3
\$\begingroup\$

I know lots of current can cause heat but what else? Can too much resistance in thin cables cause heat?

\$\endgroup\$
8
\$\begingroup\$

The mathematical formula for the Joule heating in a wire is:

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

As you can see when a large amount of current passes through a wire then it will get very hot as it is in the second power, obviously with R a constant as you have a specific wire.

But if you change the resistance (meaning that you take another wire) using a standard voltage, then using Ohm's Law:

$$V=I\cdot R$$

the current I that passes through it must fall, as to continue satisfy the Law.

So if you use another wire with greater resistance (assuming you keep the voltage constant, meaning you don't change the voltage source) then it is not going to be heated more, as you can see in the Joule heating formula both the resistance and the current change (resistance increases while current decreases).

In fact if you use the last Joule heating formula you will be able to see that the inconclusive previous result in which two parameters change, one up and one down, can be resolved now as the voltage is constant and the resistance is increased the result is that P decreases.

Hereby if you change the wire with one having more resistance then it will be heated less.

Caution

This answer only gives the physical background that ties resistance current and voltage. It doesn't try to show that in a specific real circuit is safe to assume that with wires of greater resistance the heat produced across the wire will be less! There are a lot of parameters in a real circuit to be taken into account before giving an answer regarding the heat produced. Also the analysis I made is obviously only elementary as the question seems to ask what happens in a theoretical general case, which again it isn't the same case with a real circuit.

\$\endgroup\$
9
  • \$\begingroup\$ In almost every circuit your conclusion is false as you don't have a constant voltage across a wire, but usually the voltage is across some wire and a load, as per Spehro answer, and increasing the resistance of the wire in a practical circuit will increase the heating in it. Your conclusion only applies if you are shorting out a battery with a wire vs a larger resistor. \$\endgroup\$ – Pete Kirkham Nov 4 '14 at 11:31
  • \$\begingroup\$ In a circuit you have constant voltage in every different part of that circuit. That voltage is just not the same. In different parts there will be a different voltage drop which depends on the resistance of that part. As my answer states a physical law which applies everywhere it is correct, and obviously it applies in every practical circuit. The only more accurate answer would be to use Maxwell's equations which obviously is not the best approach for this question. \$\endgroup\$ – Adam Nov 4 '14 at 11:44
  • \$\begingroup\$ If you increase the resistance of a wire supplying a load, the power loss in that wire increases until the resistance matches the load. Ignoring the load resistance and then saying that increasing resistance will not cause the wires to heat up and melt is irresponsible and unsafe. \$\endgroup\$ – Pete Kirkham Nov 4 '14 at 13:25
  • \$\begingroup\$ The purpose of the question as you can see from its tags was the physics behind the phenomenon of the heating of a wire. Not to take a specific circuit! If you check in any textbook or academic circuit source you will find the same answer as the one I gave. It is the only clear way to explain that phenomenon. There is no reason to extrapolate into a new question and perplexing its purpose. No one said that if you take a specific circuit and change the wires with ones having larger resistance the heating will be less. That would be dangerous. I just answered giving the physical explanation. \$\endgroup\$ – Adam Nov 4 '14 at 13:35
  • \$\begingroup\$ I edited my question so to avoid any misunderstanding. \$\endgroup\$ – Adam Nov 4 '14 at 13:52
5
\$\begingroup\$

Normally wires are sized so that most of the voltage ends up across the load and a much lesser voltage drop (maybe a few percent or less) is across the wires. You can think of the load resistance and the cable resistance being in series across the power supply.

In such a situation, increasing the resistance (thinner wires) will always increase the heating of the wires, right up to the point where half the voltage is dropped across the wires. Above that, it drops again, but at that point your load is only seeing half the supply voltage (eg. 60V on a 120V circuit) and is probably not working well.

See this answer for the mathematical basis. Consider the wire resistance as the variable resistance.

\$\endgroup\$
3
\$\begingroup\$

The quantity of heat P dissipated in a wire is given by $$ P = I^2R $$ where I is the current and R is the resistance of the wire (not load resistance). So yes, excessive resistance and high current both contribute to heat production.

This is true for AC and DC. For AC or any kind of variable load, I should is the root mean square of the current.

Most electrical cables are rated for a maximum current and maximum distance (because resistance is a function of resistivity, wire thickness, and wire length), listed in an ampacity chart. Charts are available for all wire sizes, from small "hobby project" wires to utility lines. The maximum current is set to keep the cable safely below the point at which any permanent damage might occur to the cable. Consult these tables when selecting wire diameters, because the actual amount of temperature increase (the real danger) is dependent on a number of parameters.

\$\endgroup\$
2
  • \$\begingroup\$ This is usually misleading--I=E/R. Assuming E is constant, increasing R will decrease I. Since the I term is squared and the R term isn't, increasing R with E constant will decrease P. This answer would/will apply if your source is constant current instead of constant voltage though. \$\endgroup\$ – Jerry Coffin Nov 4 '14 at 18:24
  • \$\begingroup\$ That is true in very general terms but not "usually". Since the resistance of the wire is small (0.0083 Ohms/m for 14 AWG) it is usually negligible compared to the load resistance, and so the wire resistance has a negligible effect on the current in the circuit. \$\endgroup\$ – jbarlow Nov 5 '14 at 23:16
0
\$\begingroup\$

Non-math answer:

I know lots of current can cause heat

is the same as

Can too much resistance in thin cables cause heat?

A cable can carry up to X amps, depending on it's size. Thin cables have higher resistance, therefore a lower amp limit, therefore "lots of current" can be quite small.

You will have the same overheating problems with wire the size of a hair and wire the size of your arm. It just depends how much current you push through it.

\$\endgroup\$
0
\$\begingroup\$

I believe that the temperature of a conductor is related to these 2 formulas:

(a) \$R_{ohm} = R_{oh} \dfrac{L}{S} \$

(b) \$P_1 V_1 T_1 = P_2 V_2 T_2\$ (= Constant)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.