This seems very counter intuitive to me. How can increasing capacitance value make the circuit more inductive ? The math tells me so :

If \$C\$ increases then \$X_c = \dfrac{1}{\omega C}\$ decreases, so the circuit becomes more inductive.

enter image description here

But I've heard that factories that use heavy motors have a large \$X_L\$. For this reason they need to increase \$X_C\$ so that \$X_L = X_C\$. To increase \$X_C\$ they seem to "add" capacitors. But adding capacitors actually decreases \$X_C\$ right ? \$X_C \propto \dfrac{1}{C}\$. How does this work ?

Definitions :
\$X_C\gt X_L\$ : more capacitive circuit

\$X_L\gt X_C\$ : more inductive circuit

  • 1
    \$\begingroup\$ Depends how you add them! Think about how capacitance gets bigger. \$\endgroup\$ May 14 '18 at 14:30
  • 1
    \$\begingroup\$ Then your circuit is inductive. Power is going back and forth between source and inductor. Add any capacitor and now some of that power goes between capacitor and inductor. \$\endgroup\$ May 14 '18 at 14:34
  • 1
    \$\begingroup\$ With heavy motors, you aren't increasing series capacitance. \$\endgroup\$ May 14 '18 at 14:35
  • 3
    \$\begingroup\$ Always in parallel for power factor correction \$\endgroup\$ May 14 '18 at 14:38
  • 2
    \$\begingroup\$ For pawer factor correction? No they don't, as I said - the capacitors are connected in parallel. \$\endgroup\$ May 14 '18 at 14:38

With series components the component that presents the highest impedance IS the dominant factor that determines overall impedance. For instance 1 ohm in series with 1000 ohms is still largely 1000 ohms.

A large value capacitor has a small value impedance hence, in a series circuit, as the capacitor value rises, it's dominance becomes less and less.

Opposite for a parallel combinations of components; lowest dominates.

  • \$\begingroup\$ Ohk.. As the capacitance decreases, the capacitor acts more closely to an "open" circuit, so more voltage drops across it. So far I've been wrongly replacing the capacitor with a "short", so the confusion. Your answer is perfect. Thank you :) \$\endgroup\$
    – AgentS
    May 14 '18 at 15:17

Try adding your capacitor here, instead. This is how facilities with large inductive loads implement power factor correction. (at least, passive PFC. Active PFC is another question entirely.)


simulate this circuit – Schematic created using CircuitLab


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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