Current through the battery immediately after key $K$ is closed and after a long time interval is:

\frac{E}{R_{1}}, \frac{E}{R_{1} + R_{3}}

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\frac{E}{R_{1} + R_{3}}, \frac{E}{R_{1} + \frac{R_{2} R_{3}}{R_{2} + R_{3}}}

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Zero

, \frac{E}{R_{1}}

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\frac{E}{R_{1} + \frac{R_{2} R_{3}}{R_{2} + R_{3}}}, \frac{E}{R_{1}}

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Solution

The correct option is A $\frac{E}{R_{1}}, \frac{E}{R_{1} + R_{3}}$
at $t = 0$ conductors will behaves as straight wires.
$\begin{matrix} ∴ i = \frac{E}{R_{1}} a t t = 0 t h e y w i l l b e h a v e a s b r e a k i = \frac{E}{R_{1} + R_{3}} ∴ O p t i o n (A) i s c o r r e c t . \end{matrix}$

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Similar questions

\frac{d i}{d t}

i = \frac{E}{R} \{1 - e^{- (R / L) t}\}

k

R_{1} = R_{2} = R_{3} = R

K

R = 6 Ω

C = 4 μ F

(K)

R = 6 Ω

C = 4 μ F

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