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.


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,

CR=C1C1=C2+6 mF3C22=C2+6 mFC22=6 mFC2=12 mFC1=18 mF

Hope this information will clear your doubts about the topic.

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