# Magnetizing current

1. Jan 7, 2013

### tonyjk

Hello.. I have a question : why when a transformer is loaded on the secondary, in the primary coil there's still the magnetizing current? I know that's the magnetizing current is the current for the no load condition and it's used to establish the flux in the soft iron. But when a load conditon applies there's a bigger current in the primary coil so why there's also a magnetizing current? Thanks

2. Jan 7, 2013

### I_am_learning

Magnetizing current is present all the time, at no-load or in full load. During No-load whole of the primary current is Magnetizing current but during full Load, magnetizing current is only a fraction of the primary current.
In terms of equation
I_primary = I_magnetizing + I_secondary*K (Full-Load) (K is turns ratio)
However, I_magnetizing and I_secondary will not be in phase and won't add algebraically (needs phasor addition)

3. Jan 7, 2013

### tonyjk

Yeah i know this but why there will always be a magnetizing current in load condition? Plus i have another question: the magnetizing inductance is also the inductance of the primary coil no? So why we put it in parallel in the equivalent circuit? And if this inductance is high how come there's an increasing current on load condition? Thanks again

4. Jan 7, 2013

### tonyjk

Can we say that the inductance of the primary coil is the magnetizing inductance and when there's a current in the secondary it produce a counter emf in the
Primary coil so that's why the current increase?

5. Jan 7, 2013

### tonyjk

So we can say that the equivalence impedance view by the source is the magnetizing impedance and impedance due to the counter emf produced by the secondsry coil in the primary coil?
PS: SORRY IM TALKING TOO MUCH

6. Jan 8, 2013

### I_am_learning

For any coil, E = N*d(phi)/dt. Always.
If E is the applied voltage phi will be the flux produced.
If phi is the applied flux to the coil, E will be the emf produced.
Lets look at this step by step.
1. When there is no load on the secondary side (i.e. No load condition)
E is the emf applied to the primary coil, then the coil will produce phi flux in the core.
You can see that, if E is sinusoidal, phi will be cosinusoidal.
The coil will consume that much current as required to produce the phi flux.
since, phi = N*I / reluctance_of_core we can calculate what will be the current required to produce the flux phi.
This current is the magnetizing current.
But since there is cosinusoidal flux in the core, the same rule applies to the secondary coil and there will be Emf induced in the secondary coil given by
E2 = N2*d(phi)/dt.

2. When there is Load in Secondary.
When you connect load to the secondary, then there will be secondary current. The secondary current flowing in the secondary coil will produce flux phi2 which will be in opposition with the flux previously being produced by the primary coil (by virtue of the primary magnetizing current). So, in effect the net flux in the core will reduce to phi-phi2. But Since Emf E is still being applied to primary coil, it demands that the flux linkage of primary coil still be phi. So what happens is the current in primary coil increases so that it now produces the flux: phi + phi2 so that the net flux linkage of the coil (that is flux in the core) becomes: phi + phi2 - phi2 = phi again. The additional current required in primary coil to restore the flux will not be equal to the current in the secondary unless the no. of turns are same in both of the coil. Hence, the secondary current that flows when load is connected to the secondary coil will be reflected in the primary coil on top of the already present magnetizing current (not as a replacement for it).

I hope you can now see the inner workings and understand why the magnetizing current is always present. Too answer in short, the flux produced by the secondary and the flux produced by primary would exactly cancel each other but because there is additional magnetizing current in primary, there will still flux in the core (hence the electromagnetic coupling)

I hope that wasn't too messy. I wish I knew LaTex.

7. Jan 8, 2013

### tonyjk

Thank you but can anyone please explain wjy we p
ut the magnetizing inductance in parallel?

8. Jan 8, 2013

### I_am_learning

Because thats the only way the equivalent circuit diagram would correctly represent the physical model.
If it is put in series then magnetizing current will vary according to load current.
Only by putting it in parallel, the magnetizing current will be constant regardless of the load current.

9. Jan 8, 2013

### tonyjk

thank you but even the current due to load is moving through the primary coil that's why i asked the question and still i didnt understand why