If a transformer is rated for a maximum winding current of 10A, would it be permissible to operate it with 10A of input current when the secondary voltage to the load is 40V? Can you explain to me why?The information on the auto transformer plate wasInput 240V/ 50/60 cy 3HPoutput 0-240/280V 20 ampsThe parallel load resistance is 34.5 ohms
It's simpler than you think. If an auto transformer has a rating of 10A, it doesn't matter if it's 10A at 280V or 10A at 2.8V. The current rating is based on the the size of the wire used to wind it along with the wiper contact characteristics. The core doesn't care about the voltage, only the current. So if you want maximum power transfer from an autotransformer, feed it across the winding ends with the full rated voltage (in this case, 280VAC), which will allow you to draw 10A * whatever voltage you have it set at in watts. And of course, yes you can set it to 40V at 10A, as long as your input voltage is anything higher than 40V (if it is not, the step-up action will cause higher than 10A to flow through the portion of the winding that the input is across).
At 40 V and 10 A the VA is 400 on the secondary side. At 240 V on the primary side the current is 400/240 which is 1.66 A. If the load on the secondary side is 34.5 ohms at 40 V then the secondary current is 40/34.5 1.16 A and the the VA on the secondary side is 40 x 1.16 46.4. T he VA on the primary side then is 46.4 and the primary current is then 46.4/240 0.19 A. 3 hp is 746 x 3 Watts 2238 W and its rated current on the primary side is 2238/240 9.325 A. In other words the VA loaded on to the secondary side is the same VA (more or less) as that on the primary side.
