What would happen if I wind another winding (call tertiary winding with the same number of turns) on the transformer core and apply an AC voltage to it which is in phase with the primary voltage but is a different value in magnitude? Let's say that the primary voltage is V1 and the impressed tertiary voltage is V3. The flux with only the primary would be V1/(4.44fN). What value would the new flux settle for? In effect, what would be the flux value if there are two primaries?
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\$\begingroup\$ Instead of adding comments - edit your original question. \$\endgroup\$– Solar MikeCommented May 21, 2018 at 7:38
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\$\begingroup\$ @winny Let's assume that V1=V3=1000V. Since E1 is close to V1, let's assume E1 to be 990V and let the initial flux value be x. Now, with the third winding, new flux value=2x (Since flux adds up according to what you said). Correspondingly, E1 now equals 1980V (since its proportional to flux). However, this doesn't seem plausible as E1>V1 and hence current direction would be negative. In short, these cannot be steady-state values according to me. Please correct me if I am wrong. \$\endgroup\$– user172541Commented May 21, 2018 at 8:40
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\$\begingroup\$ Draw a schematic with dots for each winding. "hence current direction would be negative" Although you may see it in terms of current direction, I find it much easier to just use Vt=NAB and think in terms of voltage and phase and leave the current out of the equation. \$\endgroup\$– winnyCommented May 21, 2018 at 8:54
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With two identical windings both driven in phase, the winding with the highest supply voltage will produce the magnetization current for the core. In addition to that magnetization current, it will also supply a potentially excessive current to the winding that is driven with a lower voltage.
This scenario will cause a significant volt drop across the leakage reactances and inevitably reduce the voltage that excites the core magnetization inductance. Because of the leakage component volt-drops, the magnetization current would reduce.
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\$\begingroup\$ Yup, the excessive current would burn the windings! Now, what if I keep the ratio of voltage and turns in both the windings same? (For instance V1=1000V, N1=1000 turns, V2=100V, N2=100 turns) \$\endgroup\$– user172541Commented May 21, 2018 at 10:37
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1\$\begingroup\$ In that situation, each winding supplies precisely 50% of the magnetization current. Imagine you wound a coil with two wires simultaneously and joined them at the ends - that is what you are describing - have you ever heard of Litz wire? \$\endgroup\$– Andy akaCommented May 21, 2018 at 11:20
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\$\begingroup\$ Thanks a lot for your answer. I want to ask about one more possibility. What if the voltage to turns ratio is same but the phase difference between the voltages is not zero? Can you please shed some light on this, preferably with a phasor diagram? \$\endgroup\$– user172541Commented May 21, 2018 at 11:41
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1\$\begingroup\$ I think that if you used a simulator (microcap or LTSpice are both free) and modelled a transformer you would get all the answers you want and also be able to try out other scenarios. I know your original question and understand where you are trying to get but, you won't be able to regulate a transformer by controlling the magnetization flux. If you had, on the other hand, a transformer with a saturable core you can control regulation of the output voltage with a DC current into a tertiary winding but this is very inefficient. Try looking up saturable reactors or magnetic amplifiers. \$\endgroup\$– Andy akaCommented May 21, 2018 at 11:48