Consider the circuit below. The bulb will light up if:

A. ${S_1}$ ${S_2}$ and ${S_3}$ are all closed
B. ${S_1}$ is closed but ${S_2}$ and ${S_3}$ are open
C. ${S_1}$ and ${S_3}$ are closed but ${S_2}$ is open
D. None of these.

Hint: Rate of flow of charge is current. The resistors can be connected in series or in parallel with the battery. In case of series connection the current flowing through all the resistors will be the same. In case of parallel connection current flowing will be divided into branches according to their resistances.

Formula used:
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$

Complete answer:
A material which allows current to pass through it is known as a conductor. No conductor will be perfect. It will have some resistance. The property to hinder the flow of current is called resistance and a device which does that is known as a resistor.
If the same current is passing through all resistors then we tell those are connected in series. If potential difference is the same for all resistors then those resistors are told to be in parallel.
When we had connected the iron and toaster and refrigerator in parallel, the potential difference would be the same across all the devices. Only the current flowing through the circuit will be divided. When resistors are connected in parallel effective resistance will be $\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$
In the below diagram

If switch one and switch two are closed then both will be having zero resistances and they will be in parallel. Effective resistance will be
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$
$\eqalign{
  & \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{1}{0} + \dfrac{1}{0} \cr
  & \therefore {R_P} = 0 \cr} $
So when resistance is zero current will go through that path we call it as short circuited.
Now if the third switch is closed then that wire will be short circuited and the entire current passes through that short circuited wire and no current passes through the bulb. So the third switch must be open.
So when ${S_1}$ is closed but ${S_2}$ and ${S_3}$ are open the bulb will glow
This condition is only satisfied by the second option.
So option B be the answer.

Note:
The main property of current is it always chooses the least resistant path. So when we short circuit any wire, its resistance will become zero i.e the least we can get. Hence current will flow through that path and don’t flow through the bulb. Then the bulb will not glow. Even if we close the first and second switches and open the third switch then the bulb will glow.