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

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

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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

  • $\begingroup$ Also when $C=0$, we have $X_c = \infty$, does this mean the circuit is more capacitive when the capacitance is $0$ ? Hmm.. $\endgroup$ – AgentS May 14 '18 at 14:19
  • $\begingroup$ They might be adding capacitors in series. Thats the only possible way to increase capacitive reactance at constant frequency. $\endgroup$ – Mitchell May 14 '18 at 15:07
  • $\begingroup$ Yup, the larger a capacitance, the weaker a capacitor is. Weird but true. It’s defined backwards relative to everything else. $\endgroup$ – knzhou May 14 '18 at 17:46
  • $\begingroup$ Be careful with your definitions - since reactance is the imaginary part of impedance, $Z = R + jX$, capacitive reactance is negative. $\endgroup$ – Alfred Centauri May 14 '18 at 18:43
  • $\begingroup$ Also, note that in this series RLC circuit, the total impedance is either inductive (positive imaginary part), resistive (zero imaginary part), or capacitive (negative imaginary part). $\endgroup$ – Alfred Centauri May 14 '18 at 18:45

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