Textbook error or am I just confused?

[Barclay]

The high school textbook I’m reading says this:

Statement A

“Many household appliances contain an electrically heated resistor. A resistor is a device that opposes the flow of current and as it does this, heat energy is produced. E.g. electric kettles, electric fires, light bulbs.

If we touch the electric flex/cord attached to any of these electric heaters when it’s switched on, we will notice that the flex is either at room temperature or slightly warm to touch. The electric current is increasing the temperature of the heater, by giving energy, but is not having the same effect on the flex.

This is because the heater has higher resistance, to make it difficult for the charge to flow through it. The flex contains copper wires to feed the electricity to and from the heater. Copper has a very low resistance".Then it goes on to say this which is opposite to the statement made above:

Statement B

“If two heaters are each connected to the mains power, the one with lower resistance will allow more current to flow, and will become hotter than the other one”.

Above it said that copper has lower resistance than the heater so does not heat up compared to the heater. Yet in statement B it says that the heater with the lower resistance will heat up more.
Then the section end with this statement that I don’t understand so please help here too:

Statement C
“This is why it’s so important that the live wire and the neutral wire in the flex don’t touch. If they do touch, they make a very low resistance circuit; the flex may burst into flames if the correct fuse is not fitted to break the circuit”.

Thank you

 

[mfb]

You have to distinguish between series and parallel resistors.
Statement A is for two resistors in series, where current is the same in both but voltage is different.
Statement B is for two resistors in parallel, where voltage is the same for both but current is different.

C is a short circuit: very high current, high voltage => high power => can start a fire

 

[russ_watters]

No, I think you are reading that wrong, mfb.

Statement B is about two different heaters, connected to a circuit at different times. The one with the lower resistance has the higher heat dissipation.

But for Statement A, the power cord isn't alone in the circuit: the heating element is the the larger resistance and therefore the larger part of the load.

But if you combine Statement A and Statement B, you can conclude that the power cord on the lower resistance/higher power heater gets hotter thant he power cord on the higher resistance/lower power heater.

 

[mfb]

Statement B is about two different heaters, connected to a circuit at different times. The one with the lower resistance has the higher heat dissipation.

It does not really matter if you connect them at the same time or separated in time.

But for Statement A, the power cord isn't alone in the circuit: the heating element is the the larger resistance and therefore the larger part of the load.

That's what I said.

But if you combine Statement A and Statement B, you can conclude that the power cord on the lower resistance/higher power heater gets hotter thant he power cord on the higher resistance/lower power heater.

Sure, but that should not be surprising. If you have no heater at all (extremely large resistance) you get the lowest amount of heat in the cables.

 

[Barclay]

Thanks for the replies. I'm still puzzling over statement B.
Isn't it because of resistance to the flow of electrons that the heat is produced?

Then why does statement B say:
“If two heaters are each connected to the mains power, the one with lower resistance will allow more current to flow, and will become hotter than the other one”.

Isn't current the flow of electrons? Then the easier it is for electrons to flow through the heater (low resistance) the cooler it will remain

 

[Khashishi]

The textbook is confusing but not wrong.

Better stick with the math. The equations make it clear what is happening here.

The resistive power in a circuit is given by
$P = I V$
or any of the equivalent forms:
$P = I^2 R = V^2/R$

In other words, for a fixed current, a higher resistance gives you more heating. For a fixed voltage, a lower resistance gives you more heating. That's because a lower resistance means more current will flow.

Think about how this applies to the space heater and wall supply.

 

[NascentOxygen]

Then why does statement B say:
“If two heaters are each connected to the mains power, the one with lower resistance will allow more current to flow, and will become hotter than the other one”.

Isn't current the flow of electrons? Then the easier it is for electrons to flow through the heater (low resistance) the cooler it will remain

When quantifying the heat produced in any circuit, the current's ease of passage is only half the story. The amount of current is at least equally important.

 

[russ_watters]

It does not really matter if you connect them at the same time or separated in time.

True -- I just thought it confusing and unnecessary to say they were connected in parallel. If they are not connected at the same time, they aren't in parallel with each other or anything else, necessarily.

My preference here is to stick with comparison of two completely separate series circuits, each circuit with a wire and a heater in it (modeled as two resistors in series).

 

[russ_watters]

Thanks for the replies. I'm still puzzling over statement B.

It is counter-intuitive, so I think ultimately you have to just look at and trust what the math is telling you.

Isn't it because of resistance to the flow of electrons that the heat is produced?

That's an incomplete explanation. Heat is energy. Energy (per unit time, or power) is dissipated by a combination of current and voltage (p=vi). Voltage is held (roughly) constant in these circuits, so the current is most directly related (proportinoal) to the power. And higher currents come from lower resistances (i=v/r).

 

[Barclay]

Thanks for the replies. Have been dreading returning to see the answers to my question but it does sort of make sense. I'll return to it a few times further (with pen and paper) and hope it all clicks into place.

 

[William Jackson]

I don't think it's counter-intuitive at all. If you picture it correctly, it should become obvious. Maybe Barclay is either unfamiliar with series vs parallel circuits, or is trying to deal with them mathematically instead of in "hands-on" easily visualized terms.

In the case of the heater and cord, the electrons flow through one wire of the cord, then through the heater and then through the other wire of the cord.
Don't worry about it being alternating current. At 60 cycles, the electrons will travel miles before they turn around and repeat the journey in the other direction. Just visualize one pass through the system.
The resistance is the combined resistance of the cord and heater. The resistance of the cord is more or less negligible in this case. The reason you don't get more current through the lower resistance of the cord is, it doesn't have another path to follow. It has to get through the resistance of the heater before it can pass back through the other wire of the cord.

When the textbook talks about two heaters, it means two heaters plugged into two different outlets. The current path through each heater is independent of the other heater. So the heater with less resistance will have more current flowing through it. If you cut the resistance wire, the resistance will be infinite (sort of) and current will stop. If you short across the elements with a screwdriver, the current will rise excessively and the breaker will trip.

You can do the math if you want numbers, but you don't need it to get the concept.

 

[William Jackson]

Thanks for the replies. I'm still puzzling over statement B.
Isn't it because of resistance to the flow of electrons that the heat is produced?
Isn't current the flow of electrons? Then the easier it is for electrons to flow through the heater (low resistance) the cooler it will remain

As for this comment, that would be true if the rate of electron flow didn't change. For that to be the case, the voltage would have to be lowered.

Your heater is connected to a given voltage. When you reduce the resistance, that voltage will cause a greater flow of current, and that is what increases the heat.
It's called resistive heating, but the heat is created by the interaction of the resistance with the flow of current.

If you kept the amperage the same while increasing the resistance (increase the voltage) you'll get more heat.
If you keep the voltage the same while decreasing the resistance (increase the amperage) you will also get more heat.

 

[Dale]

Thanks for the replies. Have been dreading returning to see the answers to my question but it does sort of make sense. I'll return to it a few times further (with pen and paper) and hope it all clicks into place.

You have two simple equations:
P=IV
V=IR

That is two equations in four variables, so if you are given any two then you can solve for the relationship between the other two. It is easy to verify that if you are given voltage and resistance then:
P = V^2/R

And if you are given current and resistance then:
P = R I^2

In determining the total power drawn by a given device the important equation is the first. The wall power fixes V, and the device fixes R.

In determining the power drawn by a given circuit element within the device the second equation is the important one. Each series element shares the same I, so the element with the largest R will pull the largest share of the total power.

 

[Barclay]

Thanks for your continued help. I feel that I'm dragging this post out longer than such a basic question deserves.
I think I've got it now so this may be the end of it if my understanding is correct:

Using the formulae P=IV and P = I²R and V = IR and P = V²/R

P is the POWER and here I'm assuming P also is the HEAT (?)

Statement B (above) said:“If two heaters are each connected to the mains power, the one with lower resistance will allow more current to flow, and will become hotter than the other one”.

In statement B the two components (heaters) are in the same circuit but being tested separately in a parallel circuit therefore voltage constant (?) but current is different (?) for each component. So P = V²/R applies and so as the resistance decreases the power (heat) increases. This shows statement B is correct.

Statement A (above):
"This is because the heater has higher resistance, to make it difficult for the charge to flow through it. The flex contains copper wires to feed the electricity to and from the heater. Copper has a very low resistance".

Statement A is discussing a series circuit so current is the equal across the components and voltage varies.
Therefore P = I²R applies and so as the resistance increases the power (heat) increases.
Copper has a very low resistance so the heat (power) will be proportionally low.

Phew! Finished. Has taken me hours to explain this to myself then explain it here too. I know its simple physics but I'm relatively simple when it comes to physics too. Anyway THANK YOU ALL. (May be premature in my thanks if I've not explained myself correctly).Bits and pieces from all posts helped me get to the end point but in particular the equations and comments below:

Statement A is for two resistors in series, where current is the same in both but voltage is different.
Statement B is for two resistors in parallel, where voltage is the same for both but current is different.

Statement B is about two different heaters, connected to a circuit at different times.

$P = I V$
or any of the equivalent forms:
$P = I^2 R = V^2/R$

You have two simple equations:

P=IV V=IR P = V^2/R P = R I^2

The wall power fixes V, and the device fixes R.