What will happen to the bulb of the circuit after closing the switch? 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.
 A: Yes, you are about right. According to resistors in parallel & series
Before closing switch
$R_1$ and $R_2$ are in series, current can't divide between them, so total current will be :
$$ I = \frac {V}{R_1 + R_2} \tag 1$$
After closing switch
Current will go though $R_1$, but then it will split between $R_2$ and $R_3$, meaning that your $R_1$ is in sequential connection of circuit, but $R_2$ and $R_3$ - in parallel circuit part, so total current will be :
$$ I = \frac {V}{R_1 + \frac {R_2 R_3}{R_2+R_3}} \tag 2,$$
Let's assume that $R_3=R_2$, so (2) will become as :
$$ I = \frac {V}{R_1 + \frac 12 R_2}  \tag 3,$$
And
$$ R_1 + \frac 12 R_2 < R_1 + R_2 \tag 4,$$
So in effect after closing the switch you will increase current, due to reducing total resistivity and hence bulb will go brighter. The overall result is when you add parallel circuit part,- it acts as if you would be reducing second resistivity $R_2$ in the previous circuit instead.
Of course, if your $R_3$ is very high, say $100R_2$, then reduction of total resistance will be very small, like $R_1 + 0.99R_2$, but principle idea holds.
