A storage battery of emf 10 V and internal resistance 2\ is being...

The net electromotive force (EMF) in the circuit is the difference between the two opposing voltage sources:
$E_{\text{net}} = 30\text{ V} - 10\text{ V} = 20\text{ V}$

The total resistance in the series circuit is:
$R_{\text{total}} = 3\ \Omega + 2\ \Omega = 5\ \Omega$

The current $I$ flowing through the circuit is:
$I = \dfrac{E_{\text{net}}}{R_{\text{total}}} = \dfrac{20}{5} = 4\text{ A}$

During the charging process, the current flows into the positive terminal of the $10\text{ V}$ battery. The terminal voltage $V$ across the battery is given by:
$V = E + Ir$
Substituting the given values:
$V = 10 + (4 \times 2)$
$V = 10 + 8 = 18\text{ V}$

Answer: $18\text{ V}$