Answer
418.2k+ views
Hint: The given problem can be solved by using the concept of the charging and discharging of the capacitors. Capacitors are devices that store the electrical energy in their electric field. This energy depends on the potential difference across the plates of the capacitor and the capacitance of the capacitor.
Complete step by step solution:
Let the two capacitors be ${C_1}$ and ${C_2}$ at some potential differences ${V_1}$ and ${V_2}$ respectively. We are given that both the conductors are joined by a wire. As soon as they are connected by the wire, charge starts to flow from higher potential to lower potential. This flow of charge continues till they reach a common potential difference.
The common potential difference will be:
${V_{common}} = \dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
Here, {V_{common}} is the common potential difference.
The initial energy stored in each capacitor is:
${E_1} = \dfrac{1}{2}{C_1}{V_1}^2 and {E_2} = \dfrac{1}{2}{C_2}{V_2}^2$
Here, ${E_1}$ and ${E_2}$ are the initial energy stored in both the capacitors.
After joining the wire, both the capacitors will reach a point of common potential difference, therefore the final energy in both the capacitors will be given as
${E_{final}} = \dfrac{1}{2}{C_1}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2} + \dfrac{1}{2}{C_2}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow {E_{final}} = \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
Here, ${E_{final}}$ is the final energy stored in the capacitors.
The loss in energy will be given by the difference between initial energy and final energy;
$\left( {{E_1} + {E_2}} \right) - {E_{final}} = \left( {\dfrac{1}{2}{C_1}{V_1}^2 + \dfrac{1}{2}{C_2}{V_2}^2} \right) - \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{2}\left( {{C_1}{V_1}^2 + {C_2}{V_2}^2} \right) - \left[ {\left( {{C_1} + {C_2}} \right){{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)}^2}} \right]$
On solving the equation, we have
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left[ {\left( {{C_1}^2{V_1}^2 + {C_2}{C_1}{V_2}^2 + {C_2}{C_1}{V_1}^2 + {C_2}^2{V_2}^2} \right) - \left( {{C_1}^2{V_1}^2 + {C_2}{V_2}^2 + 2{C_1}{C_2}{V_1}{V_2}} \right)} \right]$
$\Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left( {{C_1}{C_2}{V_2}^2 + {C_1}{C_2}{V_1}^2 - 2{C_1}{C_2}{V_1}{V_2}} \right)$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}{C_1}{C_2}{\left( {{V_1} - {V_2}} \right)^2}$
This is the required formula for loss of energy on joining of two charged conductors by a wire.
Note: In the derived formula, the capacitances cannot be negative individually. Also, the value of ${\left( {{V_1} - {V_2}} \right)^2}$ will be non-negative as it is a square value, this implies that the final value will be positive. Thus, initial energy was greater than final energy. Also, for ${V_1}$ = ${V_2}$ , the difference in energy will be zero. This means that no loss of energy takes place as there is no potential difference initially, there will be no flow of current.
Complete step by step solution:
Let the two capacitors be ${C_1}$ and ${C_2}$ at some potential differences ${V_1}$ and ${V_2}$ respectively. We are given that both the conductors are joined by a wire. As soon as they are connected by the wire, charge starts to flow from higher potential to lower potential. This flow of charge continues till they reach a common potential difference.
The common potential difference will be:
${V_{common}} = \dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
Here, {V_{common}} is the common potential difference.
The initial energy stored in each capacitor is:
${E_1} = \dfrac{1}{2}{C_1}{V_1}^2 and {E_2} = \dfrac{1}{2}{C_2}{V_2}^2$
Here, ${E_1}$ and ${E_2}$ are the initial energy stored in both the capacitors.
After joining the wire, both the capacitors will reach a point of common potential difference, therefore the final energy in both the capacitors will be given as
${E_{final}} = \dfrac{1}{2}{C_1}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2} + \dfrac{1}{2}{C_2}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow {E_{final}} = \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
Here, ${E_{final}}$ is the final energy stored in the capacitors.
The loss in energy will be given by the difference between initial energy and final energy;
$\left( {{E_1} + {E_2}} \right) - {E_{final}} = \left( {\dfrac{1}{2}{C_1}{V_1}^2 + \dfrac{1}{2}{C_2}{V_2}^2} \right) - \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{2}\left( {{C_1}{V_1}^2 + {C_2}{V_2}^2} \right) - \left[ {\left( {{C_1} + {C_2}} \right){{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)}^2}} \right]$
On solving the equation, we have
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left[ {\left( {{C_1}^2{V_1}^2 + {C_2}{C_1}{V_2}^2 + {C_2}{C_1}{V_1}^2 + {C_2}^2{V_2}^2} \right) - \left( {{C_1}^2{V_1}^2 + {C_2}{V_2}^2 + 2{C_1}{C_2}{V_1}{V_2}} \right)} \right]$
$\Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left( {{C_1}{C_2}{V_2}^2 + {C_1}{C_2}{V_1}^2 - 2{C_1}{C_2}{V_1}{V_2}} \right)$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}{C_1}{C_2}{\left( {{V_1} - {V_2}} \right)^2}$
This is the required formula for loss of energy on joining of two charged conductors by a wire.
Note: In the derived formula, the capacitances cannot be negative individually. Also, the value of ${\left( {{V_1} - {V_2}} \right)^2}$ will be non-negative as it is a square value, this implies that the final value will be positive. Thus, initial energy was greater than final energy. Also, for ${V_1}$ = ${V_2}$ , the difference in energy will be zero. This means that no loss of energy takes place as there is no potential difference initially, there will be no flow of current.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)