# why doesn't voltage increase

Discussion in 'Electricians' Talk' started by orbital, Jan 16, 2007.

I do honestly wonder, Mr Ha, whether you actual want someone to try and explain a point you are genuinely interested in, or are just trying to pick arguments for the hell of it.

Yes, fuse wire will have greater resistance than the supply cable because it is thinner, (the resistance of a wire is in inverse proportion to its cross-sectional area). But a 3A fuse wire, for example, still has a very, very low resistance. If it had a significant resistance, it would cause a significant voltage drop across it, which would affect whatever appliance it was supplying.

Yes, the resistance of a wire will also increase as it gets hotter. No contest.

It's all pretty academic, tho' - I'm just trying to explain why greater resistance doesn't create greater 'heat', which is what you seem to think. Did you follow my point about the electric element; if you halve it's length, you will halve (lower) its resistance, and this will double the current flowing through it, and it will glow twice as brightly.

So, reducing the resistance is increasing the element temp.

I'm trying - and am about to give up - to explain why your understanding of this: "Now you KNOW that isn't correct. The filament doesn't glow BECAUSE it is a smaller gauge wire. It glows because it is a different metal and therefore has a different resistance. For example, tungsten." is misguided.

"Otherwise, all our 3A and 13A fuses in our plugs would be glowing." Indeed they do - when they begin to reach their rated current.

2. ### Clark KentNew Member

Nice response D.A.............

However...

Did you follow my point about the electric element; if you halve it's length, you will halve (lower) its resistance, and this will double the current flowing through it, and it will glow twice as brightly.

(for a resistive load that's getting hot)

Is luminous intensity a function of (or proportional to) the current flowing through it (the load)?

3. ### Mr. HandyandyScrewfix Select

"Otherwise, all our 3A and 13A fuses in our plugs would be glowing."

current.

And pass 3A through those two fuses, which one would glow brighter, and hotter.

Don't tell me, the 3A one because it is smaller csa, therefore more resistance, therefore higher resistance gets hotter.

Wrong again, am I ?

Mr. Handyandy - really

4. ### Lokkars DaisyNew Member

Do they still explain lectricity by likening it to water in a pipe,pressure,flow etc ,,, that set me back weeks as a young un

Should have given it to me straight.

Electricity is electricity Handy, we can measure it in many different ways , it's just a different way for a different figure ,they are all measurements of the same quantity
resistance Andy , are you joking!
OHM's Law states ,that the current flowing between any two points in a metalic conductor is directly proportional to the voltage applied providing the temperature remains constant

that's from school many years ago

it says it all END of STORY

extenso ,you must get this thread removed ,trawling through it is a pain

5. ### xerxesxuNew Member

And pass 3A through those two fuses, which one would
glow brighter, and hotter.

Don't tell me, the 3A one because it is smaller csa,
therefore more resistance, therefore higher
resistance gets hotter.

Wrong again, am I ?

But if you're not wrong there must be some sort of explanation.

If the supply voltage is fixed, say 230V and the 3A fuse has a greater resistance (r1) than the 13A fuse (r2), how do you get the same current to pass through both fuses? You have to connect a load in series with r1, say R1 and a load in series with r2, say R2, such that r1+R1 = r2+R2. Then I = 230/(r1+R1) = 230/(r2+R2) = 3A.

if 3 = 230/r1+R1 r1+R1 = 76.66 Ohms = r2=R2'

For the sake of argument assume the 3A fuse, r1, = 16.66 Ohms and the 13A fuse = 6.66 Ohms.

All thats left to do is work out the power output of the 2 circuits and the power outputs of r1,r2,R1 and R2.

What are the formulae you would use?

6. ### Clark KentNew Member

For the sake of argument assume the 3A fuse, r1, =
16.66 Ohms and the 13A fuse = 6.66 Ohms.

I can tell you it is proportional..

A 3A fuse has around 4.3 times more resistance..

Hi Ext15.

"Is luminous intensity a function of (or proportional to) the current flowing through it (the load)? "

I'd have thought in this case it was a function of the current flowing through it. Is it not?

(I'm not suggesting it's directly proportional - my references to 'halve', 'double', 'twice' etc are generalisations.)

8. ### xerxesxuNew Member

I can tell you it is proportional..

A 3A fuse has around 4.3 times more resistance..

OK. r1 = 4.3r2, substitute and do it that way.

700 posts so soon!!!!

9. ### Clark KentNew Member

Hi Ext15.

"Is luminous intensity a function of (or
proportional to) the current flowing through it (the

I'd have thought in this case it was a
function of the current flowing through it. Is it
not?

Obviously you have to reach the point of which the element starts to glow..

10. ### Mr. HandyandyScrewfix Select

And pass 3A through those two fuses, which one
would
glow brighter, and hotter.

Don't tell me, the 3A one because it is smaller
csa,
therefore more resistance, therefore higher
resistance gets hotter.

Wrong again, am I ?

But if you're not wrong there must be some sort of
explanation.

If the supply voltage is fixed, say 230V and the 3A
fuse has a greater resistance (r1) than the 13A fuse
(r2), how do you get the same current to pass
through both fuses? You have to connect a load in
series with r1, say R1 and a load in series with r2,
say R2, such that r1+R1 = r2+R2. Then I = 230/(r1+R1)
= 230/(r2+R2) = 3A.

if 3 = 230/r1+R1 r1+R1 = 76.66 Ohms = r2=R2'

For the sake of argument assume the 3A fuse, r1, =
16.66 Ohms and the 13A fuse = 6.66 Ohms.

All thats left to do is work out the power output of
the 2 circuits and the power outputs of r1,r2,R1 and
R2.

What are the formulae you would use?

Look.

Cable run. 2.5mm-fuse 3A-cable 2.5mm-fuse 13A-cable 2.5mm -Appliance using 3A. That's a circuit.

Which of the above is gonna get hottest ?

Which has the highest resistance ?

??????????

Now use Ohm's law to explain that.

Mr. Handyandy - really

Ha ha, Ext15 - I see what you mean!

Mr Ha, isn't it the whole point that you can't use Ohm's law to explain it? Which is why I am saying that it isn't the higher resistance, as such, that's making the 3A wire hotter.

Ohm's law will tell you that the current flowing through the 3A wire will be LESS due to its HIGHER resistance. Ergo, the power consumed will ALSO be LESS and the wire will be COOLER (except, of course, it isn't due to it being thinner)!

So, it IS down to wire THICKNESS that's making it heat up - NOT its resistance

Hopefully this will clarify it further:

Take two thinnish wires of IDENTICAL cross-sectional area. One wire is made from copper, and the other, say, Nichrome (the alloy used in many heating elements) which has a significantly GREATER resistance than copper.

Remember, they are the same LENGTH and THICKNESS. The only difference is the copper wire has LESS resistance.

Plug 'em in, turn 'em on, and...

...which one will glow hotter?

(I'll give you a clue - it'll glow red and then melt...)

('cause copper has a lower melting point...)

12. ### oscar21New Member

I would like to add my 2ps worth. About the bad connection, The reason a good connection stays cool and a bad connection heats up is because there is another load factor preventing this.
Take a shower circuit with a bad connection in the pullcord, this will heat up and melt because the capacity of the circuit has been reduced at that point, not because of the increased resistance. In fact if you checked the difference in resistance between a good and bad joint it would probably be minimal. Now take the same circuit with a good connection and bypass the shower from the circuit. When you switch on, the pullcord connections will last for less time than if there was a bad connection in the circuit (discounting the mcb tripping of course). The bad connection would have a higher resistance in the circuit (miniscule) thus reducing the current flow and keeping the circuit cooler for longer. Probably not very well put but I belive right.

13. ### cliffy brownNew Member

who's your favourite big brother housemate up to now?

C'mon, if you're big enough, explain the conundrum: how comes a heating element needs to have a resistance before it can heat up, but then the one with least resistance is the one that gets hotter?

Eh? EH??!! EH???!!!

Hah - got you.

(and me...)

15. ### cliffy brownNew Member

i meant this years housemates doufus

pffttt

16. ### roymondoMember

Orbital, the distribution grid supply system is constantly ajusted for voltage, increased for more current/power is that what your on about?

17. ### The DormouseNew Member

This topic is great fun but the questioner should have specified the complete circuit. No current can flow in components that have no connections.

Y'mean Goody isn't on it?

Ok, here it is in Mickey Mouse terminology: a heating element has to be 'given' a resistance, or else it would be a short circuit.

That resistance is chosen to provide the overall rating of the heating appliance. Eg: a 3kw element, powered by a 230V supply would draw just over 13A, so the internal resistance of the element would be in the order of 18 ohms. (This is M-M, remember - if someone mentions impedance or anything else, I'll come and get you.)

Now, you can get that resistance in a number of ways, using conductors of various 'qualities'. You could use thin copper wire, but because copper is such a good conductor, it would need to be (a) very thin and/or (b) very long. This would be impractical. ALSO, it would melt...

Ok, must dash - taking the kids to school...

So, they commonly use an alloy such as Nichrome, which has all the properties a heating element would need such as high melting point, larger resistance (so less of it needed), etc.

BUT, GIVEN THAT THE ELEMENT HAS A RESISTANCE SO'S IT CAN 'WORK' IN THE FIRST PLACE, IF YOU THEN REDUCE THAT RESISTANCE THEN A GREATER CURRENT WILL FLOW AND IT WILL BECOME HOTTER.

Jeepers, even I - a doufus - understands it.

Now.

20. ### Mr. HandyandyScrewfix Select

Ha ha, Ext15 - I see what you mean!

Mr Ha, isn't it the whole point that you can't
use Ohm's law to explain it? Which is why I am
saying that it isn't the higher resistance, as
such, that's making the 3A wire hotter.

Ohm's law will tell you that the current flowing
through the 3A wire will be LESS due to its HIGHER
resistance. Ergo, the power consumed will ALSO be
LESS and the wire will be COOLER (except, of course,
it isn't due to it being thinner)!

So, it IS down to wire THICKNESS that's making it
heat up - NOT its resistance

Hopefully this will clarify it further:

Take two thinnish wires of IDENTICAL cross-sectional
area. One wire is made from copper, and the other,
say, Nichrome (the alloy used in many heating
elements) which has a significantly GREATER
resistance than copper.

Remember, they are the same LENGTH and THICKNESS.
The only difference is the copper wire has LESS
S resistance.

Plug 'em in, turn 'em on, and...

...which one will glow hotter?

(I'll give you a clue - it'll glow red and then
melt...)

('cause copper has a lower melting point...)

So, as I said, it is the METAL and it's resistance that makes a heater. NOT the size of the wire.

Mr. Handyandy - really