Consider the circuit shown in figure, the current `l_(3)` is equal to

Options

A. `5 A`
B. `3 A`
C. `- 3 A`
D. `-(5)/(6)A`

Answer

D

Explanation

For `l_(1)`, by Kirchhoff's voltage law
`28l_(1)=-6-8 implies l_(1)=(-14)/(18)=-(1)/(2)`
Here, `l_(1)` is -ve means opposite in direction as assumed.
` therefore l_(3)=-(l_(1)+l_(2))=-((1)/(2)+(1)/(3))`
` l_(3)=-((5)/(6))A`