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Consider a transformer with no load,

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Given \$\phi= \phi_m \sin (\omega t)\$

\$e_1 = N_1 \dfrac{d \phi}{dt} = N_1\omega \phi_m \cos (\omega t)\$

\$E_1 = e_{1rms} = N_1\omega \phi_m \ \frac{1}{\sqrt{2}} = 4.44 \phi_m N_1 f\$

Since the supply voltage \$v_1\$ is fixed, in order to satisfy Kirchhoff's law: \$v_1=e_1\$

\$\implies V_1 = E_1 = 4.44 \phi_m N_1 f\$

Therefore, the maximum value of \$ \phi\$ is determined by the supply voltage and frequency (given the remaining terms are constant)

\$\phi_m = \dfrac{V_1}{4.44 N_1 f} \$

Therefore even when the current is the cause of flux, the magnitude of flux is determined by the supply voltage.

  1. I understand this fact mathematically. But why is this fact true, intuitively or physically? (I'm not really sure what I mean by intuitively, but if there's a different perspective on this, please do share)
  2. The second question is kind of arbitrary: When we consider non-linearities of the B-H curve, it's the current that becomes "peaky" to keep the flux sinusoid, why is this true, isn't current the cause of \$\phi\$, why doesn't \$\phi\$ alter to draw a sinusoidal current?
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  • \$\begingroup\$ The current is the sum of magnetizing current (for the flux) and to the primary normalized secundary current. The value of the flux is calculated through integration of the primary voltage. \$\endgroup\$
    – HarryH
    May 22, 2021 at 16:21

3 Answers 3

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  1. Mutual coupling is dependent on the core material only that is shared between windings.but it does require an excitation voltage and current to energize this magnetic coupling. But that is independent of load current.
  2. Inductance is the ability to store charges by the change in magnetic orientation. When they have reach the limit of alignment they can no longer move , store energy and inductance drops rapidly while harmonic content rises at the same rate and eventually the wire behaves just the low wire resistance at peak voltages
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  • \$\begingroup\$ Ah I agree with the second answer, before saturation, any change in current causes a corresponding change in flux and therefore the induced emf shows uo, so it acts as an inductor, after saturation, it acts as a simple resistor (since it's just a wire, current peaks), correct? But I don't get the first answer. I understand that it is independent of load current, but I don't get why that's true \$\endgroup\$ May 22, 2021 at 4:08
  • \$\begingroup\$ It is when the core is excited with voltage in the linear region to use about 10% of it’s rated current that the core can share primary to secondary voltage efficiently without loss according the to relative area used to share and the area used by each winding. When it is not excited around this hysteresis band, the coupling is weak. But when sufficient flux cycles behave as a carrier between windings the N^2k transformation can occur of voltages or currents. \$\endgroup\$ May 22, 2021 at 4:55
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Why is the magnitude of mutual flux in a transformer independent of the current in primary?

In a transformer with no leakage inductance, and no coil resistance, the emf induced by the changing flux in the core must equal the voltage applied to the primary., at all times.

If, in such a transformer, the voltage in each cycle is fixed, then the flux must also be fixed, because it is the rate of change of flux that determines the induced emf. This is how "classical" transformer action works.

However, in a current transformer, it is not the voltage across the primary that is externally fixed, but the current through the primary. A current transformer obeys the same fundamental laws as a "classic" or voltage driven transformer. It is just operated in a different mode. In a current transformer, the induced emf is still determined by the rate of change of the flux. However, the induced emf is not externally fixed. In a current transformer, the externally fixed current determines the flux in the core.

The takeaway is that the apparent independence of the flux from the current is true only for transformers when their voltage is externally fixed, and is not a truth about transformers in themselves.

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  • \$\begingroup\$ Ow, I haven't read about CTs yet, thank you, this will surely save a ton of time for me in future \$\endgroup\$ May 22, 2021 at 12:08
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I understand this fact mathematically. But why is this fact true, intuitively or physically? (I'm not really sure what I mean by intuitively, but if there's a different perspective on this, please do share)

With the transformer unloaded on the secondary, all you have is one winding, that being the primary. That primary winding is just an inductor and your formulas are for a simple inductor. Hence, intuitively, the primary is just an inductor.

The second question is kind of arbitrary: When we consider non-linearities of the B-H curve, it's the current that becomes "peaky" to keep the flux sinusoid, why is this true, isn't current the cause of ϕ, why doesn't ϕ alter to draw a sinusoidal current?

Current has no other option than to grow non-linearly as saturation is approached (given that the "source" is a pure sinusoidal voltage).

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