Why does a thin conductor heat up more easily than a thick conductor just because it has a higher resistance?

Answer 1

Because of the way your question is phrase, I'd assume you already know why thin conductors have a higher resistance - (smaller cross sectional area, electron flow etc)

Neglecting heat capacity, realting to "JUST because of higher resistance". The explanation would be as follows:

First you need to know the definitions well.
Potential difference is defined as the energy dissipated per unit charge across 2 points. Unit is V (or Joule as it is energy)
Current is the rate of flow of charge (Amperes or C s-1)
Resistance is the ratio of potential difference across the resistor to the current flowing through it

Resistance = potential difference/current
R = V/I
V = RI

Heat dissipated across a conductor is proportional to the power across the conductor. Higher power means more heat produced within a given time.

Energy dissipated per time = Energy dissipated per unit charge x No. of charge flowing through per second
= V x I

P = VI
At this point, I would say the question is ambigious because there are 2 possible scenarios which would yield different results:

First case: Wires are connected separately/ in parallel to same power source - Not the result of the question
Assume your battery to have an emf of V and negligible internal resistance. As the power source for both is the same, V across the conductors are both equal. It is the current & resistances which will differ.

V = RI
I = V/R

Subbing I=V/R into into P = VI
P = V^2/R
Since V is same, higher R would yield less power, hence thin wire is less easily heated up.

Second scenario (related to the question) The two wires are connected in series with the same power source.
Again take the electromotive force to be V and negligible resistance.
Let Combined Resitance = R'
V = RI
In series sircuit, current is constant. Potential difference across each of the resistance however differs.

Sub V = RI into P = VI
P = I^2R
Hence the resistor with higher resistance (the thin wire) will heat up more easily as more energy is dissipated across it per second.

In real life of course, the internal resistance of power sources are usually not negligible, hence you'd take the wire to be connected in series to another resistor. Hence scenario 2, and the formula P = I^2R is more commonly used.

Answer 2

You asked why a thin conductor heats up "more easily". i don't know exactly what that means, but the resistance alone DOES NOT determine how hot it gets nor how quickly it heats. That depends on the current sent through the conductor. The thin conductor will have a higher electrical resistance than the thick one, but that tells you nothing about its heating. If you supply the conductors with a constant voltage source like a battery or the electric house voltage, the current will be lower in the thinner conductor. The heating effect is given by the total power supplied, which is E^2/R, and the conductor with the higher R (the thinner one) will get LESS power and get LESS hot. However, if both conductors are driven with voltages that provide the same power (by putting more voltage on the thinner conductor), the thinner conductor will always heat up FASTER and get HOTTER because it has a lower thermal capacity. The smaller mass and volume means there is less stuff to heat up and so it gets hot faster.

Answer 3

I'm not an expert but this is the best of my knowledge on the subject.

A thinnner conductor heats up more than a thick conductor because the electricity has less room to move creating more friction in the molacules. The more friction the hotter the object gets.

Answer 4

Because it has a higher resistance which is a measurement of how easily electrons flow through the conductor. So if you push the same amperage through a smaller conductor it will produce heat as more of the atoms will be excited.

Answer 5

Sure. A thick conductor has greater resistance and absorbs more heat. Just like a thick wire is less prone to heat up than a thin one.

Answer 6

Resistance = resistivity*length of conductor/cross sectional area.

Cross sectional area = pie* d^2/4 where d is the diameter of the conductor.

Answer 7

the more vol. you are trying to heat be it conductors or water , the more heat it takes or the more time it takes to heat up.