Question

What do you understand by the internal resistance of a cell? The potential difference of a cell becomes $1.8V$ when $0.5A$ current is drown and when $1.0A$ current is drawn, it becomes $1.6V$. Find the internal resistance of the cell and its e.m.f.

Answer 1

A cell consists of chemicals which react with each other and produce e.m.f.

When a circuit is connected externally at the +ve and -ve terminals of the cell a current flows through the circuit.

Since a current cannot move through an incomplete circuit, the cell's chemical-solution is also a part of the circuit and current flows through that also.

The opposition to the current offered by this solution is known as internal resistance of a cell.

$Given:$

Let the internal resistance be $r$

And e.m.f. be $E$

As per Ohm's law,

$\frac{V}{I}=R$esistance of external circuit

$And,$

$\frac{E}{I}=R+r$

$\therefore \frac{E}{I}=\frac{V}{I}+r$

$\therefore \frac{E-V}{I}=r$

In both the cases, $E$ and $r$ of the cell remain the same

$\therefore \frac{E-1.8}{0.5}=\frac{E-1.6}{1}$

By multiplying and dividing the LHS by 2, we get $:-$

$\therefore 2E-3.6=E-1.6$

$\therefore E=2volts$

$r=\frac{E-V}{I}$

$r=\frac{2-1.6}{1}=0.4$

$\therefore r=0.4ohms$