How does V go up and I go down?

[iScience]

I understand that power is IV, and that power should be conserved during EM- induction for example in a transformer.

$$I_pV_p=I_sV_s$$



But potential is (correct me if I'm wrong) like the driving force in a circuit. and the stuff that it drives is the charges. So say i have a step up transformer, and the voltage on the secondary is huge. Well this means the E-field will be strong. So then, what is preventing the charges from flowing freely? or, in other words, if V_s is 50kV, and i short the secondary circuit, what prevents the current from being 50kA, and instead be [I(p)V(p)/V(s)]amps?

i'm just looking for a qualitative understanding for my intuition

 

[256bits]

I can think of three reasons why or why not - how much power can the source output, what size is the transformer, and what gauge wiring is used in the transformer windings. Do you know how each has an affect upon the output?

 

[iScience]

what gauge wiring is used in the transformer windings

$$R
\alpha
A^{-1}$$

what size is the transformer

not sure what you mean by size

how much power can the source output

This seems to be the issue. Say the power input is P_i, and the output Voltage at the secondary is V_out, what is the power constraint physically doing to the current in the secondary to make it have an inverse relation to the generate voltage in the secondary?

 

[NascentOxygen]

The iron or other material chosen for the core cannot transfer unlimited power in the magnetic field. Its characteristic is not linear beyond a certain point, you are aware that the B-H loop flattens out more and more as current increases.

 

[jim hardy]

in other words, if Vs is 50kV, and i short the secondary circuit, what prevents the current from being 50kA, and instead be [I(p)V(p)/V(s)]amps?

Are we speaking here of an ideal transformer?
If so then the answer to your question is:

"Nothing but the edit: [STRIKE]ability[/STRIKE] inability of your power source to deliver into the primary a current of 50kA X transformer turns ratio. "

A real source has resistance and so does a real transformer. Those will limit the current.
There can be no power dissipated in a short circuit because voltage is zero. So the power is all dissipated in the wires that comprise your source and transformer.
That's why a shorted transformer gets hot.

 

[sophiecentaur]

I understand that power is IV, and that power should be conserved during EM- induction for example in a transformer.

$$I_pV_p=I_sV_s$$



But potential is (correct me if I'm wrong) like the driving force in a circuit. and the stuff that it drives is the charges. So say i have a step up transformer, and the voltage on the secondary is huge. Well this means the E-field will be strong. So then, what is preventing the charges from flowing freely? or, in other words, if V_s is 50kV, and i short the secondary circuit, what prevents the current from being 50kA, and instead be [I(p)V(p)/V(s)]amps?

i'm just looking for a qualitative understanding for my intuition

It's because you would need (and be using) a much higher resistance load for the same Power at a higher voltage. If you put the same resistance load then yes - you would get loads of current. The supply 'sees' the resistance of your actual load transformed down to a lower one. The VA in both cases will be the same in an ideal transformer. I used the term VA (Current times volts) because in a real transformer with a real load, VA will be greater than the actual Power that's transferred - not to worry about that, first time through.