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1. From the voltage readings in no-load condition

2. From the current reading in loaded condition

We used an auto transformer (VARIAC) to supply voltage to a single phase transformer. Supplied voltage was 220 V. Then we measured voltages across the high voltage side and low voltage side in no-load condition of the single phase transformer. V1/V2 = M is used to find the turns ratio.

Then, we connected a load to the low voltage side and then measured currents and voltages both across the high voltage and low voltage side. I2/I1 = M is used to find the turns ratio.

Now, voltage ratio gives more accurate result than the current ratio. But why's that?

In no-load condition there's no current flow in the secondary side which results in a very low current flow on the primary side, that is why voltage drop because of the internal impedance in the source is very small and we get almost same result as the ideal transformer.

In loaded condition, there's current flow in the secondary as well as in the primary, so the voltage drop is larger than the no-load condition. But what I don't understand is that how we are relating current readings in the determination of turns ratio and why the discrepancy is larger here? I understand that the voltages both on the primary and secondary side are being affected because of the load but how's it affecting the current flow?

We connected one AC ammeter between the auto transformer and single phase transformer and another between the single phase transformer and the load. That means we are measuring Ip and Is. But the ideal transformer equation is N1/N2 = I2/I1, where I1 is not equal to Ip in practical transformer. I can't understand the situation here.

The discrepancy is in both voltage and current ratio because of the source impedance, load impedance and losses in the primary and secondary side - is my understanding correct here? If it's right, then why current ratio shows more discrepancy?

Please be kind enough to explain the operation here. Thank you.