# A capacitor of capacity C is charged fully by a cell of emf V/2 and then it is disconnected and...

## Question:

A capacitor of capacity C is charged fully by a cell of emf V/2 and then it is disconnected and again connected with a cell of emf V (+ve plate of the capacitor with +ve terminal of cell and vice versa). The heat developed in connecting wire during charging by the second cell is ?

## Electrical battery :

An electrical battery consists of electrochemical cells that react and create a flow of electrons in a circuit to produce electrical energy. Electrical batteries are used to power electrical devices such as mobile phones, flashlights.

Initially, the charge on capacitor is as follows.

{eq}{q_i} = \dfrac{{CV}}{2} {/eq}

After maintaining the same polarity of the battery, the final charge on same capacitor is as follows.

{eq}{q_f} = CV {/eq}

Now, the magnitude of charge transferred to cell of emf V is as follows.

{eq}\begin{align*} q &= \dfrac{{CV}}{2} - CV\\ &= - \dfrac{{CV}}{2} \end{align*} {/eq}

Now, write the formula to calculate the work done by the cell.

{eq}\begin{align*} W &= q \times emf\\ &= \left( { - \dfrac{{CV}}{2}} \right) \times V\\ &= - \dfrac{{C{V^2}}}{2} \end{align*} {/eq}

Calculate the change in electrostatic energy of the capacitor.

{eq}\begin{align*} \delta U &= \dfrac{1}{2}CV_f^2 - \dfrac{1}{2}CV_i^2\\ &= \dfrac{C}{2}\left( {{{\left( {\dfrac{V}{2}} \right)}^2} - {{\left( V \right)}^2}} \right)\\ &= \dfrac{C}{2}\left( { - \dfrac{{3{V^2}}}{4}} \right)\\ &= - \dfrac{3}{8}C{V^2} \end{align*} {/eq}

Here, {eq}{V_f} {/eq} and {eq}{V_i} {/eq} are the final and initial emf in the cell.

Write the condition for energy conservation for heat developed in connecting wire.

{eq}W = \delta U + {\rm{heat}} {/eq}

Substitute value in above expression.

{eq}\begin{align*} - \dfrac{{C{V^2}}}{2} &= - \dfrac{3}{8}C{V^2} + {\rm{heat}}\\ {\rm{heat}} &= - \dfrac{{C{V^2}}}{8} \end{align*} {/eq}

Thus, the heat generated in connecting wire is {eq}- \dfrac{{C{V^2}}}{8} {/eq}.