RC phase angle condition: In a series circuit containing a resistor and a capacitor, which relationship between R and XC produces a phase angle magnitude greater than 45° (current leading the voltage)?
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AR < XC
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BR = XC
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CR > XC
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DR = 5XC
Answer
Correct Answer: R < XC
Explanation
Introduction / Context:For a series RC circuit, current leads voltage by an angle φ where tan|φ| = XC / R. The phase angle quantifies how reactive the circuit is and informs filter design, phase shifters, and impedance matching. Determining when |φ| exceeds 45° is a common design check.
Given Data / Assumptions:
- Series RC only (no inductance).
- XC = 1 / (2π f C) is the capacitive reactance.
- Current leads the source voltage (capacitive behavior).
Concept / Approach:For series RC, tan|φ| = XC / R. The condition |φ| > 45° occurs when tan|φ| > 1, i.e., XC / R > 1 → XC > R. Thus the resistor must be smaller than the capacitive reactance to achieve a phase shift greater than 45° in magnitude.
Step-by-Step Solution:
Start from tan|φ| = XC / R.Require |φ| > 45° → tan|φ| > 1 → XC / R > 1.Hence XC > R → equivalently R < XC.Verification / Alternative check:Example: Let R = 1 kΩ and XC = 2 kΩ. Then tan|φ| = 2 → |φ| ≈ 63.4°, which is greater than 45°, satisfying the condition.
Why Other Options Are Wrong:
- R = XC: Gives |φ| = 45° exactly.
- R > XC: Results in |φ| < 45° because the tangent ratio is less than 1.
- R = 5XC: Makes |φ| very small (highly resistive), not larger than 45°.
Common Pitfalls:
- Confusing sign and magnitude of phase; RC has current leading but we compare |φ|.
Final Answer:R < XC