In steady-state DC conditions, what is the capacitive reactance value (X_C) of a capacitor connected in a direct-current circuit?

Introduction:This question examines how a capacitor behaves under steady DC conditions. Understanding the limit of capacitive reactance as frequency approaches zero is fundamental in circuit analysis.

Given Data / Assumptions:

• Direct-current (DC) steady state, i.e., frequency f = 0 Hz.
• Ideal capacitor with no leakage.
• No switching transients considered (t → ∞).

Concept / Approach:The capacitive reactance is X_C = 1 / (2 * π * f * C). In steady-state DC, f = 0, so X_C → ∞. Physically, a fully charged ideal capacitor blocks DC current, acting like an open circuit after transients decay.

Step-by-Step Solution:Start from X_C = 1 / (2 * π * f * C)As f → 0, denominator → 0Therefore X_C → ∞Hence, no continuous DC current flows through an ideal capacitor at steady state.

Verification / Alternative check:Current through a capacitor is i = C * dv/dt. For steady DC, dv/dt = 0, thus i = 0 A, which is consistent with X_C = ∞ (open circuit behavior).

Why Other Options Are Wrong:

• 0 Ω: That would short DC, which contradicts capacitor behavior.
• A moderate finite value: Only at nonzero frequency.
• Cannot be determined without frequency: At DC, f is 0 by definition.
• Equal to series resistance: Reactance is not determined by resistance.

Common Pitfalls:

• Confusing initial transient behavior with steady-state conditions.
• Forgetting that reactance depends on frequency.

Final Answer:Infinite (open circuit)