Example 6.3: Equivalent capacitance

Question: A $1\,\mu{\rm F}$ and a $2\,\mu{\rm F}$ capacitor are connected in parallel, and this pair of capacitors is then connected in series with a $4\,\mu{\rm F}$ capacitor, as shown in the diagram. What is the equivalent capacitance of the whole combination? What is the charge on the $4\,\mu{\rm F}$ capacitor if the whole combination is connected across the terminals of a $6$ V battery? Likewise, what are the charges on the $1\,\mu{\rm F}$ and $2\,\mu{\rm F}$ capacitors?
 
Answer: The equivalent capacitance of the $1\,\mu{\rm F}$ and $2\,\mu{\rm F}$ capacitors connected in parallel is $1+2= 3\,\mu{\rm F}$. When a $3\,\mu{\rm F}$ capacitor is combined in series with a $4\,{\mu}{\rm F}$ capacitor, the equivalent capacitance of the whole combination is given by

$$\frac{1}{C_{\rm eq}} = \frac{1}{(3\times 10^{-6})} + \frac{1... ...es 10^{-6})} = \frac{(7)}{(12\times 10^{-6})}\,\,{\rm F}^{-1},$$

and so

$$C_{\rm eq} = \frac{(12\times 10^{-6})}{(7)} = 1.71\,\mu{\rm F}.$$

The charge delivered by the $6$ V battery is

$$Q = C_{\rm eq} V = (1.71\times 10^{-6})\,(6) = 10.3\,\mu{\rm C}.$$

This is the charge on the $4\,\mu{\rm F}$ capacitor, since one of the terminals of the battery is connected directly to one of the plates of this capacitor.

The voltage drop across the $4\,\mu{\rm F}$ capacitor is

$$V_4 = \frac{Q}{C_4} = \frac{(10.3\times 10^{-6})}{(4\times 10^{-6})}=2.57\,{\rm V}.$$

Thus, the voltage drop across the $1\,\mu{\rm F}$ and $2\,\mu{\rm F}$ combination must be $V_{12}=6-2.57 = 3.43\,{\rm V}$. The charge stored on the $1\,\mu{\rm F}$ is given by

$$Q_1 = C_1\,V_{12} = (1\times 10^{-6})\,(3.43) = 3.42\,\mu{\rm C}.$$

Likewise, the charge stored on the $2\,\mu{\rm F}$ capacitor is

$$Q_2 = C_2\,V_{12} = (2\times 10^{-6})\,(3.43) = 6.84\,\mu{\rm C}.$$

Note that the total charge stored on the $1\,\mu{\rm F}$ and $2\,\mu{\rm F}$ combination is $Q_{12} = Q_1+Q_2 = 10.3\,\mu{\rm C}$, which is the same as the charge stored on the $4\,\mu{\rm F}$ capacitor. This makes sense because the $1\,\mu{\rm F}$ and $2\,\mu{\rm F}$ combination and the $4\,\mu{\rm F}$ capacitor are connected in series.