This time we will learn about capacitors in series and parallel examples. This thing is crucial for us like learning the series and parallel resistors. We know from resistive circuits that a series-parallel combination is a powerful tool for reducing circuits.

**Contents**show

This technique can be extended to series-parallel connections of capacitors, which are sometimes encountered.

## Parallel Capacitors Formula

In order to obtain the equivalent capacitor *C*_{eq} of *N *capacitors in parallel, consider the circuit in Figure.(1a).

The equivalent circuit is in Figure.(1b). Note that the capacitors have the same voltage *v *across them. Applying KCL to Figure.(1a),

But *i*_{k }= *C*_{k }*d**v*/*dt*. Hence,

where

The

equivalent capacitanceof N parallel-connected capacitors is the

sum of the individual capacitances.

We observe that capacitors in parallel combine in the same manner as resistors in series.

## Series Capacitors Formula

We now obtain *C*_{eq} of *N *capacitors connected in series by comparing the circuit in Figure.(2a) with the equivalent circuit in Figure.(2b).

Note that the same current *i *flows (and consequently the same charge) through the capacitors. Applying KVL to the loop in Figure.(2a),

But,

Therefore,

where

The initial voltage *v*(*t*0) across *C*eq is required by KVL to be the sum of

the capacitor voltages at *t*0. Or according to Equation.(5),

Thus, according to Equation.(6),

The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

Note that capacitors in series combine in the same manner as resistors in parallel. For *N *= 2 (i.e., two capacitors in series), Equation.(6) becomes

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## Capacitors in Series and Parallel Examples

1. Find the equivalent capacitance seen between terminals *a *and *b *of the circuit in Figure.(3).

**Solution:**

The 20- *μ*F and 5- *μ*F capacitors are in series; their equivalent capacitance is

This 4- *μ*F capacitor is in parallel with the 6- *μ*F and 20- *μ*F capacitors; their combined capacitance is

This 30- *μ*F capacitor is in series with the 60- *μ*F capacitor. Hence, the equivalent capacitance for the entire circuit is

2. For the circuit in Figure.(4), find the voltage across each capacitor.

**Solution:**

We first find the equivalent capacitance *C*_{eq}, shown in Figure.(5). The two

parallel capacitors in Figure.(4) can be combined to get

40 + 20 = 60 mF.

This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. Thus,

The total charge is

This is the charge on the 20-mF and 30-mF capacitors because they are in series with the 30-V source. (A crude way to see this is to imagine that charge acts like current, since *i *= *dq*/*dt*.) Therefore,

Having determined *v*_{1} and *v*_{2}, we now use KVL to determine *v*_{3} by

Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage *v*_{3} and their combined capacitance is

40 + 20 = 60 mF.

This combined capacitance is in series with the 20-mF and 30-mF capacitors and consequently has the same charge on it. Hence,