Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:In alternating-current (AC) circuit analysis, understanding how phase angle varies with frequency is fundamental for filters, sensor interfaces, and power systems. A parallel RC network exhibits a frequency-dependent current lead relative to the applied voltage. This question tests whether you know the direction of that change as frequency increases.
Given Data / Assumptions:
Concept / Approach:The parallel RC is best analyzed using admittance Y. The total admittance is Y = G + jB where G = 1/R and B = ωC. The total current phasor leads the voltage by angle θ such that tan(θ) = B / G = (ωC) / (1/R) = ωCR. Because B increases linearly with frequency, the current’s lead angle increases with frequency.
Step-by-Step Solution:
Write Y = G + jB with G = 1/R and B = ωC.Compute the phase: θ = arctan(B/G) = arctan(ωCR).As frequency f (and ω = 2πf) increases, ωCR increases monotonically.Therefore θ increases (the total current leads the voltage by a larger angle).Verification / Alternative check:At very low frequency (ω → 0), B → 0 and θ → 0°, so the circuit behaves resistively. At very high frequency, B ≫ G and θ approaches 90°, dominated by capacitive current. This confirms that θ increases with frequency, not decreases.
Why Other Options Are Wrong:
Correct: contradicts admittance-phase behavior; θ does not decrease with frequency.Valid only at resonance: RC networks do not resonate without L; no resonance applies.Applies only when R = XC: that is a series condition for a specific phase in series RC; not relevant here.Zero ESR clause: ESR slightly perturbs G, but the trend (θ increasing with f) remains.Common Pitfalls:Confusing series vs. parallel behavior; mixing voltage-lead with current-lead language; assuming capacitor current behavior from series circuits applies identically to parallel circuits.
Final Answer:Incorrect
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