Question

9 equal capacitors are connected purely in parallel to a 8 volt battery. The same capacitors are then connected to the same battery purely in series. The total stored energy in the parallel configuration is 626 micro-joules greater than the series configuration. What is the value of each capacitor in micro-farads?

Answer 1

When the capacitors are connected in parallel.

Suppose capacitance of each capacitor is C.

Then its equivalent capacitace, Ceq = 9C

So, energy stored E1 = (1/2)*Ceq*V^2 = (1/2)*(9C)*V^2

In the second case, when the capacitors are connected in series.

Ceq = C/9

So, energy stored E2 = (1/2)(C/9)*V^2

Now as per condition -

E1 - E2 = 626 micro-joules

we will consider the conversion of units later for simplicity.

So, (1/2)*(9C)*V^2 - (1/2)(C/9)*V^2 = 626

=> 9C - C/9 = 2 x 626 / 8^2 = 19.56

=> 8.89C = 19.56

=> C = 19.56 / 8.89 = 2.20 micro - farads.