Question

Two capacitor connected in series with 100v battery and potential difference across them is 60 $50v. If a capacitor of 0.006mfF is connected in parallel and potential across 2nd capacitor becomes 50v.Find capacitance of both capacitor.

Answer 1

First, according to Kirchhoff’s voltage law, potential difference across two capacitors must be 60 and 40 V. For the first case (capacitors C_1 and C_2 are in series) and for the second case (C_2 across 0.006 μF with C_2 in series) we can write respectively:
Q=60C_1=40C_2 ⇨ C_1/C_2 =2/3,
q=50〖(C〗_1+0.006·〖10〗^(-3))=50C_2 ⇨ C_1=C_2-0.006·〖10〗^(-3),
Thus
C_2=1.8·〖10〗^(-5) F,
C_1=1.2·〖10〗^(-5) F.