Given data

• Two identical cells, each of emf $E = 1.5\,\text{V}$
• Connected in series, so total emf $E_{\text{total}} = 1.5 + 1.5 = 3.0\,\text{V}$
• External resistance $R = 10\,\Omega$
• Voltmeter across the $10\,\Omega$ resistor reads $V_R = 1.5\,\text{V}$
• Let internal resistance of each cell be $r\,\Omega$

Find current in the circuit Since the voltmeter reads the potential difference across the $10\,\Omega$ resistor, $V_R = IR$ So, $1.5 = I \times 10$ $I = \frac{1.5}{10} = 0.15\,\text{A}$

Write the total circuit equation The two cells are in series, so total internal resistance is $r_{\text{total}} = 2r$ Total resistance in circuit is $R_{\text{total}} = 10 + 2r$

Using Ohm’s law for the whole circuit, $I = \frac{E_{\text{total}}}{10 + 2r}$

Substitute values: $0.15 = \frac{3}{10 + 2r}$

Solve for $r$ $10 + 2r = \frac{3}{0.15} = 20$ $2r = 20 - 10 = 10$ $r = 5\,\Omega$

Final answer The internal resistance of each cell is $\boxed{5\,\Omega}$