In the given circuit, two batteries of EMFs 20\ V and 8\ V with...

The two batteries are connected in opposition. The net EMF in the circuit is:

$E_{\text{net}} = 20 - 8 = 12\ \text{V}$

The total resistance of the series circuit is the sum of all external and internal resistances:

$R_{\text{total}} = 4 + 1 + 5 + 2 = 12\ \Omega$

The current in the circuit is:

$I = \dfrac{E_{\text{net}}}{R_{\text{total}}} = \dfrac{12}{12} = 1\ \text{A}$

The current flows clockwise, driven by the stronger $20\ \text{V}$ battery, and enters the positive terminal of the $8\ \text{V}$ battery. Thus, the $8\ \text{V}$ battery is being charged.

The potential difference across the terminals of a battery during charging is given by $V = E + Ir$.

For the $8\ \text{V}$ battery:

$V = 8 + (1)(1) = 9\ \text{V}$

Answer: $9\ \text{V}$