Michael Brown

[Michael Brown]

Thank you sensei 👍

[Michael Brown]

– What’s the distinction between voltage supply and voltage drop?

Voltage supply is a cause, voltage drop is an effect of work being done in a load. Kirchoff’s law tells us that the sum of voltage drops in a circuit will equal the supply voltage.

– Why would you measure voltage supply?

We would measure voltage supply because P=I*E. E meaning “electromotive force” aka voltage (supply). If we want to solve for P, we need the values for current and voltage.

– Why would you measure voltage drop?

We would measure voltage drop to ensure the load is working as designed. For example, if we have 1 heating element in a parallel circuit with no other loads, then the voltage drop should be 240 VAC. If we did not measure 240VAC, (Say 220 VAC for example) then that would be an indication of the existence of resistance elsewhere in the circuit when there should not be.

[Michael Brown]

Wait, you said to measure voltage supply, not voltage drop. To test voltage supply, I could pull the oven out and test the power cord terminals? (L1 to L2, should read 240 VAC right?)

[Michael Brown]

I used a clamp meter to measure the amps. I used a different meter to try and measure volts at the heating element. I tried inserting the test leads on the same side that the element is inserted to but I never was able to get a reading. I know it had voltage though bc the amp meter told me. Plus the element got hot. I couldn’t get positioned to see the opposite side of where the element is connected very well though. What process would you use to measure voltage drop across the element?

[Michael Brown]

Thank you

[Michael Brown]

Yes! I can see that. Thank you so much. 😀

[Michael Brown]

Yes ma’am, I have been watching the videos on full screen. When I try to zoom in on the slide, the text becomes too pixelated for me to read. I can’t clearly make out the illustrations either. I was hoping maybe a higher resolution image of the slide existed so I could see the illustrations also. But as long as it’s not super critical information I think I’ll be ok with just taking notes on the audio. Thank you.

[Michael Brown]

So
R=E/I
Z=E/I
R=Z
So R (In Ohms law) always really means impedance. It’s just that sometimes (non-ohmic loads?) the only aspect of impedance actually is resistance?
In loads with significant reactance, would Ohms law become

I=E/Z

E=IxZ

P=I^2xZ

?

[Michael Brown]

If you can measure the voltage supply and the current in a circuit, you know the wattage, P=I*E.

Yes but isn’t it also true that if you can measure the voltage supply and the current in a circuit, you know the resistance, R=E/I?

[Michael Brown]

Oh, my brain was hurting so bad there for a minute. Thank you for clarifying that. I’ll look forward to learning more about Ohmic and non-ohmic loads as the course progresses. In the meantime, would I be able to determine that my dryer heating element is operating in spec with the following tests and information?

I set to dmm Amps AC and clamp the red wire which leads to the heating element (Current first travels through a thermal cutoff and 2 thermostats but these are not loads though and shouldn’t affect the current draw). Timed dry, high heat setting on the rotary switch. Push to start. Amp reading is 23 Amps.
P=IxE P=23×240 P= 5,520 Watts. (If the heating element wattage was listed in the schematic, I would compare my product to that. If I was within +/-10% of the listed wattage, we’d call the element good?)
But, in the dryer schematic, the wattage specs for the heating element is not listed. It only lists a range of 7.8-11.8 Ohm resistance for the heating element. (I realize Watts would be a better measurement but I have to use the specs that I am given).
P=E^2/R P=240^2/7.8 P=7,385 watts
P=E^2/R P=240^2/11.8 P=4,881 watts
If I did this correctly, then the power expended on the heating element should be between 4,881 watts and 7,385 watts. 5,520 watts is in that range so it appears as if the heating element is functioning according to specifications. Thank you sensei (Apologies for the long questions).

[Michael Brown]

From blogpost:

“ohmic loads. These are things like light bulbs and heating elements”

However, you’re saying that light bulbs and heating elements are actually non-ohmic loads?

Should I interpret this to mean that light bulbs and heating elements could be either ohmic or non-ohmic depending on their level of reactance?
Or is one statement true and the other not?
Thank you.

[Michael Brown]

I see. So a motor is non-ohmic but a heating element is ohmic. I set to dmm Amps AC and clamp the red wire which leads to the heating element (Current first travels through a thermal cutoff and 2 thermostats but these are not loads though and shouldn’t affect the current draw). Timed dry, high heat setting on the rotary switch. Push to start. Amp reading is 23 Amps. P=IxE P=23×240 P= 5,520 Watts. In the dryer schematic, the wattage specs for the heating element is not listed. It only lists a range of 7.8-11.8 Ohm resistance for the heating element. (I realize Watts would be a better measurement but I have to use the specs that I am given).
P=E^2/R P=240^2/7.8 P=7,385 watts
P=E^2/R P=240^2/11.8 P=4,881 watts
If I did this correctly, then the power expended on the heating element should be between 4,881 watts and 7,385 watts. 5,520 watts is in that range so it appears as if the heating element is functioning according to specifications. Is my thinking correct here? Thank you sensei.

[Michael Brown]

“You almost always want to use LoZ for VAC testing, including this situation.”

What would be a situation where you would want to use VAC instead of LoZ?

I have both a fluke and a field piece digital multimeter but neither one has a LoZ setting on the meter.

[Michael Brown]

So, you’re saying sweep the leg?
Jk. Thank you sensei.

[Michael Brown]

Sorry, this is another purely hypothetical scenario.

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