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Detuning Reactors


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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!




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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,

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