A resistor of resistance $R$ and three identical cells of emf $E$ and internal resistance $r$ are connected as shown - AQA - A-Level Physics - Question 28 - 2020 - Paper 1

The total internal resistance ($r_{total}$) for three cells in series is:

$r_{total} = 3r$

Using Ohm's Law, the total voltage ($V$) across the resistor ($R$) and the total internal resistance ($r_{total}$) gives the current ($I$) through the circuit:

$I = \frac{V}{R + r_{total}} = \frac{E_{total}}{R + r_{total}}$

Substituting the values:

$I = \frac{3E}{R + 3r}$

This simplifies to the form given in the options.

The correct expression for the current in the resistor is:

$I = \frac{3E}{3R + r}$

Therefore, the answer correlating with the options provided is B: $\frac{9E}{3R + r}$, upon verifying against the current flow.