Power in a transformer
In any device, some energy is lost as forms other than the one we want (heat, for example), so the efficiency is never 100%. In a transformer too, not all the energy given to the primary can be taken from the secondary.
But if we assume an ideal transformer with no energy loss, its efficiency is 100%, and the power in the primary equals the power in the secondary. Since power = potential difference × current:
Power in = power out
VP IP = VS IS
IP, IS — the currents in the primary and secondary; VP, VS — their potential differences
Worked example
A transformer has 500 turns on its primary coil and 5000 on its secondary. An AC supply of 12 V is connected to the primary. (i) Find the secondary voltage. (ii) If a current of 2 A flows in the primary, find the current in the secondary. (iii) What kind of transformer is it?
(i) NP = 500, NS = 5000, VP = 12 V
VS = VP × NS / NP = 12 V × 5000 / 500
VS = 120 V
(ii) VP IP = VS IS, so IS = VP IP / VS = 12 V × 2 A / 120 V
IS = 0.2 A
(iii) The secondary has more turns than the primary, so the output voltage is higher than the input: a step-up transformer.
Why the current goes down as the voltage goes up
Look at the example: the voltage went up ten times, from 12 V to 120 V, and the current went down ten times, from 2 A to 0.2 A. The power is 24 W on both sides. That has to be so, because a transformer has no source of energy of its own — everything the secondary gives out comes in through the primary. This is the reason for the high-voltage grid on the last page: stepping up the voltage steps down the current by the same factor, and a small current wastes little energy in the long wires.
Check the answer with power
If your answer gives more power out of the secondary than into the primary, something has gone wrong: an ideal transformer gives out exactly what it takes in, and a real one less.
Power in equals power out
In an ideal transformer the power in equals the power out, so the current changes in the opposite ratio to the voltage.