# Are the physical processes behind electrical resistance and electrical heat generation different?

I asked a similar question in electronics stackexchange and was told to come here.

I have been told seemingly conflicting information about electrical resistance and heat generation and am looking for guidance.

I have been told that:

• An electrical resistor works by applying friction to the electrons passing through it and radiating the removed energy as heat, so the higher the resistance of the resistor, the more heat is generated.

• The more resistance a resistor has, the less current will flow through a circuit. Imagine a simple circuit with a 9v battery and a single resistor, the higher the resistance of that resistor, the less current will flow through that circuit and the longer that battery will last.

So in one case more friction equals less current and in the other case, more friction equals more heat.

I imagine two circuits. Each with a resistor and a 9v battery.

Circuit #2 has a resistor with 2x the resistance of Circuit #1

I imagine that Circuit #2 will NOT generate 2x as much heat as Circuit #1.

If that is the case, the stronger resistor is slowing the flow of the current without converting the difference entirely into heat.

If that is the case, then the physical process behind electrical resistance must be different than the physical process behind electrical heat generation.

What am I missing here?

An electrical resistor works by applying friction to the electrons passing through it and radiating the removed energy as heat, so the higher the resistance of the resistor, the more heat is generated.

This is essentially true. The power loss through a resistor (that is, the rate at which thermal energy is generated) can be written $$P = I^2 R$$, where $$I$$ is the current through the resistor and $$R$$ is the resistance. If the current through the resistor is held fixed, then the power loss is proportional to the resistance. However, if by adding more resistance you also change the current, then this will no longer be true.

The more resistance a resistor has, the less current will flow through a circuit. Imagine a simple circuit with a 9v battery and a single resistor, the higher the resistance of that resistor, the less current will flow through that circuit and the longer that battery will last.

This is also true. The current being drawn from the battery in this case will be $$I = \frac{V}{R}$$, where $$V$$ is 9 volts. Increasing the resistance while holding $$V$$ fixed will decrease the amount of current being drawn from the battery, which will prolong its life.

In your example, you have a constant 9 volt battery and two circuits with resistors $$R_1 < R_2$$. The current flowing through the first circuit is $$I_1 = V/R_1$$, and the current flowing through the second circuit is $$I_2 = V/R_2$$. Therefore, the power loss through the first circuit is $$P_1 = I_1^2 R_1 = \left(\frac{V}{R_1}\right)^2 R_1 = \frac{V^2}{R_1}$$ and similarly $$P_2 = V^2 / R_2$$. Since $$R_1 < R_2$$, it follows that $$P_1 > P_2$$, meaning that the first circuit (with a smaller resistor) generates more heat.

This isn't a contradiction. Remember that $$P=I^2 R$$; the second circuit has a higher $$R$$ but a smaller $$I$$, and the latter has a larger effect on the power loss. As a result, in this particular example increasing the resistance has the effect of reducing the rate at which heat is generated.

To answer your question about the mechanism being the same, consider what happens when you rub your hands together for warmth. If you press your hands together with more force, then the frictional force (which is responsible for the heat generation) will go up. As a result, as long as you keep rubbing your hands together with the same speed, the amount of heat generated will also increase.

However, if by increasing the frictional force you also decrease the speed with which you rub your hands together, then the net effect might be a reduction in generated heat rather than an increase. This is what happens in your example of the two simple circuits.

In either of the two cases, the "potential difference" V is the same, i.e. 9V.

Now, if we try to understand in any general case, where is the heat "energy" coming from? what is it being converted from? The answer is the potential energy difference of the flowing electrons as they complete each cycle of current.

You might have heard of "drift velocity". Basically, when not connected to any EMF source, the free electrons in the material are moving around in random directions with a net momentum of 0. However, upon applying an external electric field, they attain a net group velocity "proportional to the applied field", and this is the drift velocity. The proportionality constant is "electron mobility" which is inversely proportional to "resistivity".

Now once the electrons attain this velocity, which is what causes the "current flow", it doesn't change throughout. Right from the start from the -ve of the cell to the +ve, that v stays v. But, there is a constant field acting on the electron right? how come its velocity isn't increasing? apparently whole of the kinetic energy that was supposed to be gained due to the flow down the potential is being converted to heat (via whatever means and external agents there are down there).

So for any given potential difference "V" when a group of electrons constituting a small charge packet "dq" completes one cycle (from -ve to +ve), the loss in potential energy is "Vdq" and this is being converted to heat. Now if it took "dt" amount of time for this cycle to complete, then the "power loss" which is just the rate of heat, is equal to Vdq/dt = VI.

Now something crucial to note here is that,the amount of heat "energy" dissipated for one "cycle", depends just on V and not on I, in contrast to heat "power" which depends both on V and I.

In either of the two cases you mentioned, keeping in ind that the potential difference is the same "V", whenever a small packet "dq" completes a cycle, the heat "energy" loss is the "same". Its just that in case of the higher resistance the drift velocity is lower and hence "time" taken to complete that cycle is more. So, if you compare with the same "number of cycles" completed, the heat loss is the same. However, if you compare with the same amount of "time" the current has been flowing, naturally the one with lesser resistance would have completed more cycles and hence its heat loss would be more.

• Thank you! The description of a small packet completing a cycle and different speeds was very helpful for my intuition. – markv12 Dec 6 '20 at 23:42