A capacitor C1 = 1.0 μ F is charged up to a voltage V = 60 V by… | JEE Main 2018 (15 Apr, Shift 2)

Q32JEE Main 2018 · 15 APR · SHIFT 2Difficult

Charge Sharing & Loss of Energy

A capacitor C1 = 1.0 $\mu$ F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2=3.0\ \mu\text{F}$ and C3 = 6.0 $\mu$ F through switch (2), as shown in the figure. The sum of final charges on C2 and C3 is:

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Answer: (a)

$C_2$ and $C_3$ in series: $C_{23} = \frac{3 \times 6}{9} = 2\ \mu$ F.
$Q = C_1V = 1 \times 60 = 60\ \mu$ C is shared between $C_1$ and $C_{23}$ : $V' = \frac{60}{1 + 2} = 20$ V.
Charge on the series branch $= 2 \times 20 = 40\ \mu$ C (each of $C_2$ and $C_3$ carries this same $40\ \mu$ C).
So the answer is (a) $40\ \mu$ C.

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Charge Sharing & Loss of Energy question from JEE Main 2018 (15 Apr, Shift 2)

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