The equivalent capacitance of the combination between A and B in the given figure is 4 μF.

(i) Calculate the capacitance of the capacitor C.
(ii) Calculate charge on each capacitor if a 12V battery is connected across terminals A and B.
(iii) What will be the potential drop across each capacitor?

Solution

The capacitors given are arranged in a series combination.
i) Given, equivalent capacitance of two combination is 4 μF.
Therefore,

\frac{1}{C_{AB}} = \frac{1}{C_{A}} + \frac{1}{C_{B}}

\frac{1}{C_{AB}} = \frac{1}{C_{A}} + \frac{1}{C_{B}} \frac{1}{4} = \frac{1}{20} + \frac{1}{C_{B}} ∴ C_{B} = 5 μ F

ii) Charge on capacitor is given by C=qV
Capacitors in series have the same charge .
Equivalent capacitance of the system is 4 μF.

∴

charge on each capacitor, q =CV

4 \times 10^{- 6} \times 12 = 48 \times 10^{- 6} C

iii) Potential drop across the capacitor

V_{AB} = V_{a} + V_{B} = \frac{q}{C_{a} + \frac{q}{C_{b}}} = \frac{48 \times 10^{- 6}}{20} + \frac{48 \times 10^{- 6}}{5} = 2.4 \times 10^{- 6} + 9.6 \times 10^{- 6} = 12 \times 10^{- 6} V

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