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 the 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 it is Q. Again, we know that

Here, we can consider that the voltage developed across the capacitor C_{1} is V_{1}. The same across the capacitor C_{2} is V_{2}. Similarly, the voltage across the capacitor C_{3} is V_{3}. Lastly, the voltage across the capacitor C_{n} is V_{n.} Therefore, we can write the expression of 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) flowing through the series combination is nothing but the current flowing the individual capacitors. So, the final charge of the entire series capacitors is also Q.

Finally, we can write,

## Derivation of Capacitance of Parallel Capacitor

This figure below, 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 the same. Because it depends on the impedance of individual capacitors. So, we can write the entire current of the parallel combination of capacitors as,

Since the currents are not equal, charges are not equal. But, we can consider q as the total charge 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 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.

- Capacitance – What is Capacitance?
- Capacitor What is a Capacitor? How does it work?
- Capacitors in Series and Capacitors in Parallel
- Multiplate Capacitor and Variable Multiplate Capacitor
- Energy Stored in a Capacitor
- Permittivity (Absolute Permittivity and Relative Permittivity)
- Charging of a Capacitor and Time Constant
- Discharging a Capacitor and Related Expressions