Greetings,

The the webpage http://www.lmphotonics.com/pwrfact.htm states:

(under "Supply Harmonics")
> Adding the inductance in series with the cpacitors will reduce their
> effective capacitance at the supply frequency. Reducing the resonant
> or tuned frequency will reduce the the effective capacitance further.

(under "Detuning Reactors")
> Using detuning reactors results in a lower KVAR, so the
> capacitance will need to be increased for the same level of correction.

I'm not convinced! Please consider the reasoning below. I have used
'w' to represent system frequency in radians per second.

The detuned reactor is connected in series with the
capacitors. The impedance of the reactor is:

Zr = jwL

The impedance of the capacitor is:

Zc = 1/jwC = -j(1/wC)

The total impedance of the reactor and capacitor is:

Zt = jwL - j(1/wC)


Note that the effect of adding the detuned reactor is to
reduce the combined impedance, which in turn results
a higher current. The current still "looks" purely
capacitive, since (1/wC) is larger than wL.

I conclude that:

1. Adding a detuned reactor makes the capacitor bank
draw more capacitive current.

2. A higher capacitive current means a greater supply of
reactive power.

3. The apparent kVAr is higher!

Thanks,
Submonkey

Hello submonkey

Welcome to the forum.

Yes you are correct, the impedance is reduced and therefore the current is increased. This will of course increase the KVAR of the circuit. - I will correct the comment on the web page. Thank you for picking that up.
No one else has seen this!! It is always a concern that no matter how often you read something, you can miss the obvious.

Have a great day,
Best regards,