# Transformer circuit ratio-need help

Discussion in 'Homework Help' started by kevin monroe, Nov 7, 2014.

1. ### kevin monroe Thread Starter New Member

Nov 7, 2014
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0
I'm having trouble grasping current ratios, and thus power transfer, in a transformer. Assuming 100% of the primary voltage is transmitted across the transformer, the voltages are the same in the primary and secondary circuits if the turn ratio is 1:1. It seems like the current in the secondary circuit would vary with the resistance load in the secondary circuit, and thus, not be held to the cited fixed ratio of secondary current/primary current=primary voltage/secondary voltage. It seems like the secondary circuit current would just obey Ohms law, based on the voltage and resistance in that circuit.
I would appreciate any insights to this matter. I'm sure I am missing something obvious.

2. ### crutschow Expert

Mar 14, 2008
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4,477
I'm not sure I understand you confusion since you seem to understand how the transformer works. If an ideal transformer has a 1:1 turns ratio the primary and secondary voltages and currents are identical as determined by the secondary load impedance. Voltage/current for the primary equals voltage/current for the secondary.

3. ### WBahn Moderator

Mar 31, 2012
20,267
5,763
What you are failing to recognize is that the current in the secondary is what determines the current in the primary. In an ideal transformer, if you have put a larger resistance into the secondary circuit and the current drops, then the current draw of the primary will drop by a proportionate amount. If the secondary is open circuited, then the primary will draw no current at all. In reality, of course, this is not quite true and some current will be drawn, but not much. This is where the transformer losses come from.

4. ### kevin monroe Thread Starter New Member

Nov 7, 2014
11
0
Thanks a lot!

5. ### kevin monroe Thread Starter New Member

Nov 7, 2014
11
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Why would primary current go to near zero if the secondary current is open? Thanks in advance.

6. ### WBahn Moderator

Mar 31, 2012
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Here's somewhat of a hand-wavy explanation. With no load applied to the secondary, the secondary coil might as well not be there and what you have is an inductor that typically has a very large inductance and, hence, a very large impedance even at the low power line frequencies used in most countries. For instance, a transformer with a 10 H primary has an impedance of nearly 4 kΩ at 60 Hz, so with 120 Vac applied, only about 30 mA will flow. This is because you are establishing a back-emf that largely cancels the applied emf to the coil. But when you have a second coil coupled to the first, you set up magnetic flux in it and, if current flows in it, that current sets up magnetic flux in the first coil that tends to counter the back-emf that the first coil is generating in itself and this permits more current to flow.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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One could also argue as follows...
Faraday's law of induction requires a certain flux to exist in the transformer magnetic circuit to satisfy the applied primary voltage. This is normally formalized in the well known fundamental transformer equation. Neglecting any losses or non-ideal flux leakage but allowing for a finite primary inductance, requires a certain magnetic excitation current to be present - irrespective of any secondary current.
When the secondary carries load current, Faraday's law still requires that the same magnetic flux exists in the magnetic circuit, provided the applied primary voltage remains unchanged. This 'constant' flux condition can only be the case if the primary and secondary load related mmfs ('ampere turns') exactly cancel.

8. ### kevin monroe Thread Starter New Member

Nov 7, 2014
11
0
Thanks. I don't think my book really covers this well. There are 3 sentences on reflective impedance which I think is the subject of your post.