How do transformers obey ohm's law?

[Entanglement]

Yes, I think you could say that, but I'd slightly reword it:
"… a part of it drives a current against the resistance of the primary coil, and another part drives the rate of change of current in both coils as inductors."

I don't get why you reworded it?
Why do you prefer saying "rate of change" ?

 

[Entanglement]

Hello ElmorshedyDr.

Suppose a varying p.d. with instantaneous value V_P is connected across the primary coil. Per coulomb passing through the primary, V_P joules of work will be done. An amount I_P\ R_P will be done heating the primary coil. But work will also be done per coulomb against back emfs \varepsilon_{1, 1}, the emf induced in the primary due to the rate of change of primary current and \varepsilon_{1, 2}, the emf induced in the primary due to the rate of change of secondary current.

So, using the principle of conservation of energy (per coulomb flowing),

V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ +\ I_P\ R_P.

I'm afraid that signs vary according to the sign convention used.

Now here's the punch-line… Usually R_P is small enough for us to forget I_P\ R_P, so

V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ .

But if the secondary coil is not connected to a closed circuit
there will be a V_S but there won't be a current "to report a feedback to the primary coil" to decrease V_P's share to I_P\ R_P
why isn't that a violation to law of conservation of energy?

 

[Philip Wood]

Re post 31
Because the back-emfs only arise when the currents are changing. Faraday's law.

 

[Philip Wood]

Re post 32
If secondary not connected then no energy is being supplied by the transformer, so the absence of feedback to the primary doesn't violate energy conservation, but is required by it!

Don't forget that voltages, such as Vs, tell us how much energy would be converted/transferred per coulomb flowing, so if there is no current (that is no charge flowing) no energy will be converted, even if there is a non-zero voltage.

 

[Entanglement]

Re post 32
If secondary not connected then no energy is being supplied by the transformer, so the absence of feedback to the primary doesn't violate energy conservation, but is required by it!

Don't forget that voltages, such as Vs, tell us how much energy would be converted/transferred per coulomb flowing, so if there is no current (that is no charge flowing) no energy will be converted, even if there is a non-zero voltage.

Why isn't there any energy supplied?

Inducing a voltage in an open circuit, as I understand, is like kicking a ball upwards against gravity then preventing it from falling by some barrier, once the barrier is removed the ball falls way back again consuming all its potential energy.
The removal of the barrier is just like closing the circuit and allowing the current to consume its voltage.
Even though the ball was prevented from falling but energy was consumed to raise it against gravity,
even though the circuit is open energy is consumed to create a potential.

 

[Philip Wood]

Suppose the secondary is open circuit. A very small amount of energy is taken (from the primary circuit) when the changing current in the primary induces an emf in the secondary forcing a small number of electrons through the secondary, depleting one end/terminal of the coil of electrons and giving the other end a surplus. But relatively few electrons flow in this way before the electric field set up in the coil by the deficit and surplus (positive and negative charges) at either end of the coil stop any more electrons flowing. So the flow is 'one-off' and the amount of energy required is negligible. What's more it's given back on the next part of the cycle.

In your football analogy, the energy needed to raise one football is negligible. But in a complete circuit the footballs keep being raised, requiring non-negligible energy per unit time. The energy is dissipated when they fall through a viscous medium (through the external circuit).

 

[Entanglement]

But relatively few electrons flow in this way before the electric field set up in the coil by the deficit and surplus (positive and negative charges) at either end of the coil stop any more electrons flowing. So the flow is 'one-off' and the amount of energy required is negligible.

I don't understand why only A FEW electrons will flow that way ??

 

[Philip Wood]

It doesn't take much charge to set up an electric field in the wire of the secondary coil which is equal and opposite to that due to the induced emf from the changing magnetic flux. After the electric field has reached that size, no more charge will flow (or wouldn't flow if the emf stayed the same!). THAT is the key thing to understand.

The reason why it doesn't take much charge is that, thinking of the wire (open circuit) as a capacitor, the plates (formed by the ends of the wire) will have a very small area, and are well-separated , so the capacitance will be extremely small. This is one way of looking at it. There will be others.

 

[Philip Wood]

ElmorshedyDr: You don't seem to have posted your daily transformer question...

 

[Entanglement]

I've bothered you a lot lately, I'm so sorry.

 

[Philip Wood]

You shouldn't have apologised: I was only teasing.

Some of your recent questions have made me think hard. Not a bad thing.