In series RLC circuit, does decreasing the capacitance value really make the circuit more capacitive circuit?

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.

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

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