Question

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

Answer 1

Let 'E' be the e.m.f. and 'r' the internal resistance.

Case 1: \[ r = \frac{\mathrm{R} \left( \mathrm{E} - \mathrm{V} \right)}{\mathrm{V}} \]

\[ \therefore r=\frac{4.5(\mathrm{E}-1.35)}{1.35} \] ....(i)

Case 2: \[ r = \frac{\mathrm{R} \left( \mathrm{E} - \mathrm{V} \right)}{\mathrm{V}} \]

\[ r=\frac{2.5(\mathrm{E}-1.25)}{1.25} \] ....(ii)

Comparing (i) and (ii)

\[ \frac{2.5(E-1.25)}{1.25}=\frac{4.5(E-1.35)}{1.35} \]

\[ 2(E-1.25)=\frac{10}{3}(E-1.35) \]

\[ 2E-2.5=\frac{10}{3}E-4.5 \]

\[ \therefore \frac{4}{3}E=2\]

\[\Rightarrow E=\frac{2}{4/3}\]

\[E=1.5\text{ V} \]

Substituting the value of E in (ii)

\[ r=\frac{2.5(1.5-1.25)}{1.25}\]

\[=0.5\ \Omega \]