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I am honestly confused, A transformer "Steps up the voltage and steps down the current" or vice versa. So how does it obey Ohm's law? for example, a transformer with a winding relation of 2:3, should make 4V 30mA become 6V 20mA, right? but I=V/R, so is there some internal resistance on the secondary winding? I know that impedance could be the answer, but does varying the input resistance varies the inductance of the secondary coil?

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    \$\begingroup\$ A transformer doesn't have to obey Ohm's law. Ohm's law says an ideal resistor has a certain I-V relationship. It doesn't say anything about what a transformer does. \$\endgroup\$ – The Photon Jan 18 at 16:35
  • \$\begingroup\$ Related: Ohm's Law of a circuit which have both Voltage Source and Current Source \$\endgroup\$ – The Photon Jan 18 at 16:36
  • \$\begingroup\$ The Photon thanks, but that doesn't really answer my question, putting in other words, if I have: An AC source with RMS voltage being 2v, connected to the primary coil of a transformer with W2/W1 = 1.5, in series with a 1k resistor The secondary coil being connected only to a 10k resistor. The voltage in the first coil would be 2v, and the current would be 2mA, so the output would be 3v and about 6mA, but the 10k resistor should make the current be 0.3mA, so what is the current in the secondary coil? \$\endgroup\$ – Locked Jan 18 at 17:16
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    \$\begingroup\$ Looking in to the primary, the impedance of the secondary's load will be transformed by a factor \$(W_1/W_2)^2\$, so the primary will "look like" a 4.4 kohm load. Now you have a voltage divider between your 1 kohm source resistance and the 4.4-kohm equivalent load of the primary. So the voltage across the primary won't be 2 Vrms, it will be about 1.6 Vrms. \$\endgroup\$ – The Photon Jan 18 at 18:04
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    \$\begingroup\$ And the current won't be 2 mA, it will be 2 V divided by the total load of 1+4.4 kohm, or about 0.37 mArms. (Secondary current will then be about 0.25 mA) \$\endgroup\$ – The Photon Jan 18 at 18:07
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Ohm's law describes the behavior of a resistor; it doesn't say anything about an ideal transformer.

It is true that the resistance of the windings causes a power loss and a real transformer doesn't behave like an ideal transformer.

The resistance and inductance are not directly rated, so no, changing the resistance does not necessarily change the inductance.

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  • \$\begingroup\$ Yes, but how is the current 20mA in the secondary? And I asked about the inductance because of the impedance being 1/(2*pifrequencyinductance), since 2*pi is a constant and frequency doesn't change, the only way this could change is varying the inductance. \$\endgroup\$ – Locked Jan 18 at 17:21
  • \$\begingroup\$ Well it certainly is true that the inductance of the primary and secondary are related in a way that reflects the ratios of primary-to-secondary current and voltage. But you seem to be confusing the first-order behavior of an ideal transformer with the second-order, non-ideal behavior of a real transformer. \$\endgroup\$ – Elliot Alderson Jan 18 at 17:31
  • \$\begingroup\$ Its also worth noting that, because no real transformer is ideal, there are transformer efficiencies which tell you how much is lost due to heat, resistances, etc. So for example, a transformer of 90% efficiency will give you out 90W if it had 100W input. \$\endgroup\$ – QuickishFM Jan 18 at 17:48
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There is a no-load excitation current and associated conductive loss and Mutual coupling ratio from 99.9 to 90% which in smaller units can account for 10% loss and drop where they rate the output voltage at rated power. So no load can be 10% higher voltage.

There are models for this and Ohm's law may be applied with Vin(f)=I*f)*Z(f) + losses for output V,I (f).

I won't repeat them here, but they exist.

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