HOW DO YOU FIND THE PRIMARY CURRENT IN AN IDEAL TRANSFORMER?
Work out the secondary side completely first, then cross to the primary using conservation of power, because an ideal transformer transfers all of it. Voltage scales with the turns ratio and current scales with its inverse, so a transformer that steps voltage down steps current up.
An ideal transformer has 200 turns on the primary coil and 50 on the secondary. The primary is connected to a 240 V supply and the secondary supplies a 12 Ω resistor. What is the current in the primary coil?
Finish the secondary side before you touch the primary. The load tells you the secondary current; only then does power carry you back across.
- Vs = 240 × 50⁄200 = 60 V
- Is = 60⁄12 = 5 A
- ideal, so Pp = Ps = 60 × 5 = 300 W
- Ip = 300⁄240 = 1.25 A
20 A is what you get if you scale the current the same way as the voltage multiplying 5 A by 200/50 instead of dividing. Current goes the opposite way to voltage: the side with more turns carries the higher voltage and the lower current. Getting the ratio the right way up is most of what this question tests.
5 A is what you get if you stop at the secondary and quote the current through the resistor as the answer. It is the right value for the wrong coil.
Power is the bridge between the two sides here, not the turns ratio. Once you know the secondary delivers 300 W, the primary must draw 300 W, and 300 ÷ 240 = 1.25 A follows immediately. Working through the current ratio instead gives the same answer but hands you an extra chance to invert it.
THE GENERAL RULE
An ideal transformer conserves power, so anything you know completely on one side crosses to the other through P = VI. Treat the load as part of the secondary circuit, solve that side fully, and only then cross. The trap in every transformer question is the same: the turns ratio applies to voltage directly and to current inversely, and the wrong-way-up answer is always on the list.
Updated · written to the published ESAT content specification