Question

A cell of $e.m.f.1.8V$ is connected to an external resistance of $2\Omega$, when$p.d$. recorded at the ends of resistance is $1.6\text{volts}$. Calculate the internal resistance of the cell.

Answer 1

Step 1: Given data:

1. e.m.f. $=1.8\text{volts}$
2. External resistance $=2\Omega$
3. The potential difference across the external resistance is $1.6\text{volts}$

Step 2: Find the internal resistance of the cell r,

By Ohm's law, the current through the external resistance,

$I=\frac{V}{R}$

$I=\frac{1.6}{2}$

$=0.8A$

The relation between emf and potential difference in terms of the internal resistance is,

$E-V=Ir$

$1.8-1.6=0.8r$

$r=\frac{0.2}{0.8}$

$r=0.25\Omega$

Thus, the internal resistance of the cell is $0.25\Omega$.