Capacitive reactance comparison at 60 Hz: given two capacitors, 10 µF and 20 µF, which one has the larger capacitive reactance magnitude Xc in a 60 Hz AC circuit?

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Capacitive reactance determines how strongly a capacitor opposes AC. Understanding how Xc depends on C and frequency helps in filter design and impedance matching. Given Data / Assumptions: Frequency f = 60 Hz. Capacitances: 10 µF and 20 µF. Ideal components; ESR and ESL neglected. Concept / Approach:Capacitive reactance is Xc = 1 / (2 * pi * f * C). For fixed f, Xc is inversely proportional to C. Therefore, the smaller capacitor (10 µF) exhibits the larger reactance magnitude; the larger capacitor (20 µF) has half that reactance. Step-by-Step Solution: 1) Xc(10 µF) = 1 / (2 * pi * 60 * 10e-6) ≈ 265.3 Ω. 2) Xc(20 µF) = 1 / (2 * pi * 60 * 20e-6) ≈ 132.7 Ω. 3) Compare: 265.3 Ω > 132.7 Ω, so 10 µF has larger reactance. Verification / Alternative check:Doubling C halves Xc at fixed frequency, consistent with the formula. Why Other Options Are Wrong:“20 µF larger / equal / independent of C / infinite at 60 Hz” each contradicts Xc = 1/(2*pi*f*C). Common Pitfalls:Mixing resistance with reactance; forgetting to convert microfarads to farads. Final Answer:10 µF has the larger reactance

Capacitive reactance comparison at 60 Hz: given two capacitors, 10 µF and 20 µF, which one has the larger capacitive reactance magnitude Xc in a 60 Hz AC circuit?

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