Hi currently have a 3phase 50HP motor with an 80% Power factor

I need to bring it up 95% power factor.

What capacitor size would i need to achieve this.

If someone can show the formulas they used that would be great!


This is a link to the motor https://inventory.powerzone.com/item/55179/used-50-hp-vertical-electric-motor-reliance

575 volts, 48amps

If we assumed a power factor of 80%. how would i go about correcting it to 95% power factor?

  • \$\begingroup\$ You need to know the current and voltage. There are a lot of online calculators, downloadable instructions explanations etc. \$\endgroup\$
    – user80875
    Jul 3, 2020 at 0:13
  • \$\begingroup\$ @CharlesCowie Can you link to one? \$\endgroup\$
    – Drew
    Jul 3, 2020 at 0:15
  • \$\begingroup\$ This Eaton guide looks useful, and here are a zillion possible links to calculators \$\endgroup\$
    – Russell McMahon
    Jul 3, 2020 at 3:08
  • \$\begingroup\$ The very first result (of the zillion) that @RussellMcMahon provides gives the same answer i get below (129uF). Here is that specific link. \$\endgroup\$ Jul 3, 2020 at 3:33

1 Answer 1


First, use the motor nameplate data to find kva (apparent power). Subscript 3 below indicates 3 phase.

$$ S_3 = \frac{HP * 746}{power factor * efficiency}$$

But, in your case the nameplate gives rated current as 48A, so

$$ S_3 = 48A * 575 * \sqrt3 = 47.8kva$$

$$ P_3 = S_3 * power factor = 47.8 * 0.80 = 38.24kW $$


$$ Q_3 = \sqrt{S_3^2-P_3^2} = 28.7kvar $$

Now, to get to your desired 0.95 power factor you need to provide some of that kvar from your capacitor.

$$ {desired power factor angle} = cos^-{^1}(0.95) = 18.2⁰$$

$$18.2⁰ = tan^-{^1}(Q_3/P_3)$$

$$18.2⁰= tan^-{^1}(Q_{target}/38.24kW) $$ So, $$Q_{target} = 12.6kvar $$

That means you need a capacitor that will supply an additional (28.7-12.6) = 16.1kvar 3-phase to the motor so that only 12.6kvar comes from the source.

Since, $$X_C = \frac{V^2}{Q}$$ $$X_C = \frac{575^2}{16.1kvar} = 20.5Ω $$

From this, we can find C (assuming 60Hz) as $$ C = \frac{1}{2*π*f*X_C} = 129uF $$

EDIT: I noticed that bxjockey had given us nameplate amps so we can directly calculate apparent power. That changed the subsequent results which i have now corrected.

  • \$\begingroup\$ By the way, that value of C would be for wye connection. If you connect the caps in delta you need to use 1/3 that value (43uF) to produce same kvar. \$\endgroup\$ Jul 3, 2020 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.