}}=\mathrm{t}_0=\frac{\mathrm{L}}{\mathrm{R}}=\frac{400 \times 10^{-3}}{2 \Omega}=0.2 \mathrm{~s} \\ & \therefore \mathrm{I}_2=6\left(1-\mathrm{e}^{-\mathrm{t} / 02}\right) \end{aligned}}

Potential drop area \mathrm{\mathrm{L}=\mathrm{E}-\mathrm{R}_2 \mathrm{I}_{\mathrm{b}}}

=12-2 \times 6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right)=12 \mathrm{e}^{-\mathrm{t} / 0.2}=12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}

Posted by

Sanket Gandhi

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