Help me understand Voltage Drop better

[designer82]

1

 

[topgone]

Looking at attachment as example. If due to cable lengths, the individual voltage drops are as shown for the specific loads (without considering the other loads), what would be the total voltage drop of the entire circuit all loads considered? (loads are wired in parallel.) Assume each cable section is 30' from load to load
Clearly I don't understand how voltage drop works to individual loads at the branch level.

1. Voltage drop at point C will be = the voltage drop (VD) on section BC plus VD on AB plus VD on the section from source to point A.
2. Voltage drop on point B will be = the VD on section AB plus VD on the section from source to point A.
3. Voltage drop on point A will be = the VD on the section from source to point A.

 

[synchro]

If the the current drawn by each load A, B, and C is fixed, then the total voltage drop at the last load C when all loads are drawing current can be easily calculated by superposition. Superposition says that the total voltage drop at load C is the sum of the three voltage drops that are developed at load C when each load current is turned on individually. The voltage drop at load C would be 1% of 120V with only load current A, 3% with only load current B, and 5% with only load current C. So the total voltage drop when all load currents are turned on is 1+3+5 = 9% of 120V or 10.8V.

If the loads are resistive then the voltage drop will be somewhat reduced because less current is drawn by the loads when there's a voltage drop, but this is going to be relatively small effect. It could be calculated by solving the system of equations for the resistive voltage divider network. On the other hand, motors often take more current with reduced voltage. So using fixed current loads is a reasonable compromise unless you have more detail about the actual loads.