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  • In the given circuit at t = 0, switch S is closed.  

In the given circuit at t = 0, switch S is closed.

 

Question

The current through 10\Omega resistor at any instant \mathrm{t} ; 0<\mathrm{t}<\infty will be
 

Option: 1

\mathrm{\frac{1}{6} \mathrm{e}^{\frac{1000}{3} \mathrm{t}}}


Option: 2

\mathrm{\frac{5}{6} \mathrm{e}^{-\frac{1000}{3} t}}


Option: 3

\mathrm{\frac{1}{6} \mathrm{e}^{\frac{1000}{3} t}}


Option: 4

\mathrm{\frac{6}{5} \mathrm{e}^{\frac{1000}{3} \mathrm{t}} \cdot}


Answers (1)

best_answer

From loop (i) applying Kirchoff’s law

12 \mathrm{I}-10 \mathrm{I}^{\prime}=10 \ldots \text { (i) }

From loop (ii) -\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}+10 \mathrm{I}-10 \mathrm{I}^{\prime}=0

\mathrm{I}-\mathrm{I}^{\prime}=\frac{\mathrm{L}}{10} \frac{\mathrm{dI}^{\prime}}{\mathrm{dt}}

Solving simultaneously (i) and (ii) we have 

\mathrm{I}^{\prime}=5-5 \mathrm{e}^{-\frac{5 \mathrm{t}}{3 \mathrm{~L}}} \ldots \text { (iii) }

\mathrm{\begin{aligned} & \text { and } I=5-\frac{25}{6} e^{-\frac{3 t}{3 L}} \ldots \text { (iv) } \\ & I-I^{\prime}=\frac{5}{6} e^{-\frac{1000 t}{3}} . \end{aligned}}

Posted by

Suraj Bhandari

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