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Question: A condenser of capacity \[{C_1}\] is charged to a potential \[{V_0}.\] The electrostatic energy stor...

A condenser of capacity C1{C_1} is charged to a potential V0.{V_0}. The electrostatic energy stored in it is U0.{U_0}. It is connected to another uncharged condenser of capacity C2{C_2}in parallel. The energy dissipated in the process is:
A. C2C1+C2U0\dfrac{{{C_2}}}{{{C_1} + {C_2}}}{U_0}
B. C1C1+C2U0\dfrac{{{C_1}}}{{{C_1} + {C_2}}}{U_0}
C. (C1−C2C1+C2)U0(\dfrac{{{C_1} - {C_2}}}{{{C_1} + {C_2}}}){U_0}
D. C1C22(C1+C2)U0\dfrac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{U_0}

Explanation

Solution

To solve this question, we will start with taking the initial energy and initial charge of condenser C1.{C_1}. Now, it is given that condenser C1{C_1}is connected parallelly to another uncharged condenser of capacity C2.{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 C1{C_1} which is charged to a potential V0.{V_0}.The electrostatic energy stored in it is U0.{U_0}.The given condenser is connected to another uncharged condenser of capacity C2{C_2} in parallel. We need to find the energy dissipated in the process.
So, given potential of a condenser of capacity C1= V0{C_1} = {\text{ }}{V_0}
Also given the electrostatic energy stored in condenser = U0 = {\text{ }}{U_0}
At first, we will get the initial energy of the condenser C1{C_1} by applying the formula mentioned below.
Initial energy, Ui{U_i} =QV2=CV22 = \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 condenserC1,{C_{1,}} Ui=U0=12C1V02{U_i} = {U_0} = \dfrac{1}{2}{C_1}{V_0}^2
And the initial charge of the system, Qi=C1V0{Q_i} = {C_1}{V_0}

Now, it is given that condenser C1{C_1}connected parallelly to another uncharged condenser of capacity C2.{C_2}.
So, after connecting another condenser, the final charge of both C1{C_1} and C2{C_2},$$$$${Q_f} = ({C_1} + {C_2}){V_c}where, where,{V_c} = $ common potential after connecting.

As total charge is conserved, Qi=Qf=C1V0=(C1+C2)Vc{Q_i} = {Q_f} = {C_1}{V_0} = ({C_1} + {C_2}){V_c}
∴Vc=C1V0(C1+C2)\therefore {V_c} = \dfrac{{{C_1}{V_0}}}{{({C_1} + {C_2})}}
Now, the final energy, Uf=12(C1+C2)Vc2{U_f} = \dfrac{1}{2}({C_1} + {C_2}){V_c}^2
=12(C1+C2)(C1V0C1+C2)2=C1U0C1+C2= \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 =Ui−Uf = {U_i} - {U_f}
=U0−C1U0C1+C2= {U_0} - \dfrac{{{C_1}{U_0}}}{{{C_1} + {C_2}}}
=C2U0C1+C2= \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.