# Found some errors in some equations of electric power. What is the actual problem?

I think the following two equations are incorrect when $$I, R$$ refers to current and resistance respectively and $$V$$ refers to potential in the difference of potential of the two sides of the resistor

• $$P = I^2R$$
• $$P = \frac{V^2}{R}$$

Reason: Proof of these two equations is as following:-

$$P = \frac{W}{t} \text{[Here }W, t \text{ refers to work and time respectively}$$

Now, $$W = VQ$$ (where $$V$$ is the potential of the resistor, not the difference of potential of two sides) and $$Q$$ is the amount of chage moved. So the above equation can be written as, $$P = \frac{VQ}{t}$$ Now $$\frac{Q}{t}$$ is equal to $$I$$ and so $$P = VI$$

One thing very very important that in this equation $$V$$ refers to the potential of the resistor. Not the potential difference of the two sides In the proof of the first of the two equations what is done is value of $$V$$ is inserter in the equation from Ohm's law. So this value is $$IR$$. It is as following:

$$P = VI$$

$$\implies P = IR * I$$

$$\implies P = I^2R$$ (The first equation.)

Now one thing to consider that we inserted the value of $$V$$ from Ohm's law to the equation $$P = VI$$. But $$V$$ in Ohm's law is the potential difference between two side and just the value of potential in the equation $$P = VI$$. So they are different. How can we insert it in the equation? Is there any reason?

In case of second equation: We get the value of $$I$$ which is $$\frac{V}{R}$$ from Ohm's law. Again here $$V$$ is the potential difference of two side. Which we insert into the equation $$P = VI$$ where $$V$$ is just the value of potential, not the difference of two side like below:

$$P = VI$$

$$\implies P = V * \frac{V}{R}$$

Here $$V$$ on the left of the right side is potential and $$V$$ on the right of the right side is potential difference. So they can't be of the same symbol $$V$$ and so can't be squared which is done below.

$$\implies P = \frac{V^2}{R}$$

So we can see that wrong values of $$V$$ are inserted without caring of what they are. But the two equations are being used for years. So what is the actual solution to this problem?

The crux of your question seems to be this:

One thing very very important that in this equation V refers to the potential of the resistor. Not the potential difference of the two sides.

This seems to indicate some kind of confusion or misconception. There are not two different quantities of interest here, only one. People may sometimes loosely speak of "the potential of a resistor," but a phrase like this is just shorthand for the potential difference between one side of the resistor and the other.

If you're thinking that it should be an absolute potential rather than a difference, then you need to realize that absolute potentials are not well defined. All real-world measurements and phenomena depend on potential differences.

• Well, it is stuff from my late basic school in late 70s.... Near all people here probably know the answer.. :-) – Poutnik Aug 18 at 17:31
• @MdAshrafulIslam, the reality is that you are confused. The equations you mention have been published for MANY years, and they have been viewed and used by many people. If there was an error in them, it would have been found long ago. And BTW, note that Ben Crowell is the author of a physics book, and he has a PhD in physics. – David White Aug 18 at 17:52
• @BenCrowell my intention is not to find errors of this equations. I just want to know where I have made the mistake – Md Ashraful Islam Aug 18 at 18:22
• @MdAshrafulIslam: So why do people tell that when P=VI if you increase voltage by step up transformer then current flow will become low. But what then happens with the equation P=V2/R. The equation $P=V^2/R$ holds for a resistor. A transformer isn't a resistor. – Ben Crowell Aug 18 at 18:45
• @MdAshrafulIslam my intention is not to find errors of this equations You said “I think the following two equations are incorrect”, not “Why are the following two equations correct?” And you insulted Ben Crowell after he tried to help you understand them. – G. Smith Aug 18 at 19:23

Normally, after a good and complete answer like the one by @Ben Crowell I would not bother to post my own answer. However, since @Md Ashraful Islam seems skeptical, I thought that it would be worthwhile to him to hear another similar opinion.

One thing very very important that in this equation 𝑉 refers to the potential of the resistor. Not the potential difference of the two sides

This is incorrect. In the equation $$P=IV$$, $$I$$ is the current through a component, $$V$$ is the voltage across the same component, and $$P$$ is the power delivered to or produced by the component (depending on if the active or passive sign convention is used). The voltage across a component is also called the potential difference across the component. Thus the $$V$$ in the power formula is the same quantity as the $$V$$ in Ohms law.

Further details, including a derivation from Maxwell’s equations, can be found in section 11.3 of this online textbook from MIT: http://web.mit.edu/6.013_book/www/book.html

Also, “the potential of the resistor” is not a uniquely defined concept anyway. Generally the potential is different on either side of the resistor and varies throughout its length.

• Thank you! You helped me understood! Your answer is just perfect. My intention was not to tell that Ben Crowell's answer is wrong, rather was to know why does the fact of step up transmiter happen. I thought he would reply why. The style of my question may seem negative(Actually my language is not English and I'm not expert at it so). My intention was not negative – Md Ashraful Islam Aug 18 at 23:00

I used to do this experiment with 1st year engineering student.

Let's consider a resistor that is 10cm long, not covered in insulation with resistance of 10 Ohms. Something like a piece of coal paper perhaps. We will apply a voltage of 1V across both sides of the resistor. We now measure the voltage across the resistor or parts of it using a multi meter.

First of all, the multi meter has two probes. If you put a single probe anywhere on the circuit, nothing happens. There is no such thing as an "absolute" potential: you can only measure the potential difference between two points (or between a single point and some explicit or implicit reference, like earth ground).

If we put the the probes at both ends of the resistor we get 1V. That's the total voltage drop over the resistor and equal to the source voltage.

Now we hold one probe steady at one end of the resistor and slide the other probe from the other end towards the first probe. You will see that voltage will steadily decrease from 1V to 0V. Once you are in the middle it's 500mV. Once you down to 1cm, it's 100mV.

Now we can put a probe, at, 2cm and the other at 8cm. We read 600mV. It turns out that the voltage is simply proportional to the distance between the probes with 100mv/cm.

And that's all there is to it. Assigning a single potential to an entire resistor is not possible and makes no sense. You will find all the potentials between 0 and the terminal voltage in the resistor, just depending on where you look.

Ohm's law and the power laws hold true for the entire resistor as well as any arbitrary part of the resistor. The current is 100mA, so the the overall power is $$P = (100mA)^2 \cdot 10 \Omega = (1V)^2/ 10\Omega = 100mW$$ For any section of relative length $$x$$ we get

$$Px = (100mA)^2 \cdot 10 \Omega \cdot x = (1V \cdot x)^2/ (10\Omega \cdot x) =$$ $$(1V)^2 \cdot x/ (10\Omega) = 100mW \cdot x$$

where $$x$$ is the length of the segment divided by the total length.