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In given figure at t = 0 key ‘s’ is closed then find charge on capacitor when current through cell is 50% of maximum current:

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$\frac{E C}{4}$

$\frac{E C}{2}$

$\frac{E C}{3}$

answer is D.

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Detailed Solution

$I_{max} = \frac{E}{R / 2}$

$50 % of I_{max} = \frac{E}{R}$

when current through cell is $\frac{E}{R}$ then no current will flow through capacitor at this time.

q = CE

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