Say it connects a bulb with the circuit below, with the switch $S$ open. Every cable has negligible resistance and the battery has no internal resistance.
What happens to the brightness of the bulb after closing down the circuit?
My thought is that after activating that third resistance, it will form a parallel pair of resistance with the second resistance $$\frac{1}{R}=\frac{1}{R_2}+\frac{1}{R_3}$$ means the resistance will decrease and the intensity will increase since $$I=I_1+I_2+\dots$$ causing the difference of potential in $R_2$ to be the same (since parallel resistors) and an increase in intensity, which means more brightness to the bulb? Not sure at all so I was wondering if anyone could shine a light on this.