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

(B) $\frac{V}{R_{2}}$ at $t=0$ and $\frac{V\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$ at $t=\infty$

(A) $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=\infty$

(C) $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=\infty$

(D) $\frac{V\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$ at $t=0$ and $\frac{V}{R_{2}}$ at $t=\infty$

JEE Main · 2010 · 11 Apr 2010 · Offline · Electromagnetic Induction

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In the circuit shown below, the key K is closed at

t = 0

. The current through the battery is

(A) $\frac{VR _{1} R _{2}}{R _{1}^{2} + R _{2}^{2}}$ at $t = 0$ and $\frac{V}{R _{2}}$ at $t = \infty$
(B) $\frac{V}{R _{2}}$ at $t = 0$ and $\frac{V ( R _{1} + R _{2} )}{R _{1} R _{2}}$ at $t = \infty$ ✓ Correct
(C) $\frac{V}{R _{2}}$ at $t = 0$ and $\frac{VR _{1} R _{2}}{R _{1}^{2} + R _{2}^{2}}$ at $t = \infty$
(D) $\frac{V ( R _{1} + R _{2} )}{R _{1} R _{2}}$ at $t = 0$ and $\frac{V}{R _{2}}$ at $t = \infty$

Answer: (B)

Solution

The instant the key is closed the inductor opposes any sudden change of current and behaves like a break in its own branch, so current flows only through

R_{2}

and the battery delivers

i = V / R_{2}

. After a long time the current is steady, the inductor sustains no voltage and behaves like a plain wire, short-circuiting its branch. Now

R_{1}

and

R_{2}

sit in parallel with combined resistance

\frac{R _{1} R _{2}}{R _{1} + R _{2}}

, so the battery current rises to

i = \frac{V ( R _{1} + R _{2} )}{R _{1} R _{2}}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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JEE Main · 2010 · 11 Apr 2010 · Offline · Electromagnetic Induction

Solution

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