jaus tail
- 613
- 48
Homework Statement
Homework Equations
V = IR
And R becomes 2(r of one conductor) because of return path.
The Attempt at a Solution
V minimum is at third junction. At VD
Using KVL we get:
Drop at AB = (10 + 20 + 50) * 2 * 0.1 = 16 V
Drop at BC = (20 + 50) * 2 * 0.08 = 11.2
Drop at CD = (50)*(2 * 0.06) = 6
Total Drop = 33.2 V
VD = 220 - 33.2 = 186.8V
Why have they not included the 2 factor in their calculation?
In solved examples they have used factor of 2.
Aren't all dc distribution 2 wire as the return path cannot be ground for DC? So shouldn't the factor of 2 be there in the equation?