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 $6 mF$ 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,

$$\begin{gathered} C_R=C_1\Rightarrow C_1=C_2+6 mF \\ \Rightarrow \frac{3C_2}{2}=C_2+6 mF\Rightarrow \frac{C_2}{2}=6 mF \\ \Rightarrow C_2=12 mF \\ \therefore C_1=18 mF \end{gathered}$$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards