Answer :

Given, capacitance of a, b, c, d capacitors are 10 μF each.

The voltage of the DC battery is 100V

__Formula used__

Energy stored in a capacitor can be calculated from the relation,

Or,

Where C represents the capacitance, V is the potential difference across the capacitor and Q is the charge in the capacitor.

And, effective capacitance of capacitors C_{1} and C_{2} arranged in series is

And those connected in parallel is

Considering three capacitors)

The capacitors b and c are in parallel. For simplification, we reduce it into capacitor bc as shown,

and the capacitance of bc is, from eqn.4

By substituting the values,

Now the whole arrangement is a series connection and charges in each capacitor will be the same.

To find out effective capacitance of this arrangement, we find equivalent capacitance, C_{ad} between a and d initially, by eqn.3

By substituting the values,

Now the total capacitance considering C_{ad}and C_{bc} in series, using eqn.3,

Or,

The capacitors a ,d and the parallel arrangement will have same charge,Q in it, which can be calculated as,

Where,

C_{eff}= Capacitance, V= Potential difference=100*V*

By substitution, we get , Q as

The energy stored in a and d are same due to the same capacitance value and the same charge accumulation. So energy stored in a and d are, from eqn.2

Or,

In the parallel arrangement, the charge, Q=400μC will be splitted in half as the two branches are symmetrical. So each capacitors b and c will have Q=200μC amount of charge.

Hence, by the energy relation, eqn.2 , the energy in each capacitors b and c, will be,

Or,

Hence 8mJ will be stored in the capacitors a and d , while 2mJ will be stored in b and c.

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