Consider the circuit shown in the figure. What is the value of the current `I_(3)` ?

Options

A. 5 amp
B. 3 amp
C. `-3` amp
D. `-5//6` amp

Answer

D

Explanation

At point E, the current `I_(3)` will be divided into `I_(1) and I_(2) as I_(3)=I_(1)+I_(2)`
Then applying Kirchhoff's voltage law to EFABE,
`-8-28I_(1)-6=0`
`therefore -28 I_(1)=14 therefore I_(1)= -(1)/(2)A`
and for the loop EDCBE,
`-12-54I_(2)-6=0`
`therefore 54I_(2)= 18 therefore I_(2)= -(1)/(3)A`
`therefore I_(3)=I_(1)+I_(2)= -(1)/(2)-(1)/(3)= -(5)/(6)A`