In a setting like that in an ideal transformer. An AC source connected to the primary induces an AC voltage in the secondary. Would the secondary coil have an inductive reactance in this process?
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2\$\begingroup\$ What do you mean with the secondary "having" a reactance? Do you mean whether it has a measurable effect on the primary or not? Or do you mean whether there will be a phase shift between the primary and secondary voltage? Additionally, is anything connected to the secondary? It might help if you add a schematic via the built-in schematic editor. \$\endgroup\$Jonathan S.– Jonathan S.2024-02-22 22:42:30 +00:00Commented Feb 22, 2024 at 22:42
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\$\begingroup\$ The secondary is connected to a resistor. what I mean is to say would in the secondary circuit be any \$X_{L}\$? \$\endgroup\$Jack– Jack2024-02-22 22:46:27 +00:00Commented Feb 22, 2024 at 22:46
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\$\begingroup\$ Is the transformer ideal or not? Also, as already said, please edit the question and add a schematic. \$\endgroup\$Jonathan S.– Jonathan S.2024-02-22 22:48:34 +00:00Commented Feb 22, 2024 at 22:48
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\$\begingroup\$ it is an ideal transformer. I don't have a schematic because it is just something I was thinking about recently. \$\endgroup\$Jack– Jack2024-02-22 22:52:54 +00:00Commented Feb 22, 2024 at 22:52
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Would the secondary coil have an inductive reactance in this process?
For a non-ideal inductance you will have leakage inductance. So, the secondary would have a small inductive reactance based on the coupling between primary and secondary not being 100%. If it's a 1:1 transformer and the magnetization inductance is 1 henry with coupling at 98%, the leakage inductance is 20 mH. If the transformer steps down by 10:1, the leakage inductance at the secondary would be 100 times smaller at 200 μH.
For an ideal inductance, there can be no leakage inductance and, if the primary source is a 100% voltage source then there is no inductance associated with the secondary. However, if the primary source is not ideal then, you will see a transformed impedance at the secondary and, if the source is inductive then, you will register an inductance at the secondary.
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\$\begingroup\$ I have edited the question saying it is an ideal transformer. \$\endgroup\$Jack– Jack2024-02-22 22:55:55 +00:00Commented Feb 22, 2024 at 22:55
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\$\begingroup\$ So here is what i am thinking. In a simple AC circuit, an inductor connected to an AC source for example, would have \$X_{L}\$ because of self induction. So, i am wondering would the same happens due to mutual induction as well like that case in a transformer? \$\endgroup\$Jack– Jack2024-02-22 23:01:30 +00:00Commented Feb 22, 2024 at 23:01
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1\$\begingroup\$ If the transformer is ideal, as explained in my modified answer, the coupling is 100% and no secondary inductance can exist unless the source at the primary has built-in inductance. \$\endgroup\$Andy aka– Andy aka2024-02-22 23:08:25 +00:00Commented Feb 22, 2024 at 23:08