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  • The time dependence of the current flowing through the inductance L of the circuit as shown in fig. after the switch <img alt="\mathrm S_\omega" src="/latex-image/?%5Cmathrm%2

The time dependence of the current flowing through the inductance L of the
circuit as shown in fig. after the switch \mathrm S_\omega  is shorted at the moment t = 0.

Option: 1

\mathrm { \frac{\varepsilon}{R}\left(1-e^{-t / \tau}\right)}


Option: 2

\mathrm {\frac{2 \varepsilon}{R}\left(1-e^{-t / \tau}\right)}


Option: 3

\mathrm { \frac{\varepsilon}{\mathrm{R}}\left(1-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)}


Option: 4

\mathrm { \frac{\varepsilon}{\mathrm{R}}\left(1-\mathrm{e}^{-\mathrm{t} / 2 \mathrm{\tau}}\right)}


Answers (1)

best_answer

\varepsilon-2 \mathrm{i R+I_{L} }\mathrm{R}=0 \quad \ldots \text { (i) }(\mathrm {\operatorname{loop} A B C F A) L }\frac{\mathrm{{di}_L}}{\mathrm{dt}}=\mathrm{IR}-\mathrm{i_L} \mathrm{R} \ldots(\mathrm {ii})

(loop FCDE)
substituting (iR) from (ii) in (I) we get

\begin{aligned} & \varepsilon-\mathrm{i}_{\mathrm{L}} \mathrm{R}-2 \mathrm{~L} \frac{\mathrm{di}_{\mathrm{L}}}{\mathrm{dt}}=0 \\ & \Rightarrow \int_0^{\mathrm{i}_{\mathrm{L}}} \frac{2 \mathrm{Ldi} \mathrm{i}_{\mathrm{L}}}{\varepsilon-\mathrm{i}_{\mathrm{L}} \mathrm{R}}=\int_0^{\mathrm{t}} \mathrm{dt} \\ & \Rightarrow\left(1-\mathrm{i}_{\mathrm{L}} \frac{\mathrm{R}}{\varepsilon}\right)=\mathrm{e}^{-\mathrm{t} / \tau} ; \tau=\frac{2 \mathrm{~L}}{\mathrm{R}} \\ & \Rightarrow \mathrm{i}_{\mathrm{L}}=\frac{\varepsilon}{\mathrm{R}}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \end{aligned}

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Gunjita

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