# Why higher resistance means more heat produced in a circuit

1. Feb 8, 2012

### sgstudent

They say that nichrome is a better heating element as it has a higher resistance. But if I take V=RI then ain't I decrease since R is high, so current is lesser. Thus, resulting in a lower power dissipated as heat?
Also, they aq if I use the wrong wires like thin wires, it will cause a lot of heat produced as thin wires has a lot of resistance. Using the same reasoning I don't get why higher resistance results in greater power...

If current is constant then greater resistance results in greater power but I thought that voltage is usually constant? So when R increases, I will decrease and cause low power?

Thanks for the help!

2. Feb 8, 2012

### Staff: Mentor

Who are "they"? They're wrong.

3. Feb 8, 2012

### AlchemistK

Nichrome is a good heating element.

From Wikipedia :
"A heating element converts electricity into heat through the process of Joule heating. Electric current through the element encounters resistance, resulting in heating of the element.
Most heating elements use Nichrome 80/20 (80% nickel, 20% chromium) wire, ribbon, or strip. Nichrome 80/20 is an ideal material, because it has relatively high resistance and forms an adherent layer of chromium oxide when it is heated for the first time."

4. Feb 8, 2012

### willem2

If you want to design a washing machine, or a water cooker, you know in advance what power the heating element should produce, and so you know the resistance it should have.

You can then use a short thick wire made of nichrome, or a longer or thinner wire made of some other material with a lower resistivity.
If you make the wire thinner, it will also become hotter, because the same amount of heat has to flow through the smaller surface area of the wire.
If you make the wire longer, it won't fit.

So you have to make the wire both thinner and longer, and it will have to be rolled up somehow inside the heating element, if you don't want to make the heating element itself larger.
Electrical isolation will be a pain, since plastic, rubber etc. won't stand the heat, and you probably need a ceramic around the wires.

5. Feb 8, 2012

### cjameshuff

You're forgetting the resistance of the power source and transmission lines. If the heating coil were made of a similar length of similar gauge copper, it would draw huge amounts of current, but most of the voltage would be dropped across wiring rather than the heating element.

6. Feb 8, 2012

### OJFord

$P = I^{2}R$

So as resistance increases, power does too...

7. Feb 8, 2012

### willem2

This is not the equation you use, when you design an appliance with a heating element in it.
You know that the voltage across the resistance will be equal to the mains voltage, and that the the power must be whatever is needed to drive your hair dryer, toaster or whatever.

The resistance is thus fixed by P = V^2/R, and the only question is what material, length and width to use to make such a resistance. See my earlier post.

8. Feb 8, 2012

### OJFord

Apologies, I didn't really read the question fully.

But regardless, if you do not assume that you are working in this manner, the equation P=I^2R holds true for heating elements, surely?!

Ie, I'm testing some different elements of varying resistance, and recording the power output when driven from the same source. P will be greater for elements of greater resistance - that's all I was saying.

9. Feb 8, 2012

### Born2bwire

Both equations are valid when talking about power, but the point here is that we are connecting our heating element to a voltage source. So the the voltage across the circuit is going to remain the same as long as the current draw is within reason. That's why you work from V^2/R. Now if you use I^2R, the problem is that the current is going to change due to the equivalent resistance of your circuit. So while you may increase the resistance, you will also see a decrease in the current from the same voltage source.

10. Feb 8, 2012

### tony873004

Shut the switch off, which is essentially introducing infinity ohms into the circiut. See how hot it gets then when the current drops to 0.