Question

A cell, when connected to an external resistance of $4.5\Omega$ shows a p.d of $1.35\mathrm{V}$. If $4.5\Omega$ resistance is replaced by $2.5\Omega$ resistance the p.d drops to $r$$1.25\mathrm{V}$.
Calculate (a) e.m.f (b) internal resistance of the cell.

Answer 1

Step 1: a)To determine the emf of the given cell :
Given data,
External resistance $=4.5\Omega$
$4.5\Omega$ resistance is replaced by $2.5\Omega$
The potential difference drops to $1.25\mathrm{V}$
Let ' $E$ ' be the emf and '$r$ ' the internal resistance
Case (i):
$r=\frac{R[E-V]}{V}$
$r=$ internal resistance
$R=$ External resistance
$V=$ Voltage
$E=$ e.m.f.
$r=\frac{4.5[\mathrm{E}-1.35]}{1.35}$
$=\frac{10(\mathrm{E}-1.35)}{3}$ ……. $\left(i\right)$

Case $\left(ii\right):$
$r=\frac{2.5[\mathrm{E}-1.25]}{1.25}$
$=\frac{10}{5}(\mathrm{E}-1.25)$

Comparing $\left(i\right)$ and $\left(ii\right)$,
$\frac{10[E-1.35]}{3}=\frac{10}{5}(E-1.25)$
$5[E-1.35]=3[E-1.25]$
$5E-6.75=3E-3.75$
$5E-3E=6.75-3.75$
$2E=3$
$E=\frac{3}{2}$
$E=1.5\mathrm{V}$
Thus, the emf of the given cell is $1.5\mathrm{V}$.

Step 2: (b) To find the internal resistance of the cell :

Substituting in equation (i)
$$\begin{gathered} r=\frac{10}{3}[\mathrm{E}-1.35] \\ r=\frac{10}{3}[1.5-1.35] \\ =\frac{10}{3}\times \frac{15}{100} \\ r=0.5\Omega \end{gathered}$$
Thus, the internal resistance of the given cell $r$ is $0.5\Omega$.