A network of uncharged capacitors and resistances is shown Current through the battery immediately after key $K$ 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_{2}}, \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 $\frac{E}{R_{1}}, \frac{E}{R_{1} + R_{3}}$
Let the current in the circuit is $I$ .

Just after the key is closed the capacitor can be replaced as a short circuit.

The current is given as,

$I = \frac{E}{R_{1}}$

After long time capacitor does not allow current to pass through it, so capacitor can be replaced as open circuit.

The current can be written as,

$- E + I R_{1} + I R_{3} = 0$

$I = \frac{E}{R_{1} + R_{3}}$

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Similar questions

k

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

\frac{R_{1}}{r_{1}} + \frac{R_{2}}{r_{2}} + \frac{R_{3}}{r_{3}} =

r_{1} (r_{2} + r_{3}) / a = r_{2} (r_{3} + r_{1}) / b = r_{3} (r_{1} + r_{2}) / c

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

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