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Energy of a Capacitor

In the circui...

Question

In the circuit shown in figure $E_{1} = 2 E_{2} = 20 V, R_{1} = R_{2} = 10 k Ω$ and $C = 1 μ F$ . The sum of current through $R_{1}, R_{2}$ and C when S has been kept connected to A for a long time.

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Solution

When S is closed for long time, the capacitor C is fully charged and no current will pass through it. Thus, $I_{C} = 0$
The current drawn from battery $E_{1}$ is $I = \frac{E_{1}}{R_{1} + R_{2}} = \frac{20}{(10 + 10) 10^{3}} = 1 m A$
As the resistances are in series so the current through them will be equal to $I$
Thus, $I_{R_{1}} = I_{R_{2}} = 1 m A$

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