# two capacitors A and B are connected in series with a 100V battery and it is observed that the potential difference across them are 40V and 60V respectively. A capacitor of capacitance 6mF is now connected in parallel with B and as a result the potential differnce across B falls to 50V. Find the capacitance of A and B.

Dear Student,

Please find below the solution to the asked query:

Given that the total potential difference is *100 V*. When the two capacitors are connected in series, then the potential drops across the two resistors are *40 V & 60 V* respectively. Therefore, the charge through the two capacitors will be same in series.

$\frac{{C}_{1}}{{C}_{2}}=\frac{{V}_{2}}{{V}_{1}}=\frac{3}{2}$

Now, if a capacitance of $6mF$ is connected in parallel to the second capacitor, then the potential drop across A & B becomes *50 V & 50 V* respectively. Here, the the charge flowing in *A* and in other combination will be same because those are in series. Now, the potential drops will also be same. Therefore,

${C}_{R}={C}_{1}\Rightarrow {C}_{1}={C}_{2}+6mF\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3{C}_{2}}{2}={C}_{2}+6mF\Rightarrow \frac{{C}_{2}}{2}=6mF\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{2}=12mF\phantom{\rule{0ex}{0ex}}\therefore {C}_{1}=18mF$

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