In the circuit shown below- the key K is closed at t -0- The current through the battery isA- V R 1- R 2- R 1 R 2 at t -0 and V - R 2 at t -B- V R 1- R 2- R 12- R 22 at t -0 and V - R 2 at t -C- V-R2 at t -0 and V R 1- R 2- R 1 R 2 at t -D- V - R 1 at t -0 and V R 1- R 2- R 1 R 2 at t -

The correct option is C VR_2 at t=0 and V(R_1+R_2)R_1 R_2 at t=∞After closing key K, at t=0, the coil offers infinite resistance. ⇒ Effective resistance is R_2 ...

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Byju's Answer

Standard XII

Physics

Impedance Triangle

In the circui...

Question

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

\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}

t = 0

and

\frac{V}{R_{2}}

t = \infty

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\frac{V (R_{1} + R_{2})}{\sqrt{R_{1}^{2} + R_{2}^{2}}}

t = 0

and

\frac{V}{R_{2}}

t = \infty

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\frac{V}{R_{2}}

t = 0

and

\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}

t = \infty

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\frac{V}{R_{1}}

t = 0

and

\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}

t = \infty

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Solution

The correct option is C $\frac{V}{R_{2}}$ at $t = 0$ and $\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$ at $t = \infty$
After closing key $K$ , at $t = 0$ , the coil offers infinite resistance.
$\Rightarrow$ Effective resistance is $R_{2}$
$\Rightarrow$ Current, $I = \frac{V}{R_{2}}$

At $t = \infty$ , the coil offers zero resistance.
$\Rightarrow$ Effective resistance is $(\frac{R_{1} R_{2}}{R_{1} + R_{2}})$
$\Rightarrow$ Current, $I^{^{'}} = \frac{V}{\frac{R_{1} R_{2}}{(R_{1} + R_{2})}} = \frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$

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

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