Bozo's queries

[Bozo]

Alright, so no one at all has been able to answer me this question.

Understanding that with transmission lines, to reduce power loss the voltage is "stepped up" henceforth decreasing the current and decreasing the power loss as modeled by P=I^2*R. But why is it that according to ohms law (V=IR) that when voltage increases current also increases with it. Does anyone have an explanation to this contradiction?

[b^3]

The P=I²R, I is not constant because P=VI, I must change to balance the power, R is constant and is the resistance of the transmission lines. With V=IR, if I is constant and we change the resistance to get the voltage drop, i.e. replace a load with a greater resistance and the voltage drop increases. If R is constant, then increasing the voltage decreases increases the current. It all depends on what is staying constant in the equations. I hope that makes sense, because I think I've confused myself now???

When it is being stepped up, the resistance of the wire has nothing to do with it as the power has to stay the same, hence P₁=P₂, V₁I₁=V₂I₂
With the power loss, we cannot use the voltage drop in the wire unless it is stated, so we must use I and R, so P_loss=I²R

[Bozo]

That still doesn't answer the question, cause Voltage step up reduces current, but ohms law says the opposite. And your assuming resistance is fixed in both situations?

[b^3]

That still doesn't answer the question, cause Voltage step up reduces current, but ohms law says the opposite. And your assuming resistance is fixed in both situations?

Ok think of it this way, when dealing with stepping up and stepping down, we can throw the resistance of the wire out of the window, since it won't make a difference, we have to keep power of the transformer the same. We can't apply ohms law to this here because we can't apply the resistance of the transformer. The transformer (ideal) won't have a voltage drop and so we can't use ohm's law. Someone check this please.

[Bozo]

BTW, my teacher said its beyond the year 12 physics course.

[b^3]

Does what I've said make sense or have I just confused myself?

[xZero]

in this case we assume P is constant, so P=I^2R and P=VI are both constant right? we have the equation V=IR, lets sub I=V/R into the power equation so P=V^2/R. What this equation is saying that as we increase voltage, resistance of the transmitting line will also increase(the rate of increase is higher than voltage). Now back to V=IR, if v increase and r increase at a faster rate, I must decrease to balance things out. what does this mean? it means that the resistance of the transmitted line must be very large if we want to step the voltage up by heaps

[Bozo]

Yeah i get what your saying, so basically Ohms law cannot apply to this situation because there is no voltage drop in the transformer. Is that what your saying?

[Bozo]

But the resistance stays fixed?

[b^3]

in this case we assume P is constant, so P=I^2R and P=VI are both constant right? we have the equation V=IR, lets sub I=V/R into the power equation so P=V^2/R. What this equation is saying that as we increase voltage, resistance of the transmitting line will also increase(the rate of increase is higher than voltage). Now back to V=IR, if v increase and r increase at a faster rate, I must decrease to balance things out. what does this mean? it means that the resistance of the transmitted line must be very large if we want to step the voltage up by heaps

That makes it a lot clearer.

[xZero]

well if you follow my logic (which may be wrong), its saying that the type of wire from before and after the transformer will be different, if its a step up then the wire after transforming will have a higher resistance

[Bozo]

Has anyone got a solid collection of Unit 4 Trials they can give me. PM me please, it would be greatly appreciated.

[cranberry]

yeah ive got heaps of trial exams...

[Bozo]

2011 ones?

[Bozo]

Does anyone here, know how to edit pdfs to get rid of the detailed studies your not doing. Cause I want to go to officeworks to mass print all my exams, but like 1/3 of the stuff is redundant paper that will be charging me more.