As the switch ‘S’ is closed in the circuit shown in the figure, the current passing through it is:

A) 5 A
B) 6 A
C) 3 A
D) zero

Hint : In this solution, we will use Kirchhoff's current law which helps us relate the current entering and leaving a junction. Further using ohm’s law we will be able to determine the current passing through the switch.

Formula used: In this solution, we will use the following formula:
Kirchhoff’s current law (KCL): $ \Sigma i = 0 $ at a junction
Ohm’s law: $ V = IR $ where $ V $ is the potential difference between two points between which current $ I $ is flowing through a resistor $ R $

Complete step by step answer
Let us start by assuming a potential $ V $ at the junction of the three cables as shown in the image below:

Let us also assume that the branch of 20 V current will have current $ {i_1} $ flowing through it, the branch of 5 V will have current $ {i_2} $ and the grounded branch will have current $ {i_3} $ . Then from Kirchhoff’s current law, we can write
 $\Rightarrow {i_1} + {i_2} - {i_3} = 0 $
Then from ohm’s law, we can write $ i = \Delta V/R $ where $ \Delta V $ is the potential difference between two points of the branch
 $\Rightarrow \dfrac{{20 - V}}{2} + \dfrac{{5 - V}}{4} - \dfrac{{V - 0}}{2} = 0 $
 $\Rightarrow \dfrac{{40 - 2V}}{4} + \dfrac{{5 - V}}{4} - \dfrac{{2V}}{4} = 0 $
Taking an LCM of 4 we get,
 $\Rightarrow \dfrac{{40 - 2V + 5 - V - 2V}}{4} = 0 $
 $ \Rightarrow 45 - 5V = 0 $
Solving for $ V $ , we get
 $\Rightarrow V = 9\,V $
Then the current through the switch will be
 $\Rightarrow {i_3} = \dfrac{{V - 0}}{2} $
 $ \Rightarrow {i_3} = 9/2\,A \approx 5A $
Hence the current flowing in the switch will be $ 5A $ which corresponds to option (A).

Note
Here we can intuitively observe that the potential of the junction point will be lower than the potentials at point A and B since current will be flowing from A and B towards the grounded branch and as a result, there will be a potential drop across the resistors causing the junction to have less potential than A and B. The sign of $ {i_3} $ will be opposite while using the KCL as it is flowing away from the junction and not towards it.