Series RLC behavior: In a series R–L–C circuit, the larger reactance (either inductive XL or capacitive XC) determines the sign and overall net reactance Xnet = XL − XC of the circuit. State whether this statement is true or false.

Introduction / Context:
This statement examines how inductive reactance (XL) and capacitive reactance (XC) combine in a series RLC circuit. Understanding which term dominates determines whether the circuit behaves overall as inductive or capacitive, which directly affects current phase, impedance magnitude, and resonance behavior.

Given Data / Assumptions:

• Series connection of R, L, and C.
• Inductive reactance: XL = 2 * pi * f * L.
• Capacitive reactance: XC = 1 / (2 * pi * f * C).
• Net series reactance: Xnet = XL − XC (sign matters).

Concept / Approach:

Because L and C are in series, their reactances algebraically subtract. The magnitude of the larger of XL or XC determines not only the sign of Xnet but also the overall reactive nature of the circuit. If XL > XC, the circuit is net inductive; if XC > XL, it is net capacitive. At resonance, XL = XC and Xnet = 0, leaving only the resistive part R.

Step-by-Step Solution:

Verification / Alternative check:

Plot XL and XC versus frequency. The curves cross at f0 where XL = XC. On either side of f0, the curve with the larger magnitude dictates the sign and magnitude of Xnet, confirming that the “larger reactance” determines the net reactive behavior.

Why Other Options Are Wrong:

• “False” would imply the smaller term or some other operation sets Xnet, which contradicts Xnet = XL − XC in series.

Common Pitfalls:

Confusing series and parallel combination rules; in parallel, susceptances add instead. Also, ignoring sign leads to wrong conclusions about leading/lagging current.

Final Answer:

True