BY GEORGE LANTHIER

Everybody teaches Ohm’ s Law while teaching electricity on and around oilburners and HVAC equipment, but what practical use does it really have in the field? The following is taken from our new book Fundamentals of Oilburner Wiring, Figure 1, which is only available to our Basic Oilheat School students or by special mail order. It was written to accompany the NORA OTM in teaching.

So, if you remember your basic electricity, you know that most of the electrical work we do can be verified using Ohm’ s Law. Ohm’ s Law is not a math formula, but rather a way of determining a missing unknown. In Figure 2 a mind jogger is shown for Ohm’ s Law. Using this and the following established formulas, we have what we need to calculate electrical requirements. I = E / R States that the current( I) in amps equals the voltage divided by the( R) resistance in ohms. R = E / I States that( R) resistance is equal to the voltage( E) divided by the( I) current. E = IR States that( E) voltage is equal to the( I) current times the( R) resistance.

Now if you also use the‘ pie’ in Figure 1, you will see that the missing factor can be determined by placing

Figure 1

Figure 2

your finger over the missing element. What is left is the formula you need. In Figure 3 we show a chart for the AWG and the current carrying capacities and resistances of wire. In Figure 4 we show a gaugetool for measuring the wire. Did you notice that the resistance of the wire goes up as the temperature increases? This fact is something to keep in mind around oilburners and heating equipment.

As you can also see, unless we go over 20 amps of current 14 AWG wire will do the trick. When you calculate your loads, as we will do, the amperage will determine your fuse size and your wire size.

Did you also note that for every one thousand feet of 14 AWG wire, you have 2.57 ohms of resistance? Hey, let’ s try Ohm’ s law on that.

Figure 3

Figure 4

115 volts % 2.57 ohms equals 44.75 amps.

Well, with that much carrying capacity, it looks as if you could put a lot of equipment on this line, but wait, is that all there is? No way.

What about the resistance’ s of the burner components. Let’ s try Ohm’ s Law on them and also old school components( pre-2000) versus new school.

Old oilburner motor = 115 volts % 3.2 amps equals

35.9 ohms. Ignition transformer = 115 volts

% 2.3 amps equals 50 ohms.

New oilburner motor, Figure 5 = 115 volts % 1.9 amps equals 60.5 ohms.

Electronic Ignitor, Figure 6, = 115 volts %. 724 amps equals 158.8 ohms.

These examples prove that the

PSC( Permanent Split Capacitor) and the electronic ignitor have

Figure 5 higher resistances, but use lower amperage making them much more efficient electrical devices.

To calculate the total circuit, you would also have to add any oil valves, blower motors, relays, circulators, and all other‘ loads’ to the formula. Let’ s try using Ohm’ s Law to calculate some data on a three-zone circulator type system with small circulators, Figure 7, and a typical oilburner.

The first thing we can establish is the voltage, 115 volts, right? We also know the amperages draw of the oilburner, we can get that by either totaling the amperage stampings on the loads like we did earlier, or we can get it from the manufacturer’ s literature. Next, we can take the total of the circulators; 0.54 amps per circulator X three circulators = 1.62 amps. Older units from the same company drew 0.75 each X three

26 OCTOBER 2025 | FUEL OIL NEWS | www. fueloilnews. com