We often connect a number of **capacitors in series or parallel** in a circuit. Obviously, we do it to obtain the desired capacitance value. Although, the equivalent capacitance of the **series-parallel capacitors** is different from that of series-parallel resistors.

For example, in the series connected resistors, the equivalent resistance is the sum of the resistances of individual resistors. However, it differs in the case of series connected capacitors. The equivalent capacitance is the reciprocal of the summation of reciprocal of individual capacitances.

Where n is the number of resistors in the series combination.

But in the case of series capacitors

## Derivation of Capacitance of Series Capacitors

Let us consider n number of **capacitors connected in series** as shown.

Here, we apply V volts across the **series combination of capacitors**. Then we consider the current i is flowing through the capacitors. Again, the current through any capacitor is the rate of change of electrical charge in respect of time. So, we can write,

The current i is the same through all **capacitors in series**. Therefore, the final charge across each capacitor is the same. And say this is Q. Again, we know that

Again, we come to our figure.

Here we can consider that voltage developed across capacitor C_{1} is V_{1}. These same across capacitor C_{2} is V_{2}. Similarly, across capacitor C_{3} it is V_{3}. Lastly, the voltage across the capacitor C_{n} is V_{n.} Therefore, we can write the applied voltage as,

Now, we consider the capacitance of the entire series combination is C_{e}. Hence we can write,

This is because the current ( i=dq/dt) through the series combination is nothing but the current flowing the individual capacitors. So, the final charge of the entire series capacitor is also Q.

Finally, we can write,

## Derivation of Capacitance of Parallel Capacitor

This figure shows the parallel combination of n number of capacitors.

The capacitors have capacitances of C_{1}, C_{2}, C_{3},…C_{n} respectively. The current through the entire parallel combination of capacitors is the sum of the current of individual capacitors.

Certainly, the current through each of the capacitors is not equal. Because it depends on the impedance of individual capacitors. So, we can write the entire current of the parallel combination of capacitors as,

Since currents are not equal, charges of the capacitors are also not the same. Ultimately, we can consider q as the total charges of this parallel combination. Therefore, we can write,

Again, we consider q_{1}, q_{2}, q_{3}…q_{n} are the charges of individual capacitors respectively.

Therefore,

So, from the expressions of the current of the parallel combination, we can write,

Again the charge is the product of its capacitance and voltage in a capacitor.

That means,

Therefore,

It is the equivalent capacitance of parallel capacitors. That is the algebraic sum of the capacitance of individual capacitors. Hence, this is the exact contrast of the equivalent resistance of a parallel combination of resistors.