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This is, probably, a very silly question, but I was thinking:

If you have a watt meter and you connect a step up transformer with the lower voltage side attached attached to it (both the voltage and current wiring,) and you connect your load on the higher voltage side of the transformer, will you measure a lower value?

I think you would, unless you connect the potential measuring to the high voltage side of the transformer (in parallel to the load.)

If this is so, what would happen if you connected a step down transformer to the higher voltage side of the step up transformer?

Assuming the step down has a turn ratio that exactly reverts the initial transformation, if you connect the load to low voltage side of the step down tranaformer and keep the watt meter on the initial position, what would you measure? Would you measure the correct value, or would it still be wrong? (No this is not a homework question not yet, anyways.)

Sorry if this sounds odd or just confusing, but I had this thought on the way home today, and couldn't figure out if I analysed it correctly or wrong. I think you would still get a wrong reading. Am I wrong in thinking so? enter image description here

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    \$\begingroup\$ I think you need to add a diagram of what you're thinking, or really work on clarifying your language. \$\endgroup\$ – TimWescott Mar 19 at 2:07
  • \$\begingroup\$ Just a tip, on a professional website like Stack Exchange, try to avoid pronouns like "u" and "ur". Also, we don't "analise" here. \$\endgroup\$ – Bart Mar 19 at 7:55
  • \$\begingroup\$ Ok, sorry about the language usage. \$\endgroup\$ – Jack Mar 19 at 13:12
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With real transformers, your watt meter will show more power being consumed when you use a transformer.

Say you have a device that needs 20 volts AC, and requires 40 watts of power while operating.

You have a step down transformer that goes from 240 volts AC to 20 volts AC.

If you use your wattmeter to measure the power the device is consuming (on the low voltage side,) then you will see it consuming 40 watts

If you use your wattmeter to measure the power going in (the high voltage side of the transformer) then you will measure more than 40 watts being consumed. That's the 40 watts your device needs, plus the power the transformer wastes in doing its work.

If the transformer were perfect, then you would measure the same power being consumed on the high voltage side as on the low voltage side.

If you have a device that needs 240 volts AC, and you chain two transformers as in your drawing, then you would measure more power going in (all the way at the left in your drawing) than the device is consuming (all the way at the right in your drawing.)

The losses in this case are doubled - you have two transformers wasting power instead of one.

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  • \$\begingroup\$ Perfect answer! I thought about this, and after trying to write equations for it and getting the same result you guys suggested, I tried in a simulation on multisim! The only way the error I speculated over can occur, is if the watt meter is connected wrongly! \$\endgroup\$ – Jack Apr 21 at 19:01
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An ideal transformer will not attenuate power taken by a load no matter what the turns ratio. An ideal wattmeter will measure the same power on either side of the transformer. It's a non-event that you speculate over.

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