The resistances in the four arms of a wheatstone network in cyclic order are `4Omega, 2 Omega , 6 Omega`and `15 Omega`. If a current of 2 A enters the junction of `4 Omega` and `15 Omega`, then the current through `2 Omega` resistor is:

Options

A. 2.0 A
B. 1.2 A
C. 1.6 A
D. 2.5 A

Answer

c

Explanation

Resistance of the upper arm = `6 Omega`
Resistance of the low arm = `21 Omega`

Current through the upper arm
`I_1 = (2 A xx 21 Omega)/(6 Omega + 21 Omega) = 1.6 A`