For my Circuits II final design project I am given an old transformer that needs to have its power factor corrected to 1 but I am a bit unclear what needs to be done. I'm using this post Power Factor Correction as a reference for power factor information, but I don't really understand it.

Here are the transformer specs:

  • 240V to 120 V at 60 Hz
  • Single-phase
  • Isolated
  • Low coupling
  • Single primary, dual secondaries

Using a Z-Y Bridge, it was found that Z = (0 + 300j) \$\Omega\$ and Y = (79 + 0j) \$\mu Mhos\$ which are the two measurements I was told were needed. I was also told that an LRC circuit might be the way to go (not sure why exactly). It is noted that the transformer purely inductive to the 60 Hz and therefore (correct me if wrong please) requires a capacitor be placed across transformer to cancel out the inductive reactance. Does this cancellation result in the power factor being 1?

\$ Z_C = \frac{1}{j\omega C} = Z_L = j\omega L = 300j\$ then solve for C? If so, where does the conductance come into play?

My expected layout is one of these three:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Nevermind my deleted comment, I was hallucinating. \$\endgroup\$ Nov 30, 2013 at 11:06
  • \$\begingroup\$ Could you draw a circuit diagram of how you intend to solve it? Mind you that if you put a capacitor in parallel to an inductor where the reactive current is exactly canceled out, is called resonance and I doubt you want that. \$\endgroup\$
    – jippie
    Nov 30, 2013 at 15:18
  • \$\begingroup\$ @jippie I was thinking maybe one of these two setups, or both combined? i.sstatic.net/E8yLe.jpg I assumed a capacitor was needed because of the linked question and beacuse the professor mentioned a capacitor being needed. \$\endgroup\$ Nov 30, 2013 at 16:51
  • \$\begingroup\$ We do have a nice circuit diagram editor. If you edit your question, then hit Ctrl-M (or click the Schematic) button at the top of the edit pane. \$\endgroup\$
    – jippie
    Nov 30, 2013 at 17:15
  • \$\begingroup\$ @jippie there is no physical difference between power factor correction using a capacitor and parallel resonant tuning of a coil. The two are identical in concept and the formulas are the same. \$\endgroup\$
    – Andy aka
    Nov 30, 2013 at 18:07

1 Answer 1


Power factor correction of a linear inductive load is exactly the same as tuning a parallel circuit of a capacitor and inductor. You pick a capacitor that works with the value of the transformer's magnetizing inductance to satisfy this: -

60Hz = \$\dfrac{1}{2\Pi\sqrt{L_M.C}}\$

Where \$L_M\$ is the transformer's magnetizing inductance and C is the capacitor chosen to "neutralize" the current taken by the coil.

For a pure lossless inductance and capacitance, the resultant current taken from the supply is zero.

If your transformer primary indicates an inductive reactance at 60Hz of 300 ohms, the magnetizing inductance value is this divided by \$2\Pi\times 60\$ = 0.8 henries.

The capacitive reactance required is 300 ohms and this is 8.8 uF. As a sanity check: -

F = \$\dfrac{1}{2\Pi\sqrt{0.8\times 8.8\times 10^{-6}}}\$ = 59.98Hz

  • \$\begingroup\$ Makes very good sense to me, the equation is the same for a resonant LC. How does the conductance found work into this? I was told it was needed to solve the problem. \$\endgroup\$ Nov 30, 2013 at 19:20
  • \$\begingroup\$ a parallel resistor (conductance) makes no difference at all - it represents a power loss that cannot be fixed because it's phase angle is precisely zero. \$\endgroup\$
    – Andy aka
    Nov 30, 2013 at 21:44

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