Difficulty: Easy
Correct Answer: True
Explanation:
Introduction / Context:
Capacitive reactance describes how a capacitor opposes alternating current. Knowing how XC varies with frequency f and capacitance C is fundamental for filter design, impedance matching, and time-constant calculations in electronics and electrical engineering.
Given Data / Assumptions:
Concept / Approach:
The well-known formula for capacitive reactance is XC = 1 / (2 * pi * f * C). This makes XC inversely proportional to both f and C. As frequency increases or as capacitance increases, XC decreases, allowing more AC current to flow for a given voltage amplitude.
Step-by-Step Solution:
Verification / Alternative check:
Double either f or C and observe that XC halves. Halve either f or C and XC doubles. This proportional behavior matches simulation and bench measurements with LCR meters across frequency sweeps.
Why Other Options Are Wrong:
“False” contradicts the formula. “True only for ideal capacitors” is overly restrictive; parasitics modify results slightly but the inverse proportionality remains the governing first-order law. “True only at DC” is wrong since at f = 0, XC tends to infinity, not a finite inverse. “Indeterminate without power factor” is irrelevant; XC is defined without needing power factor.
Common Pitfalls:
Confusing XC with capacitive reactance sign (−j) or with impedance magnitude in RC networks where resistance contributes additional effects. Also, mixing up units: f in Hz, C in farads, XC in ohms.
Final Answer:
True
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