In the following networ, the current folowing through `15 Omega` resistance is

Options

A. `0.8A`
B. `1.0A`
C. `1.2A`
D. `1.4A`

Answer

C

Explanation

In fig `15Omega & 3Omega` are in series
`therefore R_(s1)=18Omega`
Now `20 Omega & 4Omega` are in series
`therefore R_(s2) =24 Omega`
Now current through `15Omega` resistance is given by
`I_(1)= ((R_(S2))/(R_(S1)+ R_(S2)))I`
`I=(24)/(42) xx2.1 =1.2A`