# Representation of magnetic hysteresis loss as a resistance in equivalent circuits

1. Mar 20, 2009

### b.shahvir

Hi Guys,

Can someone please explain to me the logic behind the representation of 'Magnetic Hysteresis loss' as a resistance in electrical equivalent circuits?..... will be extremely grateful.

I have studied some info on this subject on the net. Even though the physics of Hysteresis Loss has been explained appropriately, I wish to dwell deeper into the phenomenon to get to the source of my doubt. In spite of the relevant theories, it becomes difficult to visualize a magnetic phenomenon into an electrical parameter as mentioned by me earlier. When we talk of energy expended by the power source to ‘pump up’ the magnetic field in the core, sounds a bit abstract!

I want to understand what phenomenon in the magnetic hysteresis causes the machine (Xmer or motor) to suck an additional losses (wattful) current from the power source.
Is it a result of the delay caused by the magnetic dipole friction in the magnetic core........ which prevents the magnetic field from building up quickly? I would be grateful if someone could elaborate on this part. Thanx.

Kind Regards,
Shahvir

2. Mar 20, 2009

### Bob S

In general, if when the magnetizing force H (amp-turns per meter) is increased from zero to a large value in a magnetic material, and the magnetization B (tesla) also increases, and when H is brought back down to zero, a small remanent magnetization Br remains. Suppose we go through complete cycles of H, and this continuously occurs. What does this mean electrically?

Inductance L (henrys per meter) is the property of an electrical circuit to store energy using current, and recover it when the current returns to zero. Let's look at the equivalent equations in EE and physics for energy storage using current.

(1/2) L I2 = (1/2) integral[ H * B dV], (where dV is integral over volume), or

L = (1/I2) integral [ H*B dV]

When ramping H up, a certain amount of energy is stored in the magnetic field, and when ramping H back down to zero, not all of the stored energy is recovered if Br is not zero. Recall that voltage is associated with dB/dt (Faraday's Law), so we are ramping the current up and down, and the circuit is responding by producing a voltage whenever dB/dt is non zero. In fact, B(t) is lagging behind H(t). IF H(t) is a sine wave, and B(t) is also a sine wave with the same phase, then dB(t)/dt is 90 degrees out of phase (V = L dI/dt. like an inductance should). However, if B(t) lags behind H(t), then dB(t)/dt is no longer 90 degrees out of phase. Thus the current I(t) and the voltage V(t) have an in-phase component, meaning a resistive component.

3. Mar 21, 2009

### b.shahvir

Dear Bob,

Thanx, your reply is quite appropriate and technically correct.....but i will keep the thread open to further dicussion.

Kind Regards,
Shahvir