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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

In the circuit shown below, the key K is closed at $ t = 0$. The current through the battery is

 

$\begin {array} {1 1} (A)\;\large\frac{VR_1R_2}{\sqrt{R_1^2+R_2^2}}at \: t=0\: and \: \large\frac{V}{R_2}at\: t= \infty & \quad (B)\;\large\frac{V}{R_2} at \: t=0 \: and \: \large\frac{V(R_1+R_2)}{R_1R_2} at \: t= \infty\\ (C)\;\large\frac{V}{R_2} at \: t=0 \: and \: \large\frac{VR_1R_2}{\sqrt{R_1^2+R_2^2}} at \: t=\infty & \quad (D)\;\large\frac{V(R_1+R_2)}{R_1R_2} at \: t=0 \: and \: \large\frac{V}{R_2}at\: t= \infty \end {array}$

 

1 Answer

At $t = 0$, the inductor behaves like an open circuit and at $t = \infty $, it behaves like short circuit.
Ans : (B)
answered Feb 13, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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