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So let's say you step down 120 volts with a current of 0.1 amp, to 12 volts with a current of 1 amp transformer.

What does is mean exactly when the current steps up to 1 amp in my example? Does this mean when you short circuit the secondary, you will have 1 amp of current (neglecting dc resistance of coil)? What about with the DC resistance of the coil?

Or does it mean that you can only draw 1 amp total from different loads set in parallel?

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  • \$\begingroup\$ 120V @ 0.1A is 12W. 12V @ 1A is 12W. It means power is conserved. \$\endgroup\$ – Connor Wolf Aug 19 '13 at 8:30
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In circuit theory, the definition of a short circuit is 0 V across its two terminals.

In your example, this would mean you are forcing an infinite current through the transformer's output (and, as a consequence, through the input, too), because the short is reflected to the primary side, and the network feeding the transformer will, theoretically, force an infinite current into the transformer.

In real life, the DC resistance of the copper windings (secondary and primary!) will prevent the universe from melting down, but your transformer might become too hot.

Now, the rating tells you what you can do with your transformer under proper conditions: Connect a load that doesn't take more than 1 A when being fed with 12 V. If your load uses less than 1 A at 12 V, you wil still get an output voltage of 12 V because the ratio of the windings defines the ratio of the voltages.

Real transformers take their losses into account and are designed for the rated voltage output at their rated current, so they will output a bit more than the rated voltage under no-load or light-load conditions.

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  • \$\begingroup\$ I would like to understand why (I know it is correct I just can't understand why) if the secondary is short circuited the primary winding will also reflect the short circuit despite the fact that they are isolated and the number of turns remains the same on every winding. Would the electromagnetic flux through the core be increased with a short circuit on the secondary? Thanks \$\endgroup\$ – VMMF Sep 24 '18 at 18:47

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