V=IR R=V/I I=V/R

Hi mnk,

I read the thread that you mentioned and the thing that kept going across my mind was that the longer an extension cord you have, the hotter it will be driving the same load. It would seem that the longer (more resistance) cords heat up because of increased current.

So, I think your question could be answered by asking the question,

"Why is it that longer extension cords heat up more than shorter ones? (driving the same load)"

I think I know the answer, but I don't have that much electrical background. However, I know that there are a lot of them out there that could answer this question.

Regards,

Mike

No, put simply (simply here mind you) it is due to volt drop. If your lead is half the resistance of the heater then it will be dissipating half the power the heater (which is just a big resistor) is dissipating.

There is no magic here.

<throws hands in air>

Right, lets try and solve this once and for all using the TWO basic, well-known electrical formulas.

Ohms law V=IR

Variations in the Ohms law formula (by substituting) BUT they ALL mean the same thing.

I=V/R

R=V/I

AND.......

Power formula P=VI

Variations in the power formula (by substituting) BUT They ALL mean the same thing.

P=I²R

P=V²/R

Example - 1000W heating element replacing a 500W heating element in a stove operating on 230V ac.

P=VI so I=P/V so for a 500W element, I=500/230 = 2.17A (2 decimal places)

P=VI so I=P/V so I=1000/230 = 4.34A (2 decimal places)

So, 500W element draws 2.17A in the stove. Replacing it with a 1000W element draws 4.34A = twice the current.

Additional proof - The Element resistance.

500W element rated at 230V (ie- designed for 230V operation) has a resistance of 105.8 ohms.

P=V²/R so (rearranging) R=V²/P so for a 500W element R=(230)²/500 = 105.8 ohms.

R=V²/P so for a 1000W element R=(230)²/1000 = 52.9 ohms = HALF the resistance of the 500W element.

.....Now, the practical explanation - A stove heating element is constructed out of resistance wire wound around a insulating former and protected by a metal sheath on the outside. A 500W heating element will have TWICE the length of resistance wire inside compared to a 1000W heating element, hence a 500W element will only draw HALF the current of a 1000W heating element (remember, the voltage is kept constant in your example).

An additional (but more advanced) question that may arise - "What happens if I want to operate the 1000W element on, say 460V (twice the voltage)"

Element resistance constant = 52.9 ohms. Voltage changes to 460V, current will then change.

I=V/R so I=460/52.9 = 8.68A (twice the current when running on twice the voltage).

What's the power being dissipated by the 1000W element now....

P=VI so P=460*8.68 = 4000W = 4 times the power.

It is all in the math. It works, you just need to know how it interacts. I am not sure I can explain it any simpler. NOTE - decimal place rounding errors have been ignored to prevent misunderstandings and prove that, in fact 4 is twice as big as 2.

The mistake in the reasoning of your sparky friends is that a single element is viewed as the total system.

The load in question should be considered as a load in parallel with a lot of other loads.

Please refer to your notes regarding resistance and current in a parallel system.

Hope this will lead you in understanding the problem.

P = VI even with alertnative current.

so I = P / V

So in your sample :

500 = 100 * 5 => 5 = 500 / 100 => I = 5 A

and 1000 = 100 * 10 => 10 = 1000 / 100 => I = 10 A

then 10 A > 5 A You can see http://en.wikipedia.org/wiki/Joule's_Law

Best regards

"Let 's assume we have a heating element of 500w(100v x 5A) of an electrical stove.Theoretically,when you increase/replace the element with a 1000w(100v x 10A) keeping the voltage constant ,IN THIS INSTANCE WHICH OF THESE TWO WOULD TAKE MORE CURRENT ???? If it is the 1000w then would you please explain to me a little more for me to really understand the circumstance because Mathematically, I am not able to prove this theory with Ohms Law"

The 500W element has a resistance of 100/5 = 20 ohms

The 1000W element has a resistance of 100/10 = 10 ohms

Since the current flowing inb the 500W element is 5amps then it is clear that decreasing to 10ohms of the 1000W element allows MORE current to flow(10 amps)

The current will indeed increase but it do not prove ohm's law incorrect.

The current at the power station or source will increase to balance the extra load.

Lets not get into transmission power theory shall we. Part of my job is designing and building substations, and the original poster is having enough trouble understanding and visualising basic ohms law with purely resistive elements. Lets not muddy the waters with power factor theory as well.

Lady's and Gentlement: Ohm's law cannot be used on a heating element for the purpose of determining the current drawn using known resistance (measured) and constant voltage. The ONLY use for ohm's law is to determine the resistance when the heater is fully heated and dissipating its intended power.

This is because a heating element has a non-linear temperature dependent resistance. As the heating element heats, the resistance of the element drops until a steady state current and resistance is reached.

As ohm's law states, an increase in resistance causes a decrease in current for a constant voltage. But it cannot be used to estimate current in a non-linear time varying system unless you are measuring the two other states of that system as it changes. Needless to say, you can't measure the resistance directly. You have to measure the current instead to determine the resistance, which is what you were after in the first place.

Anyway, heating elements and light bulbs are just two examples where you cannot predict current or power required based on a measurement of the resistance when the elements are cold.

Context is everything and nature's laws are unbreakable. If you think nature's laws don't work, you need to question your knowlege of how these laws apply, not the law itself (unless you are a scientist studying the laws themselves).

So, your friends are wrong. An increase in resistance does not result in increased current. A 500w element draws half the current of a 1000w and when fully heated and in a steady state will have twice the resistance of the 1000w element.

Well but one thing is still unclear. That how does the current increase with inncrease in resistance???? This sounds so wierd !!!