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Question:

A capacitor C1 = 1.0 μF is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C1 is disconnected from the battery and connected to a circuit consisting of two uncharged capacitors C2 = 3.0 μF and C3 = 6.0 μF through a switch (2) as shown in the figure. The sum of final charges on C2 and C3 is:

40 μC

36 μC

20 μC

54 μC

Solution:

Q1 = 60μC
Equivalent capacitance of C2 and C3
C' = C2C3/(C2+C3) = 3×6/(3+6)μF = 2μF
Common potential difference to C1 and C' combined, V' = C1V/(C1+C') = 60/(1+2) = 20 volts
Charge of C2, C3 system = C'V' = 2 × 20 = 40μC