# 240v Split Phase power

#### Mac Rodriguez

Joined Mar 24, 2016
140
I apologize to all for this question if has been repeated before (most likely) and is just reviving a easy answer.
Allaboutcircuits.com explains a split phase 240v circuit in the following version:

" An alternative solution would be to use a higher voltage supply to provide power to two lower voltage loads in series. This approach combines the efficiency of a high-voltage system with the safety of a low-voltage system...The current through each load is the same as it was in the simple 120 volt circuit, but the currents are not additive because the loads are in series rather than parallel. The voltage across each load is only 120 volts, not 240, so the safety factor is better. Mind you, we still have a full 240 volts across the power system wires, but each load is operating at a reduced voltage. If anyone is going to get shocked, the odds are that it will be from coming into contact with the conductors of a particular load rather than from contact across the main wires of a power system.."

Series connected 120 Vac loads, driven by 240 Vac source at 83.3 A total current.
If voltage in a SERIES circuit is additive why isn't there 240v across each load or at least on load #1

#### panic mode

Joined Oct 10, 2011
1,834
If there is 220 miles from San Diego to Santa Barbara and Long Beach is in between, why is it not 220 miles from San Diego to Long Beach AND another 220 miles from Long Beach to San Diego?

In electrical theory this is known as one of Kirchhoff's laws. (sum of all voltages...)

When loads are in series, they all get portion of the supply voltage but same current flows through both of them. When equal loads are connected in series, they will have equal voltage drops. if you connect five equal loads in series, each of them will only have 20% drop (1/5th of supply voltage). But if loads are not equal, say one of loads was to have 70% of supply voltage, other would have 30% but.... they would always add up to 100% of whatever supply voltage happens to be.

When loads are in parallel, their voltages are same but current leaving supply is sum of currents drawn by all loads.

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#### Mac Rodriguez

Joined Mar 24, 2016
140
If I use this chart

:

and add 2 or 3 more resistance R's to it I end up with 600 -720 V's in the "Total" & Volts box and 120v in each resistance R & Volts box. In a 240v circuit, it doesn't add up correct.

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#### BR-549

Joined Sep 22, 2013
4,938
What is the resistance of load #1?

Joined Jul 18, 2013
20,859
If in the pic in post #1 each load were connected in the normal way to a neutral C.T. at the junction of the loads, the current in the neutral would be zero anyway.
But unless one of the L1 or L2 was grounded there would be no earth protection.
Max.

#### Mac Rodriguez

Joined Mar 24, 2016
140
What is the resistance of load #1?
If I did my math right resistance should be 1.44 OHMS

#### Mac Rodriguez

Joined Mar 24, 2016
140
If in the pic in post #1 each load were connected in the normal way to a neutral C.T. at the junction of the loads, the current in the neutral would be zero anyway.
But unless one of the L1 or L2 was grounded there would be no earth protection.
Max.
I understand and sort of agree, but for this situation let's just PRETEND that this is the circuit.
The question is more about the voltage than the current.

#### BR-549

Joined Sep 22, 2013
4,938
Ok.....so load #1 has a resistance of 1.44 ohms and a current of 83.33 amps going thru it. What does ohms law tell you about the voltage drop across it?

The voltage drop comes from current, not the power supply.

#### Mac Rodriguez

Joined Mar 24, 2016
140
Ok.....so load #1 has a resistance of 1.44 ohms and a current of 83.33 amps going thru it. What does ohms law tell you about the voltage drop across it?

The voltage drop comes from current, not the power supply.
OHM's law tells me that the TOTAL resistance of the circuit is 2.88 ohms and IT multiplied by the series circuit total current of 83.33 amps it = 240vs.
I can see now how it works for the circuit I posted but (If I may) use this chart

:

and add 2 or 3 more resistance R's to it I end up with 600 -720 V's in the "Total" & Volts box and 120v in each resistance R & Volts box. In a 240v circuit, it doesn't add up to correct Total volts.

#### BR-549

Joined Sep 22, 2013
4,938
What happens to the current if you add another 1.44 ohms in series?

#### Mac Rodriguez

Joined Mar 24, 2016
140
Nothi
What happens to the current if you add another 1.44 ohms in series?
Nothing, current stays the same.
I now get 4.32 total series circuit ohms.
I multiply that Total by series circuit current of 83 amps and get 400Vs.
And if I used the chart I now would have 360Vs in the "Total" volts box.
I'm lost.

#### BR-549

Joined Sep 22, 2013
4,938
You sure are my friend. You will need to review basic DC and AC circuits. You need to get a feel for current. The reactance caused by current is much harder to understand.

#### panic mode

Joined Oct 10, 2011
1,834
Nothi
Nothing, current stays the same.
no...

you need to get better understanding of quantities in use here....
current is movement of charge PER unit of time (any interval of your choice). the more charged particles move through some place, per second, the higher the current is.
if you keep adding obstacles (which is what resistance is) that will make journey of those particles harder and more time consuming. in other words, in one unit of time, fewer charge particles will make the trip hence current is lower when higher resistance is used.

lets say charged particles are soldiers navigating obstacle course.
lets say course is clear (no obstacle) and time for slowest soldier to finish the course is T. this will be our reference.
without any obstacles, everyone will easily complete the course within that time (lots of people moving = large current).

the you start adding obstacles - one at a time and repeat test (give them time to rest too). obstacles could be whatever, climb over wall, crawl under the barbed wire etc.
with light obstacles MOST of them will still make in time T but few will take longer that. current is number of those that made it in time T which in this case is less than what we had without any obstacle.
the more obstacles you add, the fewer soldiers will manage to still complete course in time T (even lower current).
if you add really really difficult obstacle (i don't know, maybe a FORCE FIELD) then nobody will complete the course (current = zero).

working with electrical circuits is similar: voltage is driving force (motivation) that pushes current through.
resistance is course difficulty (higher resistance is tougher course : more length, more obstacles and obstacles of greater difficulty such as tighter crawl spaces)
current is not merely number of participants, it is only number of participants that manage to complete course (or part of it ) within given time.

in electrical circuit motivation is provided by power source, the higher the voltage the higher the motivation.
if voltage did not change (still 240V) but resistance is increased (we add another 1.44 ohms into our electrical "obstacle course"), current WILL be smaller.
individual elements are 1.44 ohm but total resistance in series circuit of three elements is 3*1.44 Ohm= 4.32 Ohms

now apply Ohms law...