Inside Story

## COMMON POTENTIAL AND LOSS OF ENERGY ON SHARING CHARGES

### COMMON POTENTIAL

When two capacitors of different potential are connected by a conducting wire, then the charges flows from capacitor at higher potential to the capacitor at lower potential. This flow of **charges** continues till their potential become equal, this equal potential is called **common potential**. [latexpage]

$$

\text{common potential},V=\frac{C_1V_1+C_2V_2}{C_1+C_2}

$$

Where $C_1$ and $C_2$ are capacitance of the two capacitors and $V_1$ and $V_2$ are the potential respectively.

So the common potential is given as-

$$

\text{common potential}=\frac{\text{total charge}}{\text{total capacity}}

$$

\begin{align*}

\therefore\; C_1V_1+C_2V_2&=C_1V+C_2V\\

Or\;\; C_1V_1-C_1V&=C_2V-C_2V_2

\end{align*}

From this equation, we can say that, charge lost by one capacitor is equal to **charge** gained by other capacitor.

But here you have to keep in mind that, this is not true for potential. i.e, **potential** lost by one capacitor is not equal to the potential gained by other. Because their capacitance are different.

### LOSS OF ENERGY ON SHARING OF CHARGES

When two charged **capacitors** are connected to each other, they start sharing their charges till they acquire a **common potential**. On sharing charges there is always some loss of energy. However, total charge of the system remains conserved.

Let’s consider two capacitors having capacitance $C_1$ and $C_2$ and their potential is $V_1$ and $V_2$. Then, the total energy stored in the capacitor before the two capacitors are connected is-

\begin{equation}

U=U_1+U_2=\frac{1}{2}C_1V_1^2+\frac{1}{2}C_2V_2^2

\end{equation}

When the two capacitors are connected together, the total charge on the capacitor is-

$$

q=q_1+q_2=C_1V_1+C_2+V_2

$$

And the total **capacitance** of the two capacitors are-

$$

C=C_1+C_2

$$

Now, the total energy stored in the capacitor after they are connected is given as-

\begin{equation}

U’=\frac{1}{2}.\frac{q^2}{C}=\frac{1}{2}\frac{\left(C_1V_1+C_2V_2\right)^2}{C_1+C_2}

\end{equation}

After subtracting equ(1) and equ(2), we get-

\begin{align*}

\quicklatex{size=10}

U-U’&=\left(\frac{1}{2}C_1V_1^2+\frac{1}{2}C_2V_2^2\right)-\frac{1}{2}\frac{\left(C_1V_1+C_2V_2\right)^2}{C_1+C_2}\\

&=\frac{C_1^2V_1^2+C_1C_2V_1^2+C_1C_2V_2^2+C_2^2V_2^2-\left(C_1V_1+C_2V_2\right)^2}{2\left(C_1+C_2\right)}\\

&=\frac{C_1C_2\left(V_1^2+V_2^2-2V_1V_2\right)}{2\left(C_1+C_2\right)}

\end{align*}

$$

\boxed{\Delta{U}=\frac{C_1C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}}

$$

We see that, this is a positive quantity. Since U – U’ is positive, so we conclude that there is always some loss of energy, when two charged capacitors are connected together, in the form of **heat radiation** die to **electric** **current** while charging.