A cell is connected to an external resistor. When the external resistance is 2 Ω, the current is 0.8 A. When the external resistance is 7 Ω, the current is 0.3 A. Find the EMF and internal resistance

We are given two scenarios for a cell connected to an external resistor. The formula relating electromotive force ($\mathcal{E}$), current ($I$), external resistance ($R$), and internal resistance ($r$) is $\mathcal{E} = I(R + r)$.

From the problem:

• Scenario 1: External resistance $R_1 = 2 \, \Omega$, Current $I_1 = 0.8 \, \text{A}$.
• Scenario 2: External resistance $R_2 = 7 \, \Omega$, Current $I_2 = 0.3 \, \text{A}$.

For Scenario 1:

$$\mathcal{E} = 0.8 \, \text{A} (2 \, \Omega + r) \quad (1)$$

For Scenario 2:

$$\mathcal{E} = 0.3 \, \text{A} (7 \, \Omega + r) \quad (2)$$

Since both equations equal $\mathcal{E}$, we can set them equal to each other:

$$0.8(2 + r) = 0.3(7 + r)$$

Distribute the current values:

$$1.6 + 0.8r = 2.1 + 0.3r$$

Rearrange the terms to solve for $r$:

$$0.8r - 0.3r = 2.1 - 1.6$$ $$0.5r = 0.5$$ $$r = \frac{0.5}{0.5}$$ $$r = 1 \, \Omega$$

Using equation (1):

$$\mathcal{E} = 0.8(2 + 1)$$ $$\mathcal{E} = 0.8(3)$$ $$\mathcal{E} = 2.4 \, \text{V}$$

The electromotive force (EMF) of the cell is $\boxed{\text{2.4 V}}$ and the internal resistance of the cell is $\boxed{\text{1 } \Omega}$.