How Does Voltage Affect Current in Magnetic Circuits?

[Shikhar]

how magnetic circuits work?

hi everyone there...this is my first post and I'm here to find d solution to a very basic but confusing doubt!

in all magnetic circuits such as a transformer (consider single phase operating at knee point), if we apply a sinusoidal voltage source, the flux is taken to be sinusoidal and the current waveform as distorted. (all books say this without any explanation )

but as far as my thinking goes, the current should be sinusoidal and the flux waveform should be dependent on the operating point on the B-H curve.
...why is it not this way?

 

[berkeman]

hi everyone there...this is my first post and I'm here to find d solution to a very basic but confusing doubt!

in all magnetic circuits such as a transformer (consider single phase operating at knee point), if we apply a sinusoidal voltage source, the flux is taken to be sinusoidal and the current waveform as distorted. (all books say this without any explanation )

but as far as my thinking goes, the current should be sinusoidal and the flux waveform should be dependent on the operating point on the B-H curve.
...why is it not this way?

As long as you are not driving the transformer into saturation, and as long as you have a load resistance (no open circit on the secondary, the currents and flux should follow the primary voltage, and be sinusoidal.

 

[Shikhar]

As long as you are not driving the transformer into saturation, and as long as you have a load resistance (no open circit on the secondary, the currents and flux should follow the primary voltage, and be sinusoidal.

normally all electrical machines are operated at the edge of saturation (alse called the knee point)...so the B-H relationship is obviously non-linear at that point...so either of current or flux have to be non-sinusoidal...
my question is why current is taken non-sinusoidal and not flux...as current is the cause(mmf) that produces the flux!

 

[Bob S]

For transformers, the primary voltage is sinusoidal, and the primary current (with no secondary load) is non-sinusoidal. This is because

V = L(I) dI/dt

Because the inductance L is dependent on the current I, the differential current change dI, given by

dI = V dt/(L(I)

becomes nonlinear and increases rapidly as the knee in B is approached and L(I) decreases.

Bob S

 

[Shikhar]

For transformers, the primary voltage is sinusoidal, and the primary current (with no secondary load) is non-sinusoidal. This is because

V = L(I) dI/dt

Because the inductance L is dependent on the current I, the differential current change dI, given by

dI = V dt/(L(I)

becomes nonlinear and increases rapidly as the knee in B is approached and L(I) decreases.

Bob S

Thanx for the reply Bob...but i still have a doubt...
if you are saying that inductance depends on current, that means even at operating point well below the knee point, the current should b non-sinusoidal...which is usually not the case!
as far my knowledge goes, inductance remains constant for a coil irrespective of the current!

 

[m.s.j]

It is very clear. The relation of emf and flux is linear ( e=N.dø/dt ),if we neglect from transformer internal resistance we can write:

V= e = N.dø/dt

Therefore while we have sinusoidal voltage wave form we face to sinusoidal flux. But the relation of flux and magnetizing current is nonlinear and specify with transformer magnetizing curve, so we front to nonsinusidual magnetizing current (transformer no load current) even we have sinusoidal voltage and flux.


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Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[Bob S]

It is very clear. The relation of emf and flux is linear ( e=N.dø/dt ),if we neglect from transformer internal resistance we can write:

V= e = N.dø/dt

The relation between emf and flux is not linear.

Lat's first write down the magnetic (inductive) energy stored in a circuit containing nonlinear magnetic material:

½LI² = ½∫B·H dV_vol

where the volume integral ∫dV_vol is over all space. The left side is the EE version of the stored energy, and the right side is the physics version.

Substituting B = μμ₀H where μ is the relative (nonlinear) permeability and μ₀ is 4 pi x 10^-7 Henrys per meter.

So we now have

LI² = μμ₀ ∫H² dV_vol

Now, H = NI/z where N = # of turns, I = current, and z = magnetic path length

So L = μ(I)·μ₀N² V/z² Henrys

where I explicitly show the dependence of the relative permeability μ(I) on current (amp-turns).

So the inductance depends on current.

Furthermore,

dL = μ₀N² V/z² [∂μ(I)/∂I] dI

V = d(LI)/dt = L dI/dt + I dL/dt

Substituting, we get

V = L(I) dI/dt + I dL(I)/dt = L(I) dI/dt + μ₀N² V/z² [∂μ(I)/∂I]·I dI/dt ={L(I) + μ₀N² V/z² [∂μ(I)/∂I]I} dI/dt

So the inductance in the usual EE equation V = "L" dI/dt becomes

"L" → L(I) + μ₀N² V/z² [∂μ(I)/∂I]·I

The second term is zero if the relative permeability is independent of current.

The inductance L depends nonlinearly on the current I. The flux depends nonlinearly on the current. The inductance and flux are independent of the emf except through their dependence on the current.

Bob S

 

[uart]

It is very clear. The relation of emf and flux is linear ( e=N.dø/dt ),if we neglect from transformer internal resistance we can write:

V= e = N.dø/dt

Therefore while we have sinusoidal voltage wave form we face to sinusoidal flux. But the relation of flux and magnetizing current is nonlinear and specify with transformer magnetizing curve, so we front to nonsinusidual magnetizing current (transformer no load current) even we have sinusoidal voltage and flux.

Agreed msj. (and Bob I think you are way over complicating a simple issue here.)

Neglecting resistance \phi = \frac{1}{N} \int v \, dt[/tex]. This implies that if v is sinusoidal then \phi is sinusoidal. It's basic maths.

 

[Bob S]

For the effect of nonlinear core saturation on current on voltage, see the thread
https://www.physicsforums.com/showthread.php?t=353333
and in particular post #12 with attachment. Here in my attachment are voltage and current waveforms from the thread for a transformer without a secondary load. The nonlinear inductance causes the primary current to spike when the knee in the permeability is reached. Excitation voltage is on top, and the primary current is on bottom.

Bob S

[added] Note that the peak (reactive) current lags the peak applied voltage by 90 degrees.

 

[uart]

Excitation voltage is on top, and the primary current is on bottom.

Yes no problems with that, but the flux is not shown. The flux is proportional to the time integral of the excitation (which is sinusoidal). The flux is not proportional to the magnetizing current in the case of a nonlinear core.

See my diagram of flux versus current attached. It's very crude (doesn't include hysteresis) but shows basically what's going on. Notice how a sinusoidal flux implies a non-sinusoidal current (and visa vera)

The whole point of this thread was to explain to the OP why it is the former (flux sinusoidal, current distorted) rather than the latter (visa versa) that is typically the case. The reason is that given by msj in post #6. It is because the flux is linearly related to the voltage and the current is non-linearly related to the flux. So for a sinusoidal voltage driven transformer it's the flux that is sinusoidal and the current that is distorted (that is, current distorted as shown in your attachment).

 

[Bob S]

The flux Φ=B·A (where A = area), and the flux density B, are both functions of the iron core magnetization H. In mks units, the flux Φ is measured in webers, the flux density B in webers per m² or Tesla, and the magnetization H in amp-turns per meter. See attachment for flux density plots for soft iron.

The attachment (left hand scale) plots the flux density (solid curve) against the magnetizing force, in amp-turns per meter. The relative permeability (dotted line), right hand scale, is also plotted.

Very specifically, B = μμ₀ N·I

where μ is the relative permeability, μ₀ is the permeability of free space, and NI = amps I times turns N. Note the knee in the flux density curve at ~ 1.2 Tesla.

Bob S

[added] The Faraday Law is

V = -N (d/dt) ∫B·n dA

∫V·dt = -N·B·A = -N·Φ

 

[m.s.j]

I think a good engineer can be a physician but just a genius physician may be an engineer.



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Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[Adjuster]

I think a good engineer can be a physician but just a genius physician may be an engineer.

I really can't agree with this, even if you really meant to say PHYSICIST (in English, the word PHYSICIAN means "medical doctor". I would like to say more - but this is going off topic.

 

[Shikhar]

thanx every one for ur posts...i really got my answer...