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
Medium
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(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$$ = 10 .......(ii)$
On solving $ (i)$ and $(ii)$ $V_A$ $= 6V$, $V_B$ $= 4V$
Also $V_A$ + $V_B$$ = 10 .......(ii)$
On solving $ (i)$ and $(ii)$ $V_A$ $= 6V$, $V_B$ $= 4V$
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