Confused about I V tradeoff in circuits

[iScience]

to skip the introductory part go straight to the question between the hyphens.

I wanted some help mending two seemingly contradictory concepts.

on one hand we have ohm's law: I= \frac{V}{Z}

stating that current is proportional to the potential between two points. makes sense.

on the other hand we have power conservation (for instance: in a transformer circuit): P_1=P_2

okay, the idea that power must be conserved makes sense. but I'm having trouble with how this is manifested:

$$P=IV$$
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I & V are inversely proportional so if the potential goes down, conceptually can someone explain how this leads to an increase in current?

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[Windadct]

You are looking at two different things - P=IV is essentially a 2 terminal statement. There can only be input or output. So looking at a Transformer P=I*V can be used, however essentially ignoring what ever is on the other side of the transformer.

For a transformer --- we have power input ...and then some nearly equivalent power output ( there are ALSO losses in a real system). The P1=P2 is really for an ideal case.

So - the P1 = I1 * V1 ... = I2 * V2 = P2 . So for the Power In to equal the Power Out - as we step up / or step down the voltage - then the current must respond inversely. 100W = 5 v * 20A = 50V * 2A = 100W.

 

[anorlunda]

There is an identical question answered in this thread.

https://www.physicsforums.com/threa...wer-volatage-and-current.783543/#post-4921579

 

[iScience]

I read through the thread; it seems people were trying to address the primary side of the circuit.

To restate my question: if i have N turns on the secondary and that gives me Power (IV), keeping my primary side input constant (I&V), and i increase the number of turns on the secondary, i will now have a Power value with a higher voltage and a lower current.

hope that clears up my question

 

[anorlunda]

Energy is conserved, so what you said is correct. Increase the turns ratio on the secondary and secondary voltage will increase and current will decrease in general.

However, this ignores the load connected to the secondary side which might impose a second relationship between voltage and current. You hypothesis is correct if the characteristic of the load is constant power.

You should look at http://en.m.wikipedia.org/wiki/Transformer#Ideal_transformer

If the load is constant impedance, note that the apparent load reflected to the primary side will change if the turns ratio changes. In that case primary voltage and/or current will also change when turns ratio is changed.

 

[sophiecentaur]

I read through the thread; it seems people were trying to address the primary side of the circuit.

To restate my question: if i have N turns on the secondary and that gives me Power (IV), keeping my primary side input constant (I&V), and i increase the number of turns on the secondary, i will now have a Power value with a higher voltage and a lower current.

hope that clears up my question

This is a very common question that people ask about transformers. Assume that the transformer is ideal. The secret is to approach the thing in the right order - the order which allows you, in effect, at each step to use an equation with only one unknown. First of all you work out the Secondary Volts that the transformer will produce at its output (given by the transformer ratio and the primary volts). Then you work out the current that those secondary volts will cause to flow in the load. I = V/R.
Then, the Power conservation tells you that V1I1 = V2I2, so you can work out the primary current. The apparent load is then V1/I1. The whole thing has been solved without any of the apparent contradictions that can arise if you leap in half way through and try to work both ways.

 

[iScience]

First of all you work out the Secondary Volts that the transformer will produce at its output (given by the transformer ratio and the primary volts). Then you work out the current that those secondary volts will cause to flow in the load. I = V/R.
Then, the Power

so it would be incorrect to calculate the input power and then calculate the output power?

if i did do it this way though (idk say, i have a low input impedance circuit (primary side) ), the current value i solve for at the secondary, would that be referring to the current capacity rather than actual current?

 

[sophiecentaur]

so it would be incorrect to calculate the input power and then calculate the output power?

if i did do it this way though (idk say, i have a low input impedance circuit (primary side) ), the current value i solve for at the secondary, would that be referring to the current capacity rather than actual current?

Ah well - this is the whole point. How could you work out the input power before working out how much power the load is dissipating? Hence my advice to work out what you can, using the information that you have.
All you can know about the transformer, initially, is the Voltage ratio and hence the Watts per Amp in the secondary (but you don't know the Amps yet) You need to know the load resistance value to know the current - and then you can work backwards. If you want to get a grasp of this (and all the rest of 'Electricity') you need to start thinking in terms of 'what values do I actually know' and then decide where that knowledge can take you further.
You could start with a balloon diagram with all the possible quantities in separate balloons. Put in the values you actually know and then see which of the other balloon contents can be worked out from the existing full balloons (join them up with lines at each stage). If the problem is soluble (and not all problems are, of course - but homework and exam problems will be) then you will eventually be able to fill in values for all the balloons but only if you go at it in the right order. Soon enough, you will learn the best way to get through such problems and when there are available short cuts.
There are only a very few formulae involved in this sort of problem so it's not really that hard to get used to the process.

 

[iScience]

alright, and confirm this one last thing for me, because i'd like to look at this another way than just "power conservation."

so another way to explain the reason why I_p is inversely proportional to I_s (even though this explanation is probably just another side of the same coin) would be that the greater the induced current, the greater its B-field which opposes that of B_p (virtually increasing the inductance as seen from the primary side), and therefore reduces the current in the primary.

you've been so much help :)
thanks a bunch

 

[sophiecentaur]

My picture of this:
The load resistance doesn't appear as an inductance to the primary input (which wouldn't involve any power transfer. The load has to appear as a Resistive component in parallel with the (very high and ideally infinite) Inductance of the primary. But any good analysis of how a transformer actually works involves an 'Equivalent Circuit that is quite complicated. http://www.ece.msstate.edu/~donohoe/ece3614single_phase_transformers.pdf of explaining what happens but it's a big jump from the easy 'turns ratio = volts ratio' equation we all recognise. Not surprising really when you think you have two coils with their own inductance and then the mutual inductance between them, all inductances causing induced emfs due to the currents in them and all inter linked. When you get near the bottom of the rigmarole, you arrive at an equivalent, idealised circuit with Primary and Secondary volts at each end, connected by an Inductance in series with a Resistance.
The relationship between the Load Impedance and the Impedance as seen looking into the primary is derived in the link but it ends up approximately as the simple theory (power conservation) predicts. (Not surprisingly - but I don't think there's an easy arm waving explanation of where the reflected load impedance comes in, apart to say that the current in the Load will induce an emf in the primary that is less than the supply volts in amplitude and in anti phase with the supply,- thus allowing current to flow. That current is the same as would flow if the load times the turns ratio squared were put in place of the transformer plus load. There is probably a more elegant way of putting that!