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I posted a question earlier that helped me to understand Ohm's Law and how power is lost to length of wire due to resistance. So my original example was a 15A, 10ft cable surge protector plugged into a 15A 6 foot cable plugged into a extension cord 13A 10ft which then has the appliance's cable at 8 ft. The max power this device would need is 110 watts.

We worked out Ohms law: 𝑃=0.922𝐴×0.434Ω=0.367𝑊

I do not understand how the 0.367 W applies to the power needed (110 watts)? Does this mean if the TV needs to pull at max and requests 110 watts it will only receive 109.64? This is safe and will it not damage any of the TVs components, I heard under powering is very bad.

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  • \$\begingroup\$ Underpowering is not necessarily very bad. Whatever voltage the TV receives would probably still be within its specified level. What might end up happening is that the TV would draw slightly more current to make up for the lower voltage (if it needs a constant power). Eventually, if the voltage is low enough, it would try to draw too much current and wires/components could overheat, but this would likely not happen in your example. \$\endgroup\$ – Justin Nov 21 '19 at 16:37
  • \$\begingroup\$ I might be rusty on terminology.. But I don't think it is Ohms law. \$\endgroup\$ – Eugene Sh. Nov 21 '19 at 16:39
  • \$\begingroup\$ it's I²R, not IR. And that's not Ohm's Law, though I forget which one it is. \$\endgroup\$ – Hearth Nov 21 '19 at 17:14
  • \$\begingroup\$ @Hearth Some refer to it as Watt's Law, but I am not convinced by the sources. I don't think Watt was the one who came up with it. \$\endgroup\$ – Eugene Sh. Nov 21 '19 at 17:31
  • \$\begingroup\$ I've typically considered it as Ohm's Law (E=IR) that incorporates electrical power (P=IE). Once you do that, you have "Watt's Wheel" which is all of the various permutations. I'm not sure either who is credited for the relationship. \$\endgroup\$ – JYelton Nov 21 '19 at 17:36
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Most appliances have regulated power supplies, meaning that they adjust the resistance they present to the power cable to ensure that they always get the same power.

However for devices without regulation, the device really will get less power. For example, people often expect to be able to string large numbers of 5v LED strips together, only to find that the further diodes get dimmer and eventually stop lighting up all together. This is due to the resistance of the wiring limiting how much power the more distant lights get.

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  • \$\begingroup\$ So if I have a new Samsung tv from 2018-2019 it will not get damaged from my given situation? Thank you for clarifying \$\endgroup\$ – Luke F Nov 21 '19 at 17:19
  • \$\begingroup\$ @LukeF Hopefully you're not more confused. The calculations on the previous question treated the TV as a simple resistive load. The purpose was to illustrate that less voltage results in less power (with some lost to the cabling). In reality, as user1850479 says, the TV's power supply will compensate for a slight under-voltage. It would require more significant under-voltage to cause problems. (How much? It depends on the design. Check out this article from Newark about power supplies, particularly the "input" section.) \$\endgroup\$ – JYelton Nov 21 '19 at 17:53
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The TV is not a resistor so Ohm's law does not apply to it. It has a switch mode power supply so it will try to pull the power it needs, regardless of wiring resistance.

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  • \$\begingroup\$ The switch mode power supply is what user1850479 is referring to as well? \$\endgroup\$ – Luke F Nov 21 '19 at 17:13
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    \$\begingroup\$ Any regulated power supply will work like that. Switch mode is one common type of power supply, and the only one you would see used on electronics like TVs, but there are other types too. \$\endgroup\$ – user1850479 Nov 21 '19 at 18:00
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To some degree you are overthinking this problem.

Let's start with your original problem: what is the effect of multiple extension cords on a 120 VAC television which needs 110 watts.

If you assume that the television was actually drawing 110 watts, then the power dissipated in the extension cords was calculated to be .376 watts. How much voltage is this ? Well, it's .92 x .434 (since Ohm's law says that voltage is current times resistance), or.399 volts. To put it another way, you'll only get exactly 110 watts if the line voltage at the wall plug is 120 + .399, or 120.399 volts.

Is this an issue?

Well, no. I suggest you look at this site for an idea of what you can actually expect from a power line at your house. The short version is that Range A, 114 to 126 volts, is what you can reasonably expect. Also note Range B, which suggests that, for limited periods of time, Range B is acceptable, and this indicates that you can actually expect your house voltage to vary between 110 and 127 volts, although the upper and lower limits should not occur often.

Your TV manufacturer is (or should be) aware of this, and have designed the TV accordingly. For that matter, the manufacturer would be well-advised to do somewhat better, in order to avoid a bad reputation.

With a potential line voltage which varies by 12 volts, reducing that range by 0.4 volts simply isn't something which you should worry about. It's only about 3% of the variation which the TV already expects.

If, for some reason, this were to become a problem, you would have much bigger problems on your plate already, as your electricity provider would be experiencing major issues, and you would be experiencing brownouts on a regular basis.

Finally, as Justme has answered, your TV is not a resistor. Its power supply almost certainly adjusts its operation to accept voltages other than 120 volts exactly. As far as it is concerned, voltage drop (and power loss) in your extension cords is no different than a low voltage from the electric company, and it will do just fine.

What you do need to be aware of is that the power lost in the extension cords shows up as heat, and if you get the cords hot enough they can do nasty things like catch fire. Since your current is only about 1 amp, and all of your cords are 13 amps or better, you should have no worries at all.

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  • \$\begingroup\$ Well-stated to put things in perspective. Something I think I failed to do in other answers/comments. \$\endgroup\$ – JYelton Nov 21 '19 at 20:29
  • \$\begingroup\$ JYelton you did great I understand Ohms law and electricity so much better. Thank you everyone \$\endgroup\$ – Luke F Nov 21 '19 at 22:11
  • \$\begingroup\$ Just checked the cabling and it is 14AWG so does that mean that the resistance would be even less and the .376 watt calculated would be even less? Does this also mean the loss of volts would be less as well since the original calculation was made with 18AWG? \$\endgroup\$ – Luke F Nov 22 '19 at 18:00
  • \$\begingroup\$ @LukeF - Yes indeed. You do understand Ohm's law. \$\endgroup\$ – WhatRoughBeast Nov 22 '19 at 22:41

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