A real battery, having nonnegligible internal resistance, is connected across a light bulb as shown in Fig. Q 26.10. When the switch S is closed, what happens to the brightness of the bulb? Why?

Ohm's law states that the potential drop between two points of a component depends on the currents passing through the component.

V = IR

When S is closed, the voltage across the bulb and the resistance are the same and the resistance of their combination will reduce, hence the resistance in the circuit decreases and as given by Ohm’s law, the current in the circuit will increase as$R_{eq}$ decreases, this increase in the current will change the voltage drop between the terminals of the battery where the voltage drop in the battery is given as:

$V=\epsilon -lr$

Where the term lr decreases the voltage drop due to the internal resistance of the battery. So, the voltage across the bulb decreases, hence the brightness of the bulb decreases.

Thus, the brightness decreases.