Given that a current of 5.0 A passes along the branch from C to B is fig What is the voltage of points A,D,and G?
a. `V_A = …………` , b. `V_D = …………….` c. `V_G= …………..`

Explanation

Givne `I_(2) - I_(3) = 5`
in `ADC, - 7i_(1) - 4i_(3) +2 = 0 (i)`
in `FAEBC, 2 + 3(i_(1) - i_(3)) - 29 - 4i_(3) = 0`
or `-27 +3i_(1) -f 7i_(3) = 0 (ii)`

`(i)xx3 +(ii)xx7 gives i_(3) = 3A`
Putting in Eq (i) we get `i_(1) = 2A`
Putting in Eq. (A) we get `i_(2) = 2A`
in `ACB, V_(A)-V_(B) = 2 - (-3xx4) or V_(A) - 0 = 14V`
a. `|V_(A)| = 14 V, V_(A) - V_(D) = 7xx2 = 14 `
b. `V_(D) = 14 -14 = 0`
`V_(D) - V_(G) = 6xx(i_(1) - i_(2))= 6(2-2)` or `V_(G) = 0+V_(D)`
c. `V_(G) = 0V`