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

Answer

B

Explanation

Given circuit is

Here, `l_(1)=-0.5A,l_(4)=1A,l_(5)=0.5A`
Applying Kirchhoff current law at junction, we get
`l_(1)+l_(2)=l_(4)`
`l_(2)=l_(4)-l_(1)=1A-(-0.5A)=1.5A`
Applying Kirchhoff current law at junction P, we get
`l_(3)+l_(4)=l_(5)`
`l_(3)=l_(5)-l_(4)=0.5A-1A=-0.5A`
also, `l_(6)=l_(5)=0.5A`