Question

A condenser of capacity \[{C_1}\] is charged to a potential \[{V_0}.\] The electrostatic energy stored in it is \[{U_0}.\] It is connected to another uncharged condenser of capacity \[{C_2}\]in parallel. The energy dissipated in the process is:
A. $\dfrac{{{C_2}}}{{{C_1} + {C_2}}}{U_0}$
B. $\dfrac{{{C_1}}}{{{C_1} + {C_2}}}{U_0}$
C. $(\dfrac{{{C_1} - {C_2}}}{{{C_1} + {C_2}}}){U_0}$
D. $\dfrac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{U_0}$

Answer 1

Hint: To solve this question, we will start with taking the initial energy and initial charge of condenser \[{C_1}.\] Now, it is given that condenser \[{C_1}\]is connected parallelly to another uncharged condenser of capacity \[{C_2}.\] So, now we will take the total energy and total charge of the system. Then we will evaluate the energy dissipated in the whole process.

Complete step by step answer:
We have been given a condenser of capacity \[{C_1}\] which is charged to a potential \[{V_0}.\]The electrostatic energy stored in it is \[{U_0}.\]The given condenser is connected to another uncharged condenser of capacity \[{C_2}\] in parallel. We need to find the energy dissipated in the process.
So, given potential of a condenser of capacity \[{C_1} = {\text{ }}{V_0}\]
Also given the electrostatic energy stored in condenser \[ = {\text{ }}{U_0}\]
At first, we will get the initial energy of the condenser \[{C_1}\] by applying the formula mentioned below.
Initial energy, \[{U_i}\] $ = \dfrac{{QV}}{2} = \dfrac{{C{V^2}}}{2}$
where, Q \[ = \] charge
V \[ = \] potential
C \[ = \] condenser
So, on putting the value in the above-mentioned formula, we get
The initial energy of condenser\[{C_{1,}}\] \[{U_i} = {U_0} = \dfrac{1}{2}{C_1}{V_0}^2\]
And the initial charge of the system, ${Q_i} = {C_1}{V_0}$

Now, it is given that condenser \[{C_1}\]connected parallelly to another uncharged condenser of capacity \[{C_2}.\]
So, after connecting another condenser, the final charge of both \[{C_1}\] and \[{C_2}\],\[\]${Q_f} = ({C_1} + {C_2}){V_c}$
where, ${V_c} = $ common potential after connecting.

As total charge is conserved, ${Q_i} = {Q_f} = {C_1}{V_0} = ({C_1} + {C_2}){V_c}$
$\therefore {V_c} = \dfrac{{{C_1}{V_0}}}{{({C_1} + {C_2})}}$
Now, the final energy, ${U_f} = \dfrac{1}{2}({C_1} + {C_2}){V_c}^2$
$ = \dfrac{1}{2}({C_1} + {C_2}){\left( {\dfrac{{{C_1}{V_0}}}{{{C_1} + {C_2}}}} \right)^2} = \dfrac{{{C_1}{U_0}}}{{{C_1} + {C_2}}}$​
Hence, the energy dissipation $ = {U_i} - {U_f}$
$ = {U_0} - \dfrac{{{C_1}{U_0}}}{{{C_1} + {C_2}}}$
$ = \dfrac{{{C_2}{U_0}}}{{{C_1} + {C_2}}}$

So, the correct answer is “Option A”.

Note:
In the question above we have been asked about the energy dissipation. So, energy is dissipated, when there is a change in a system, the energy gets transferred and in that process some of that energy is dissipated or wasted. We can say the energy gets lost to the surroundings.