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
AThe potential difference across the plates of $A$ is $4V$ and across the plates of $B$ is $6V$
BThe potential difference across the plates of $A$ is $6V$ and across the plates of $B$ is $4V$
CThe ratio of electrical energies stored in $A$ and $B$ is $2 : 3$
DThe ratio of charges on $A$ and $B$ is $3 : 2$
Medium
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BThe potential difference across the plates of $A$ is $6V$ and across the plates of $B$ is $4V$
b (b) In series combination of capacitors, voltage distributes on them, in the reverse ratio of their capacitance i.e. $\frac{{{V_A}}}{{{V_B}}} = \frac{3}{2} ......(i)$
Also $V_A$ + $V_B$$ = 10 .......(ii)$
On solving $ (i)$ and $(ii)$ $V_A$ $= 6V$, $V_B$ $= 4V$
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