Question

Suppose you have two capacitors 8.0 μF and 4.0 μF connected in series and parallel with points a and b whose potential

difference is 20 V. Find the equivalent capacitance and charge.

Answer 1

Parallel connection:

The equivalent capacitance

$$C=C_1+C_2\\ =\rm 8.0\: \mu F+\rm 4.0\: \mu F=\rm 12\: \mu F$$

The equivalent charge

$$q=CV\\ =12*10^{-6}*20=240*10^{-6}\: \rm C$$

Series connection:

The equivalent capacitance

$$C=C_1C_2/(C_1C_2)\\ =\rm 8.0\: \mu F*\rm 4.0\: \mu F/(\rm 8.0\: \mu F+\rm 4.0\: \mu F)=\rm 2.7\: \mu F$$

The equivalent charge

$$q=CV\\ =2.7*10^{-6}*20=54*10^{-6}\: \rm C$$