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Two capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then

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The potential difference across the plates of A is 6V and across the plates of B is 4V

The ratio of charges on A and B is 3 : 2

The ratio of electrical energies stored in A and B is 2 : 3

The potential difference across the plates of A is 4V and across the plates of B is 6V

answer is B.

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Detailed Solution

In series combination of capacitors, voltage distributes on them in the reverse ratio of their capacitance and equal charge.

i.e., $\frac{V_{A}}{V_{B}} = \frac{3}{2}$ …(i)

Also V_A + V_B = 10…..(ii)

on solving (i) and (ii), V_A = 6V, V_B = 4V.

$\frac{U_{A}}{U_{B}} = \frac{27}{8}$

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