# A 10 F, 20 F, 22 F, and 100 F capacitor are in parallel. What is the total capacitance?

Jan 30, 2015

The answer is: $152 F$.

• The difference of potential ($\Delta V$) across the two capacitors is the same (the one of the generator);
• Q is divided between the capacitors;
• If ${C}_{1} \mathmr{and} {C}_{2} \mathmr{and} {C}_{3} \mathmr{and} \ldots$ are different, so ${Q}_{1} \mathmr{and} {Q}_{2} \mathmr{and} {Q}_{3} \mathmr{and} \ldots$ are different.

So:

${C}_{e q} = \frac{Q}{V} = \frac{{Q}_{1} + {Q}_{2} + {Q}_{3} + \ldots}{V} = {Q}_{1} / V + {Q}_{2} / V + {Q}_{3} / V + \ldots = {C}_{1} + {C}_{2} + {C}_{3} + \ldots$

So:

To find the equivalent capacitor of two or more capacitors in parallel it is necessary to use this formula:

${C}_{e q} = {C}_{1} + {C}_{2} + {C}_{3} + \ldots$

In this case:

${C}_{e q} = 10 F + 20 F + 22 F + 100 F = 152 F$.