Question

A cell when connected to an external resistance of 2 Ω shows a p.d. of 1.2 V. If the 2 Ω resistance is replaced by 1 Ω resistor the p.d. drops to 1 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: \[ r = \frac{\mathrm{R} \left( \mathrm{E} - \mathrm{V} \right)}{\mathrm{V}} \]

\[ \therefore r=\frac{2(\mathrm{E}-1.2)}{1.2} \]  ....(i)

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

\[ r=\frac{1(\mathrm{E}-1)}{1} \] ....(ii)

Comparing (i) and (ii)

\[ \frac{1(E-1)}{1}=\frac{2(E-1.2)}{1.2}\]

\[\Rightarrow 1.2 E-1.2=2 E-2.4 \]

⇒ \[ 0.8 \mathrm{E}=1.2\]

\[\Rightarrow E=\frac{1.2}{0.8}\]

E = 1.5 V

Substituting the value of E in (ii)

\[ r=\frac{1(1.5-1)}{1}={0.5} \Omega\]