## Voltage drop vs Power loss confusion

Frequently asked question regarding resistors, potentiometers, types, requirements, ratings etc.

### Voltage drop vs Power loss confusion

Ok, so, this has been bothering me for a while now. I have recently become interested in amp circuits, and higher voltage pedal schematics (18v instead of 9v), and I noticed a lot of schematics use a low ohm resistor on the supply (as part of a supply filter and/or to prevent shorts from frying something), mostly 10R or 100R. For simplicity, I will use the 10R for an example. Let's say our circuit is 100R, with a 10R supply resistor, at 18v. Ok, so my thinking is that the power loss in the 10R is that which the whole circuit draws, which is (18/(100+10))*18=(18/110)=0.16363...*18=2.94534 watts. But then there is the (I^2)*R=P equation, which means that I would do this:

18/100=0.18
18/10=1.8
100*(0.18^2)=3.24 watts
10*(1.8^2)=32.4 watts

So that means that either the resistors drop 2.95ish watts total, which would be split to 1.475 watts each; or 32.4 and 3.24 watts separately, which averages to 17.82 watts, or 35.64 watts for the entire circuit.

2.95W or 35.64W for total draw?

Or could it be (bare with me. I'm still not all that steady calculating voltage dividers):

(10+100)/(10*100)=...

Scratch that equation...

100/(100+10)*V=.9090...*V=16.362 volts

The 100R drops 16.362 volts. That means that the 10R drops 18-16.362=1.638 volts. That then give us:

(1.638/10)*1.638=0.263044
(16.382/100)*16.382=2.68269924

So that gives us a total dissipation of about 2.95 watts, which makes more sense with the total resistance, but that doesn't answer the dissipation of the individual resistances. So to summarize:

2.95W10R+100R or
35.65W10R+100Ror
0.263W10R and 2.683W100R or
1.475W10R+100R or
32.4W10R and 3.24W100R or
17.82W10R+100R

So my question is, how do I know which equation(s) to use to find the total draw and the individual draw?

Man, and I thought doing trig identities was fun...
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friedtransistor
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### Re: Voltage drop vs Power loss confusion

My understanding is that the circuit would be like this:
basic_ohm.png

Here are my 2 cents:

Method 1:

The current through the both resistors is the same and you'll get from

I = U / R = (U1 + U2) / (R1 + R2) = 18V / (100 + 10)ohm = 0.1636363 A

The voltage drop on each of the resistors is

U1 = I * R1 = 1.636364 V
U2 = I * R2 = 16.36364 V

Power on each resistor would be then

P1 = U1 * I = I^2 * R1 = 0.267769 W
P2 = U2 * I = I^2 * R2 = 2.677686 W

Method 2:

The total power used by the circuit is

P = U * I = U^2 / R = U^2 / (R1+R2) = 18 * 18 / 110 = 2.945455 W

We can divide the total resistance in 10 ohm unit, each eating the same power. This gives us 11 units, each one (which coincidentally is R1) using

P1 = P / 11 = 0.267769 W

Then the 100 ohm resistor is eating 10 times as much power, or 2.67769 W (don't mind the decimals).
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ggedamed
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### Re: Voltage drop vs Power loss confusion

Idk why, but my reply just deleted itself when I hit the submit button. To summarize what I said, Thank you so much! I now understand how this stuff is figured out. This is why I like talking to people instead of using google, since you can't expect google to understand a simple question. Plus, it's not worth my time to go through hundreds of pages of search results, when other people who actually know this stuff can help me out. The internet is for sharing info, but it just gets so mixed up that it no longer is helpful...
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friedtransistor
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