In this circuit, the value of `I_(2)` is-

Options

A. 0.2 A
B. 0.3 A
C. 0.4 A
D. 0.6 A

Answer

C

Explanation

`V = IR` as V = constant
`I prop 1/R`
`I_(2) = (1/R_(2))/(1/R_(eq)) xx I = (1/R_(2))/(1/R_(1) + 1/R_(2) + 1/R_(2))I`
`= [1/15/(1/10 + 1/10 + 1/30)] xx 1.2 = 0.4 A`