Answer :

Given,


Potential difference, V is 12V


Capacitance of initially charged capacitor, C1 is 2 μF


Capacitance of initially uncharged capacitor, C2 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 C1capacitor, Q1 is



Or,



Now, let’s assume that after connecting the second capacitor C2, the charge on C1 and C2 as q1 and q2 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 C1 and C2, 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 C1



Or,



Similarly,


On C2



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, Eb is



Where V=12V


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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