Resistance! In motors (1 Viewer)

[Darrow]

Ok, we all know that V=IR
SO by that R=V/I
So, as Current increases shouldnt resistance in a wire decrease?

But power loss increases with an increase in current (P=I^2R)

So Im confused

Edit: In power transmission actually

 

[munchiecrunchie]

Energy = P x t
= VI x t

now V = IR (substitute for V)

therefore, Energy loss = (IR) x I x t
= I^2R x t

 

[Jase]

The resistance of the wire does not change. It is a set value depending on the following factors: length of the wire, cross-sectional area, temperature and material.

What really happens is current increases when voltage increases because of that fact the resistance must remain constant. This is ohms law (V=IR). Anything that has a variable resistance is 'non-ohmic' meaning it doesnt obey this law.

 

[Riet]

Energy = P x t
= VI x t

now V = IR (substitute for V)

therefore, Energy loss = (IR) x I x t
= I^2R x t

Not quite, electrical power is just P=VI
V=IR
P=I^2R

This is why for transmitting long distances you want to have a very high voltage
(and relative low current) because voltage drop/power loss increases with the square of current.

 

[munchiecrunchie]

Not quite, electrical power is just P=VI
V=IR
P=I^2R

This is why for transmitting long distances you want to have a very high voltage
(and relative low current) because voltage drop/power loss increases with the square of current.

Yes I know, I was talking about energy.

 

[Riet]

Yes I know, I was talking about energy.

why, kilowatt hours is only used for measuring power-consumption? It's unecessary to worry about that, just think about it in terms of resistance (which depends on length of the wire).

 

[rips]

As Jase says, resistance is a property of the wire that we assume doesn't change in these questions (although in reality, R increases as the temperature increases with power loss from the current).
For transmission type questions, use P = VI to calculate current such as transmittng 64MW at 110,000V, what is the current (582A).
For heat loss, use P = I^2R example if R = 20 Ohm for above transmission, what is heat loss? P(loss) = 582^2 x 20 = 6.8MW which is near 10% of 64MW and too high.
What if you wanted to keep your losses to below 2% (Loss = 1.28MW) what current and voltage would you need? Remember if you raise voltage, you lower current in the power formula, so I required = (1.28 x 10^6 ÷ 20) ^½ = 253A. Now use Transmission power formula P = VI to find the voltage (64MW ÷ 253 = 253kV). this means you need to transmit the 64MW at 253000 V to keep your losses at about 2%.
ps: you could have also used the ratio formula to find the voltage, but what the heck. its the same thing.

I tutor physics, so hope this helps. The BOS usually ask explain why or what would you do type questions rather than number crunching. So make sure you can explain the square of current relationship in heat loss.

Ok, we all know that V=IR
SO by that R=V/I
So, as Current increases shouldnt resistance in a wire decrease?

But power loss increases with an increase in current (P=I^2R)

So Im confused

Edit: In power transmission actually