Answer :

Given,

Potential difference, V is 12*V*

Capacitance of initially charged capacitor, C_{1} is 2 μF

Capacitance of initially uncharged capacitor, C_{2} is 4 μF

__Formula used,__

Energy stored in a capacitor of capacitance C and charge Q is,

Energy stored in a capacitor of capacitance C across a potential difference V is,

a)

Initial charge on C_{1}capacitor, Q_{1} is

Or,

Now, let’s assume that after connecting the second capacitor C_{2}, the charge on C_{1} and C_{2} as q_{1} and q_{2} respectively.

So,

We know that for a parallel arrangement of capacitors across a single battery, the potential differences are the same. So

Or, by substituting the values for C_{1} and C_{2}, we can re-write it as,

Or,

Substituting eqn.3 in eqn.2, we get

Or,

And

b) Energy stored on each capacitor, by eqn.1 is

On C_{1}

Or,

Similarly,

On C_{2}

Or,

Hence the energy stored is 16μJ and 32μJ on 2μF and 4μF capacitors respectively.

c) For heat dissipation, we have to find the initial energy stored.

Hence from eqn.1, the initial energy with 2μF capacitor only in the circuit, E_{b} is

Where V=12*V*

So after substitution,

Hence heat produced is the difference between the initial energy and the algebraic sum of the energy stored after connection.

So,

Or,

The heat produced/dissipated during the charging is 96μJ

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