Question

If I connect a large number of 65.0 μF capacitors in parallel across a 120.0 V battery, how many capacitors do I need to store 39.3 J of energy?

Answer 1

Let the required number of capacitors be 'n'.

The total energy required is 39.3 J.

The energy between two capacitor plates is given by -

$E = \frac{1}{2}(CV^{2})$

Where C is the capacitance,

And V is the voltage between the plates.

Here, the votage value = V = 120V, and C = 65 $\mu$F.

So, the energy stored in 1 capacitor would be -

$E = \frac{1}{2}\times 65 \times 10^{-6} \times (120)^{2}$

$\Rightarrow E = 0.468J$

So, the total number of capacitor plates required would be -

$n = \frac{39.3}{0.468}$

Hence, n = 83.97.

So, approximately 84 capacitors would be required to store energy of 39.3J.