When the switch S, in the circuit shown, is closed, then the value of current I will be :

Options

A. 3A
B. 4A
C. 2A
D. 5A

Explanation

Let potential of point C is `V_c`, after closing the switch
Using KCL at point C,
`(20-V_c)/2+(10-V_c)/4=(V_c-0)/2 implies 20-V_c+(10-V_c)/2=V_c implies 40-2V_c+10-V_c=2V_c`
`implies 50=5V_c implies V_c=10 vo l t therefore I=V_c/2=10/2=5A` i.e., Option (D)