Electronic – Generalized Power Equation, Ohm’s Law and Ohm’s Power Loss

currentohms-lawresistancetheory

I'll try to state my questions precisely as I can.

I'm a beginner in Electronics and since I have a background in Physical Chemistry and basic Calculus, the book Practical Electronis for Inventors, 4th Ed. suited very well my needs. Not too deep in maths but also not superficial. And that's the book I'm using for.

On the part about the Generalized Power Law (GPL, for short), that's \begin{equation}P = IV \qquad(\text{W/A})\end{equation}
It's said that

provides a general result, one that is independent of type of material
and of the nature of the charge movement […] The generalized power law
can be used to determine the power loss of any circuit, given only the
voltage applied across it and the current drawn, both of which can
easily be measured using a voltmeter and an ammeter. However, it
doesn’t tell you specifically how this power is used up

What I don't understand is exactly what he means about power loss. Is it about producing heat or another form of energy not intended for that device? So if that's true this power calculated its not the useful power produced by that device, but the difference between the input and the output?

  1. So this GPL can be applied only and only for this purpose of power loss calculation (aka heat production), correct?

Now the Ohm's Law (OL, for short):
\begin{equation} V = IR\qquad (\text{W}/\text{A}^2) \end{equation}

  1. Which we use if the resistor or another device has a linear relationship of \$I\$ and \$V\$ while \$R\$ is constant –or have a linear area on its graph, like LEDs, I think–, and then we can calculated its useful (?) power. Correct?

IF and ONLY IF all the power is converted to heat of something like that, the OL can be substituted on GPL, giving us the Ohm's Power Loss (OPL, for short)
\begin{equation}P = VI = V(V/R) = V^2/R \end{equation}
or
\begin{equation}P = VI = (IR)I = I^2R\end{equation}

The books says

In this form, the power lost due to heating is often called ohmic heating, Joule heating,
or \$I^2R\$ loss.

  1. So the GPL and the OPL are the same thing?…
    I'm days struggling to understand these equations and it's meanings but I just can't. May be stupid questions but I just can't understand who is who and what is what.

Best Answer

What I don't understand is exactly what he means about power loss. Is it about producing heat or another form of energy not intended for that device?

Couple of points of confusion here, generalized power law tells you TOTAL power draw of that circuit. Also the 2 equations you mentioned, GPL is just derived from Ohm's law, it's not some kind of equation that calculates total heat production. Both equations can tell you total power draw only.

Let me give you an example: suppose we have battery \$V=10\,\text{V}\$ and we hook it up to a resistor say, \$R=100\,\Omega\$. Total power draw of the circuit would be \$P=\dfrac{V^2}{R}=1\,\text{W}\$

Now, that \$100\,\Omega\$ resistor can be a light bulb, maybe heater element etc. It doesn't matter, total power draw of that circuit will be \$1\,\text{W}\$, what percentage of that is useful work, we don't know.

If your resistor is a heater element, most of the energy will be transformed to heat and that would count as useful work (that we expect from that).

If your resistor is a light bulb it would still produce a lot of heat, and that's something we don't want from a light bulb, hence now percentage of useful work will be much lower.