Lets say 120 Volts at 60 Hz with a Transformer across it with 1 Henry as the primary and 1 Henry on the secondary. The secondary has no load and is an open circuit. How would I physically measure the Apparent Power of the primary coil?

Side question, if secondary is shorted, will voltage across the primary head towards zero?


1 Answer 1


Apparent power is \$V_{RMS}\times I_{RMS}\$ as shown on the phasor diagram below: -

enter image description here

To measure the apparent power you multiply the RMS measurements of voltage and current.

If you had a wattmeter you could also measure the true power and then compute reactive power using \$ \sqrt{P_{apparent}^2 - P_{true}^2}\$.

This enables you to compute iron losses in the transformer.

Side question, if secondary is shorted, will voltage across the primary head towards zero?

Don't short the secondary unless you are connected to a supply that is much, much lower than the normal intended operating voltage of the primary or you might get a fire.

Shorting the secondary (in order to determine copper losses for instance) is usually done by controlling the primary voltage with a variable transformer such that the shorted secondary current is at (and not above) full operating level. This usually means that the primary voltage is down at possibly one-twentieth of its normal operating level.

  • \$\begingroup\$ Thanks. So I should be good if I use an ammeter and a volt meter and multiply the results. Regarding the short circuit of the secondary, I didn't quite understand it. If I were to measure 5 volts across the primary of a transformer, and then short the secondary, realistically, would I still see about 5 volts? or would it go down one-twentieth as you said? Thanks. \$\endgroup\$
    – user29150
    Sep 17, 2013 at 15:29
  • \$\begingroup\$ @user29150 Yes, this will give you \$P_{apparent}\$ \$\endgroup\$
    – Andy aka
    Sep 17, 2013 at 15:37
  • 1
    \$\begingroup\$ To get this information the normal way is to ramp your supply voltage until you reach the maximum current on the secondary. Trying this any other was risks melting the windings. Using a variac is usual way. \$\endgroup\$
    – Spoon
    Sep 17, 2013 at 17:26

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