CAPACITORS IN SERIES AND PARALLEL

Capacitors are manufactured with certain standard capacitances and working voltages (Fig. 1). However, these standard values may not be the ones you actually need in a particular application. You can obtain the values you need by combining capacitors; many combinations are possible, but the simplest combinations are a series connection and a parallel connection.

Fig.1:An assortment of commercially available capacitors.

Capacitors in Series

Figure 2a is a schematic diagram of a series connection. Two capacitors are connected in series (one after the other) by conducting wires between points and Both capacitors are initially uncharged. When a constant positive potential difference is applied between points and the capacitors become charged; the figure shows that the charge on all conducting plates has the same magnitude. To see why, note first that the top plate of acquires a positive charge The electric field of this positive charge pulls negative charge up to the bottom plate of until all of the field lines that begin on the top plate end on the bottom plate. This requires that the bottom plate have charge These negative charges had to come from the top plate of which becomes positively charged with charge This positive charge then pulls negative charge from the connection at point onto the bottom plate of The total charge on the lower plate of and the upper plate of together must always be zero because these plates aren’t connected to anything except each other. Thus in a series connection the magnitude of charge on all plates is the same.

Fig.2: A series connection of two capacitors.

Referring to Fig. 2a, we can write the potential differences between points and and and and as

$V_{ab}=V_1=\frac{Q}{C_1}$

$V_{cb}=V_2=\frac{Q}{C_2}$

$V_{ab}=V=V_1+V_2=\frac{Q}{C_1}+\frac{Q}{C_2}=Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)$

and so

$\frac{V}{Q}=\frac{1}{C_1}+\frac{1}{C_2}$

Following a common convention, we use the symbols $V_1$, $V_2$ and $V$ to denote the potential differences $V_{ac}$ (across the first capacitor), $V_{cb}$(across the second capacitor), and $V_{ab}$ (across the entire combination of capacitors), respectively. 

The equivalent capacitance $C_{eq}$ of the series combination is defined as the capacitance of a single capacitor for which the charge Q is the same as for the combination, when the potential difference V is the same. In other words, the combination can be replaced by an equivalent capacitor of capacitance $C_{eq}$. For such a capacitor, shown in Fig. 2b,

$C_{eq}=\frac{Q}{V}$ or $\frac{V}{Q}==\frac{1}{C_{eq}}$

Combining Eqs. (24.3) and (24.4), we find

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

We can extend this analysis to any number of capacitors in series. We find the following result for the reciprocal of the equivalent capacitance:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...$                  (1)

The reciprocal of the equivalent capacitance of a series combination equals the sum of the reciprocals of the individual capacitances. In a series connection the equivalent capacitance is always less than any individual capacitance.

Capacitors in Parallel

Fig.3:A parallel connection of two capacitors.

The arrangement shown in Fig. 3a is called a parallel connection. Two capacitors are connected in parallel between points a and b. In this case the upper plates of the two capacitors are connected by conducting wires to form an equipotential surface, and the lower plates form another. Hence in a parallel connection the potential difference for all individual capacitors is the same and is equal to $V_{ab}=V$. The charges $Q_1$ and $Q_2$ are not necessarily equal, however, since charges can reach each capacitor independently from the source (such as a battery) of the voltage $V_{ab}$. The charges are

$Q_1=C_1V$ and $Q_2=C_2V$

The total charge $Q$ of the combination, and thus the total charge on the equivalent capacitor, is 

$Q=Q_1+Q_2=(C_1+C_2)V$

so $\frac{Q}{V}=C_1+C_2$

The parallel combination is equivalent to a single capacitor with the same total charge $Q=Q_1+Q_2$ and potential difference V as the combination (Fig. 3b). The equivalent capacitance of the combination, is the same as the capacitance $Q/V$ of this single equivalent capacitor. So from Eq. (3),

$C_{eq}=C_1+C_2$

In the same way we can show that for any number of capacitors in parallel,

$C_{eq}=C_1+C_2+C_3+...$                                         (2)

The equivalent capacitance of a parallel combination equals the sum of the individual capacitances. In a parallel connection the equivalent capacitance is always greater than any individual capacitance.

Example 1

Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 4a. All capacitances are in microfarads.

Solution

Using Equations 1 and 2, we reduce the combination step by step as indicated in the figure. The $1.0 \ \mu F$ and $1.0 \ \mu F$ capacitors are in parallel and combine according to the expression $C_{eq}=C_1+C_2=4 \ \mu F$. The $2.0 \mu F$ and $6.0 \mu F$ capacitors also are in parallel and have an equivalent capacitance of $8.0 \mu F$. Thus, the upper branch in Figure 4b consists of two $4.0 \mu F$ capacitors in series, which combine as follows:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C_{eq}}=\frac{1}{4.0 \mu F}+\frac{1}{4.0 \mu F}=\frac{1}{2.0 \mu F}$

$C_{eq}=2 \ \mu F$

Fig.4

The lower branch in Figure 4b consists of two $8.0 \mu F$ capacitors in series, which combine to yield an equivalent capacitance of $4.0 \mu F$. Finally, the $2.0 \mu F$ and $4.0 \mu F$ capacitors in Figure 4c are in parallel and thus have an equivalent capacitance of $6.0 \mu F$.

Additional example problems can be studied at the link below.

Additional Example Problems.

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