Question

Understanding EMF - TUTORIAL

- Mathslover

Answer 1

\(\large{\color{red}{\textbf{ELECTROMOTIVE FORCE}} ~~ \left( \sf{EMF} \right) }\)

\(\color{Blue}{\textbf{Electromotive ~ Force ~ }}\)- It is the potential difference between the terminals of a electrodes of the cell, when \(\bf{no ~ current}\) is being drawn or in an \(\bf{open}\) circuit.

\(\bullet\) Factors on which EMF depends : (We will be denoting EMF by E while generally, it is denoted by : \(\large \epsilon\) )

Nature of Electrodes and Electrolytes

Why is this called "Electromotive \(\bf{Force} \) " ?

Basically, EMF is not a mechanical force but because of the Potential difference between the terminals of a cell (E.M.F), the charge is forced to move from one terminal to other terminal.

OR

It is the amount of work done required to move unit (+ve) test charge from one electrode to another electrode in a circuit.

As it is a potential difference, thus, its unit is Volt. (SI UNIT)

\(\color{blue}{\bf{Terminal ~ Voltage ~/~ Terminal~ Potential~ Difference ~in ~Closed~ Circuit }}\)

Denoted by symbol \(\bf{V}\)

It is defined as the potential difference between the terminals of electrodes of the cell when it is in `closed circuit`. Closed circuit - When some current is being drawn from the cell.

\(\color{blue}{\bf{INTERNAL~ RESISTANCE ~OF ~CELL}}\)

Denoted by \(\bf{r}\)

It is defined as the resistance offered by the electrolytes and electrodes of cell (s) .

Factors on which "r" depends on -

\(\bullet\) Nature of the electrolyte/electrodes
\(\bullet\) Area of electrodes
\(\bullet\) Distance between the electrodes.

> For a freshly prepared / purchased cell, the internal resistance is very small.
> For a cell being used , the internal resistance is comparatively higher due to the chemical reactions in the cell.

The value of internal resistance is thus, not fixed.

\(\color{blue}{\bf{Relation ~between}}\) \(\bf{E ~ \& ~ V}\) :



\(I = \cfrac{E}{(r + R)}
\implies E = I(r + R)
\implies E = Ir + IR \\
\text{Since, V = IR } \\
\implies E = Ir + V \)

Since, Ir > 0 , thus, E > V (when current is being drawn from the cell). Now, when cell is being charged, then the relation will change. As current will flow in opposite direction in the cell.

Thus, the equation will become : (The current \(\bf{I}\) will become negative now )
Thus,

\( E = -Ir + V \\
\implies E + Ir = V \)

Therefore, in this case, V > E

Case when V = E

Clearly, for V to be equal to E , the current should be zero or \(\bf{r}\) has to be negligibly small (in this case : \(V \approx. E\) .

Thus, V = E when the circuit is open, that is \(\bf{no ~ current}\) is drawn from the cell.