I have a school exercise that goes like this:

Voltage is 230V. Frequency is 50Hz. Inductor is 400mH.

How big of a resistor should be added with the inductor (in a series) to reduce reactive power by 75%? (The correct answer is apparently 218 ohms.)

How do you calculate this? Is this even possible? I know only of capacitors reducing reactive power.

I can only figure out that 25% of the reactive power is be 105,24 var. Also sorry, I don't study in English, so the terms might be a bit strange. Thank you.

  • \$\begingroup\$ it's perfectly possible. You simply calculate the new allowed current through the inductor, what total impedance would pass that current from 230V 50Hz and subtract with the pythagoras rule the already existing inductive reactance from the total impedance. \$\endgroup\$ – user287001 Oct 1 at 17:17

The resistor in series with the inductor just serves to reduce the overall current draw through the inductor. Thus, the resulting reactive drop (Q) in the inductor goes down.

The reactive drop in the inductor is,

$$ Q = I^2X_L$$

You can calculate the current with the formula I give above.

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