#### electrofelon

##### Senior Member

- Location
- Cherry Valley NY, Seattle, WA

- Thread starter electrofelon
- Start date

- Location
- Cherry Valley NY, Seattle, WA

- Location
- Fairmont, WV, USA

As a first approach I take would be to sum all the loads and calculate the drop as though they were at at the farthest end of the run. If the answer is an acceptable voltage drop, then you are done.

Then I would take your approach and see if the voltage drop was clearly below the maximum allowed. If it was near the max, I might just up the wire by a size (assuming conduit installation). The cost of the wire is swamped by the cost of the devices.

To correctly calculate the voltage drop you need to calculate the drop to each load in the circuit. Take the first load, calculate the drop using total current. Then use the voltage at this point and the current for all but the first for the next drop. Rinse, Repeat, Rinse.

Then I would take your approach and see if the voltage drop was clearly below the maximum allowed. If it was near the max, I might just up the wire by a size (assuming conduit installation). The cost of the wire is swamped by the cost of the devices.

To correctly calculate the voltage drop you need to calculate the drop to each load in the circuit. Take the first load, calculate the drop using total current. Then use the voltage at this point and the current for all but the first for the next drop. Rinse, Repeat, Rinse.

Last edited:

- Location
- Cherry Valley NY, Seattle, WA

But that method would result in a vd of about twice the actual. Also if the devices are powered by both ends of the loop as I asked in the op then your method results in 4 times the vd. No small difference.As a first approach I take would be to sum all the loads and calculate the drop as though they were at at the farthest end of the run. If the answer is an acceptable voltage drop, then you are done.

Then I would take your approach and see if the voltage drop was clearly below the maximum allowed. If it was near the max, I might just up the wire by a size (assuming conduit installation). The cost of the wire is swamped by the cost of the devices.

To correctly calculate the voltage drop you need to calculate the drop to each load in the circuit. Take the first load, calculate the drop using total current. Then use the voltage at this point and the current for all but the first for the next drop. Rinse, Repeat, Rinse.

- Location
- Massachusetts

The reason we go class A is to make sure the system will still work with an open wire.But that method would result in a vd of about twice the actual. Also if the devices are powered by both ends of the loop as I asked in the op then your method results in 4 times the vd. No small difference.

That being the case it is possible that all the devices would have to be supplied from one direction. I would base my VD on that worst case.

- Location
- Cherry Valley NY, Seattle, WA

Of course. Got it thanksThe reason we go class A is to make sure the system will still work with an open wire.

That being the case it is possible that all the devices would have to be supplied from one direction. I would base my VD on that worst case.

- Location
- New Jersey

- Occupation
- Professional Engineer, Fire & Life Safety

- Location
- Cherry Valley NY, Seattle, WA

I dont buy that. Lets expand/clarify with your example and say the last device is X distance from the FACP and the return (class A) is also X. I dont see any possible way the conductor length would be 4X. I admit I do not know the specific power scheme for a class A loop: if both supply and return terminals are common or isolated supplies but either way how could we have 4X and not X for conductor length?

- Location
- New Jersey

- Occupation
- Professional Engineer, Fire & Life Safety

Please see the System Sensor application guideI dont buy that. Lets expand/clarify with your example and say the last device is X distance from the FACP and the return (class A) is also X. I dont see any possible way the conductor length would be 4X. I admit I do not know the specific power scheme for a class A loop: if both supply and return terminals are common or isolated supplies but either way how could we have 4X and not X for conductor length?