In the following network of 5 branches, the respective current are `l_(1),l_(2),l_(3)` etc. given that `l_(1)=-0.5A,l_(4)=1A` and `l_(5)=0.5A`, the remaining currents are

Options

A. `I_(2)= - 1.5 A, I_(3)=0.5A, I_(6)=0.5 A`
B. `I_(2)=1.5 A, I_(3)= -0.5 A, I_(6)=0.5 A`
C. `I_(2)=1.5 A, I_(3)=0.5 A, I_(6)= -0.5 A`
D. `I_(2)=1.5 A, I_(3)=0.5 A, I_(6)=0.5A `

Answer

B

Explanation

To find the currents in the various branches, we use Kirchhoff's current law.
At `C,I_(1)+I_(2)=I_(4)`
`therefore I_(2)=I_(4)-I_(1)=1-(-0.5)=1.5A`
and at the junction D,
`I_(3)+I_(4)=I_(5)`
`therefore I_(3)=I_(5)-I_(4)=0.5-1= 0.5 A`
and `I_(5)=I_(6)=0.5A` as the current is not divided at A
Thus `I_(2)=1.5 A, I_(3)= - 0.5A and I_(6)=0.5A`