voltage drop calculations single phase ac
The part you're missing is that the return wire (neutral) has the same resistance as the red wire in your drawing. Let's add two points: C is right below the load, and D is right below the source. Assuming your example where 2 volts drop on the red wire, you have 218V at point B. Now here's the catch: the neutral wire is only at 0V at the source. Since the same current flows through the same length of wire as the positive side, you'll get a 2V drop there too. So point C would be 2V. So you're losing 2V in the red wire, and 2V is the black wire. The potential across your load is 218V-2V = 216V, which is 4V less that the source. That's why you multiply by 2.







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