Question

Three capacitors housing capacitances of 6, 6 and 3uF are connected in series across a
12 V battery. Find (i) the charge on the 3uF capacitor, and (i) the stored energy of the
combination.

Answer 1

(i) Let's first find the equivalent capacitance of the combination of the capacitors connected in series:

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3},$$ $$C_{eq}=\dfrac{1}{\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}},$$ $$C_{eq}=\dfrac{1}{\dfrac{1}{6\ \mu F}+\dfrac{1}{6\ \mu F}+\dfrac{1}{3\ \mu F}}=1.5\ \mu F.$$

Since for a series connection of capacitors, the magnitude of charge on all the capacitors is the same, we can write:

$$q=C_{eq}V=1.5\cdot10^{-6}\ F\cdot12\ V=18\ \mu C.$$

(ii) The energy stored by the combination of capacitors can be found as follows:

$$E=\dfrac{1}{2}C_{eq}\Delta V^2,$$ $$E=\dfrac{1}{2}\cdot1.5\cdot10^{-6}\ F\cdot(12\ V)^2=1.08\cdot10^{-4}\ J.$$