Interpret the circuit

Two cells are connected in opposition, so their emfs oppose each other.

Given:

• Cell $E_1 = 8\,\text{V}$, internal resistance $r_1 = 2\,\Omega$
• Cell $E_2 = 2\,\text{V}$, internal resistance $r_2 = 4\,\Omega$

Find the net emf of the combination

Since the cells oppose each other, $E_{\text{net}} = E_1 - E_2 = 8 - 2 = 6\,\text{V}$

Find the total internal resistance

The internal resistances are in series in the loop: $r_{\text{total}} = r_1 + r_2 = 2 + 4 = 6\,\Omega$

Calculate the current in the circuit

Using Ohm’s law for the complete circuit, $I = \frac{E_{\text{net}}}{r_{\text{total}}} = \frac{6}{6} = 1\,\text{A}$

Determine the terminal potential difference of cell $E_2$

Since $E_2$ is the weaker cell, current enters its positive terminal, so it is being charged.

For a cell being charged, terminal potential difference is: $V_{\text{terminal}} = E + Ir$

Therefore for cell $E_2$, $V_2 = E_2 + I r_2 = 2 + (1)(4) = 6\,\text{V}$

Final answer

$\boxed{6}$

Comparison with stored answer

Stored correct answer = $6$

This matches our derived answer.