# Electronic – How to calculate equivalent capacitance to a battery

capacitancecapacitor

I have a 1.25V 2Ah battery and I'm trying to calculate a equivalent capacitance with rated voltage of 2.7V for each of those batteries. This is what I did:

Work of Battery = \$1.25V \cdot 2A \cdot 3600s = 9000J\$

From the capacitor work equation:

$$W = 0.5 \cdot C \cdot V^2$$

$$9000J = 0.5 \cdot C \cdot 2.7V^2$$

$$C=2469.1358F$$

Is this correct?

What you have calculated is not an equivalent capacitance but, instead, the capacitance required to store 9kJ of energy at 2.7V.

That fact that the battery may also store that much energy does not mean that there is a capacitor equivalent to a battery.

While an ideal battery maintains the voltage across its terminals until the stored energy is exhausted, the voltage across an ideal capacitor will gradually approach zero as the stored energy is depleted.

If the attached circuit will only function properly above some minimum voltage, not all of the energy stored in the capacitor is available to the attached circuit.

Thus, one must first specify the allowed drop in voltage to determine the required capacitance.

For example, stipulate that \$9kJ\$ of energy must be supplied by the capacitor before the voltage falls to \$1V \$.

Then:

$$\frac{C(2.7V)^2}{2} - \frac{C(1.0V)^2}{2} = 9kJ$$

Solve for the required C:

$$C = \frac{2}{(2.7V)^2 - (1.0V)^2}9kJ = 2.86kF$$