AC series RC phase-angle calculation for a 5.6 µF capacitor in series with a 50 Ω resistor when driven by a 1.1 kHz, 5 V (rms) sinusoidal source. Find the circuit’s phase angle θ (voltage across series RC lags/leads current?) and report its value in degrees with sign convention (capacitive → negative).

Difficulty: Medium

Correct Answer: -27.3 degrees

Explanation:


Introduction:
In a series RC circuit supplied by a sinusoidal source, the impedance has a resistive part R and a capacitive reactance Xc. The phase angle θ between the source voltage and the current indicates whether the circuit is more resistive (angle near 0) or more capacitive (negative angle). For capacitors, current leads voltage, so the net circuit angle is negative.


Given Data / Assumptions:

  • Capacitance C = 5.6 µF
  • Resistance R = 50 Ω
  • Frequency f = 1.1 kHz
  • Source amplitude 5 V (rms) — magnitude is not needed for the phase angle
  • Series RC, ideal components


Concept / Approach:

The impedance angle for a series RC is θ = -arctan(Xc / R). Capacitive reactance is Xc = 1 / (2 * π * f * C). Negative sign indicates a capacitive circuit (current leads).


Step-by-Step Solution:

Compute Xc: Xc = 1 / (2 * π * 1100 * 5.6e-6) ≈ 25.84 ΩCompute ratio: Xc / R ≈ 25.84 / 50 = 0.5168Compute angle magnitude: arctan(0.5168) ≈ 27.3 degreesApply sign for capacitive series RC: θ = -27.3 degrees


Verification / Alternative check:

If R were equal to Xc, θ would be -45 degrees. Here Xc < R, so |θ| must be smaller than 45 degrees, consistent with -27.3 degrees.


Why Other Options Are Wrong:

  • 27.3 degrees: sign is wrong; capacitive series RC must be negative.
  • -62.7 degrees / 62.7 degrees: magnitude too large for Xc < R.
  • -18.0 degrees: underestimates the actual arctan ratio.


Common Pitfalls:

  • Using θ = -arctan(R / Xc) instead of θ = -arctan(Xc / R).
  • Dropping the negative sign for capacitive circuits.


Final Answer:

-27.3 degrees

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