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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.
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  • \$\begingroup\$ Please don't keep adding units in parenthesis after every item in an equation. It makes it difficult to read. \$\endgroup\$ – Andy aka Apr 7 at 14:30
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    \$\begingroup\$ Ohm's law is silent on power and indeed does not even mention 'resistance' although that is the name we give to the proportionality of voltage and current. I am not going to derive power from first principles here but (for non reactive circuits) is indeed P = V*I \$\endgroup\$ – Peter Smith Apr 7 at 14:41
  • \$\begingroup\$ The standard, I think, is to remove the italics from the units but it generally doesn't read well. e.g., \$ P \ \text W = V \ \text V I \ \text A \$ \$ P \ \text W = V \ \text V I \ \text A \$. Using the European 'U' for voltage makes sense in this context. Note that for HTML you can use <sup>...</sup> and <sub>...</sub> for super and subscript. \$\endgroup\$ – Transistor Apr 7 at 16:29
  • \$\begingroup\$ Thank you for the tip, Andy. Really bad for reading. Peter, non-reactive component is a device that obey Ohm's law perfectly? Transistor, my bad for the huge units. Thank you for the tip on how to write it. \$\endgroup\$ – Atila Coimbra Apr 7 at 21:18
  • \$\begingroup\$ In this question's context, the terms are the same and P only refers to heat loss or power source. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 8 at 1:35
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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.

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  • \$\begingroup\$ No, GPL is not derived from OPL. The GPL is, as its name suggests, more general. And it can tell you how much heat is produced if the load is resistive. \$\endgroup\$ – Elliot Alderson Apr 7 at 19:09
  • \$\begingroup\$ Well, as I said if the load is pure resistor, then yes it will only produce heat (in this case 1W), Now if load is a light bulb, it would draw 1W but not everything will be converted to heat. Also, I remember learning Ohm's law pie chart which shows you how to derive equations from one to another, is that incorrect? Are you implying that Ohm's law is derived from GPL, or something different? \$\endgroup\$ – Healow Apr 7 at 19:31
  • \$\begingroup\$ No, I'm saying that Ohm's Law describes the relationship between current, voltage, and resistance for an ideal resistor. The Power Law describes the relationship between current, voltage, and power for any element, not just a resistor. There is no need for either of them to be derived from the other. Just because you saw a pie chart you can not assume that the chart revealed the history of Ohm's Law and the Power Law. What the OP called "Ohm's Power Law" is just a combination of the two, not a derivation. \$\endgroup\$ – Elliot Alderson Apr 7 at 19:54
  • \$\begingroup\$ I think I understood: - The GPL is for any element, not just resistors, but you need I and V; - The OL is designed for ideal resistors (since you have I and V) but you can apply it to real devices, but you need R and I; - The OPL is for any element (like resistors) but now you have the advantage of calculation with I, V and R. If that's true, using any of these three equations I should get on same result (knowing my variables). Correct? \$\endgroup\$ – Atila Coimbra Apr 7 at 21:43
  • \$\begingroup\$ Yes, you are correct. When you are dealing with linear circuits you should always get the same results. \$\endgroup\$ – Healow Apr 7 at 22:14
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At any instant, if a circuit is drawing current I from a voltage source of emf V, then this means the voltage source is delivering P = VI joules per second to the circuit at that instant.

How the circuit uses that energy is upto the circuit:
- If the circuit has an incandascent bulb, the energy is used to produce light + heat.
- If the circuit has a motor, the energy is used to do mechanical work + heat.
- If the circuit has a speaker, the energy is used to vibrate the diaphragm of the speaker + heat.

Note that power can be positive, negative or 0:
If VI is positive, then the voltage source is actually giving power to the circuit. However, if VI is negative then the circuit is giving power to the voltage source(battery charging circuit).

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  • \$\begingroup\$ Your last paragraph is confusing, as you haven't clarified how the passive sign convention is involved in determining the sign of the power. \$\endgroup\$ – Elliot Alderson Apr 7 at 19:11
  • \$\begingroup\$ That's one of my questions: why did you choose P = VI instead of P = IR? Is it because you assumed that you have V and not R in this example? \$\endgroup\$ – Atila Coimbra Apr 7 at 21:48
  • \$\begingroup\$ @AtilaCoimbra Voltage source and Resistance are different things. this khan academy short tutorial on current, voltage and power explains it way better than I ever could. Pls see if it helps... else I'll be happy to update my answer.. \$\endgroup\$ – beccaboo Apr 7 at 22:20
  • \$\begingroup\$ While studying power in circuits, it helps to revise 3 physics concepts: Force, Work, Energy. @AtilaCoimbra \$\endgroup\$ – beccaboo Apr 7 at 22:25
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    \$\begingroup\$ Loads output positive power in Watts and Generators or Power Sources are "negative" thus the sum of power x time = energy =0 as usual. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 8 at 1:23
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Your understanding is mostly correct. Power is always equal to V X I as long as the values of V and I occur at the same instant in time. If you want to determine the average power for alternating current it is a little more complex. The power into a device is V X I. It is difficult to determine how much of the power is useful. As you said, you need to measure the output.

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  • \$\begingroup\$ So the Ohm's Law is valid only for direct current circuits? And in AC circuits I have to use the GPL? \$\endgroup\$ – Atila Coimbra Apr 7 at 21:50
  • \$\begingroup\$ Ohms law is really only about DC and resistance. However there are numerous ways that it is extended to power and AC circuits, inductance, capacitance and non-linear circuits. I agree with Edin Fifić's answer about the book that you are using and the need to use a course of study that leads you from one concept to another in an orderly manner. \$\endgroup\$ – Charles Cowie Apr 8 at 3:09
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  • To answer your questions very briefly:
    1) No. Incorrect.
    2) Somewhat correct.
    3) GPL and OPL talk about different things but use the same units for power.
    • I think you got the wrong book to start learning electronics. The authors use too much unnecessary jargon and situations, making a simple description and an understanding of the basic principles hard to grasp for beginners. It's like mentioning multiplication, powers and logarithms in the same paragraph that explains numbers addition to the first graders, not even giving them a chance to process the absolute simplest basics of the concept of numbers and addition. It simply causes more confusion than clarity for someone just starting in this area.
      In short, get a book with a title like "Basic electronics" or "Basics about electricity", something to that effect. You first need to be clear about the basic concepts like voltage, current and power, and just then you can move on to power losses and efficiencies. One step at a time, or you will barely learn anything and comprehend even less.
    • FIRST, you need to start with DC (Direct Current, like the one from a battery), and I will talk in terms of DC here, to keep things simple and easy to grasp. Don't go beyond DC until you understand it.
    • SECOND, you need to understand that a power in a circuit/device (be it a motor, a heater, light bulb, whatever) is calculated by multiplying the voltage applied to that circuit/device with the current flowing through that circuit/device.
      That is the basic formula for power: P=VxI.
      This formula is used for determining how much power does a circuit/device take/use from the power supply and is only concerned with the electrical power and not other forms of energy, so just focus on one thing at a time. No need to exhaust yourself with all other forms of energy while not grasping the one you're learning about.
    • THIRD, learn and comprehend the PVI (power, voltage, current) triangle, and focus only on those 3 units while learning it.
      Do the same with the VIR (voltage, current, resistance) triangle. In both cases, get to fully understand relationship between the 3 units before trying to expand on their many possible applications. Your head will hurt less, and you will start to get the picture.
    • FOURTH, you are trying to learn and understand PRACTICAL electronics while you don't understand even the basic concepts. There is no shame in being a beginner in something and not immediately understanding even a small part of it. We have all been there. The key is "baby steps". Just because we are adults, it doesn't automatically mean that we can be faster at learning and understanding something we knew nothing about before.
      While our life's previous knowledge and experiences can help us learn and understand some things faster and easier, sometimes our preconceived notions and self-confidence can be a hindrance.
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  • \$\begingroup\$ Thank you, Edin. Yes, I'm thinking in changing books. I'm not advancing in this subject until I fully understand and comprehend these basic Laws. \$\endgroup\$ – Atila Coimbra Apr 7 at 21:55
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I think the author has basically just muddied the waters. The instantaneous power delivered to a load is simply the instantaneous voltage times the instantaneous current.

P(t) = I(t) * V(t)

where P is power, I is current and V is voltage. V is measured across the load and I is measured through (in series with) the load.

If the power is a repeating cycle (for example if the voltage is a sine wave) then the average power can be calculated over one full cycle using the average value theorem from math. If the power is very irregular with time, then you could just pick a long time frame and compute the average over that longer time frame.

The load will pretty much always convert some of the power to heat. But if you consider a motor, maybe 80 or 90% of the power will be converted to kinetic energy. For an LED, maybe 20% of the power will be converted into light. If you consider a battery charger, maybe 95% of the power will be converted to stored energy in the battery.

So power delivered to the load is not the same as power loss.

The other place I have to take issue is with sine wave power and power factor. If you follow my text above, and calculate average power from the instantaneous power, then you needn't worry about power factor. But if you use the RMS current and RMS voltage in a sinusoidal AC system, then you have to know or estimate power factor in order to calculate power. In that case, you have:

Pav = Irms * Vrms * cos(theta)

Where Pav is the average power, Irms is the rms current, Vrms is the rms voltage and theta is the angular distance between voltage and current waveforms. In other words, theta is the phase angle between current and voltage. This term, cos(theta) is also known as "power factor." As before, voltage is measured across the load, and current is measured in series with the load. A simple volt meter (well, a digital multimeter) can measure voltage and current but it cannot tell you the phase angle between voltage and current, so cannot help you figure out the actual power (also known as real power). If you have experience, it might be reasonable to estimate power factor based on what the load is. For example a small single phase motor will be around roughly 0.8 power factor. An electric heater of any kind will be 1.0. Etc.

The rules for RMS current and voltage are perfectly compatible with the instantaneous rules. It is a just a useful shortcut when you know the voltage and current are sinusoidal. But you need to include power factor.

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    \$\begingroup\$ Thank you. Gonna take this when I hit AC. \$\endgroup\$ – Atila Coimbra Apr 7 at 22:04

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