We have a battery with an unknown electromotive force (EMF) $\mathcal{E}$ and unknown internal resistance $r$. We are measuring the terminal voltage using two different voltmeters with finite resistances.

• Reading with Voltmeter 1 ($R_1$): $V_1 = 0.9 \, \text{V}$
• Reading with Voltmeter 2 ($R_2$): $V_2 = 0.6 \, \text{V}$
• Reading with both in parallel: $V_3 = 0.45 \, \text{V}$

When a voltmeter with resistance $R$ is connected across a battery with EMF $\mathcal{E}$ and internal resistance $r$, the measured voltage $V$ is the terminal voltage.

$$ V = \mathcal{E} – I r = \mathcal{E} – \frac{V}{R} r $$

Rearranging this equation gives:

$$ \frac{1}{V} = \frac{1}{\mathcal{E}} + \frac{r}{\mathcal{E} R} \implies \frac{1}{V} – \frac{1}{\mathcal{E}} = \frac{r}{\mathcal{E}} \cdot \frac{1}{R} $$

Let’s write this for the three cases:

1. Case 1 ($V_1$): $\frac{1}{V_1} – \frac{1}{\mathcal{E}} = \frac{r}{\mathcal{E}} \cdot \frac{1}{R_1}$
2. Case 2 ($V_2$): $\frac{1}{V_2} – \frac{1}{\mathcal{E}} = \frac{r}{\mathcal{E}} \cdot \frac{1}{R_2}$
3. Case 3 ($V_3$): The resistance is the parallel combination $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$. Thus $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.
   Equation: $\frac{1}{V_3} – \frac{1}{\mathcal{E}} = \frac{r}{\mathcal{E}} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$

Substitute the expressions from Case 1 and Case 2 into Case 3:

$$ \left( \frac{1}{V_3} – \frac{1}{\mathcal{E}} \right) = \left( \frac{1}{V_1} – \frac{1}{\mathcal{E}} \right) + \left( \frac{1}{V_2} – \frac{1}{\mathcal{E}} \right) $$

Simplify to solve for $\mathcal{E}$:

$$ \frac{1}{V_3} – \frac{1}{\mathcal{E}} = \frac{1}{V_1} + \frac{1}{V_2} – \frac{2}{\mathcal{E}} $$ $$ \frac{2}{\mathcal{E}} – \frac{1}{\mathcal{E}} = \frac{1}{V_1} + \frac{1}{V_2} – \frac{1}{V_3} $$ $$ \frac{1}{\mathcal{E}} = \frac{1}{V_1} + \frac{1}{V_2} – \frac{1}{V_3} $$

Inverting gives the final formula:

$$ \mathcal{E} = \frac{V_1 V_2 V_3}{V_2 V_3 + V_1 V_3 – V_1 V_2} $$