For a parallel RC network fed by a 5 Vac (RMS) source at 500 kHz, with R = 4.7 kΩ and C = 47 pF connected in parallel, what is the overall phase angle of the total current relative to the applied voltage (use the passive sign convention)?

Introduction / Context:
In a parallel RC circuit, the total current equals the vector (phasor) sum of the resistive branch current (in phase with voltage) and the capacitive branch current (leading the voltage by 90 degrees). The phase angle of the total current indicates how “capacitive” the overall admittance is.

Given Data / Assumptions:

• Frequency f = 500 kHz.
• Source voltage Vrms = 5 V (magnitude not needed for angle).
• R = 4.7 kΩ, C = 47 pF in parallel.
• Ideal components; small-signal, linear operation.

Concept / Approach:
Work with admittance: Y = G + jB. For a resistor, G = 1/R. For a capacitor, susceptance B_C = 2 * π * f * C (positive for capacitive). The phase angle of total current relative to voltage is θ = arctan(B / G). Because capacitive current leads voltage, θ is positive (leading) when B > 0.

Step-by-Step Solution:
Compute G: G = 1 / 4700 ≈ 0.0002128 S.Compute B: B = 2 * π * 500000 * 47e-12 ≈ 0.0001477 S.Find angle: θ = arctan(B / G) ≈ arctan(0.0001477 / 0.0002128) ≈ arctan(0.6939) ≈ 34.8°.Interpretation: total current leads voltage by about 34.8°, indicating a moderately capacitive net admittance.

Verification / Alternative check:
Power factor pf = cos(θ) ≈ cos(34.8°) ≈ 0.82 (leading). Simulation or an LCR meter in parallel mode would show similar angle at this frequency.

Why Other Options Are Wrong:
±55.3° imply a much stronger susceptance relative to conductance, not supported by B/G ≈ 0.694.

Negative angles denote lagging (inductive) behavior, which is not present here.

0° would mean purely resistive with B = 0, not the case with a capacitor in parallel.

Common Pitfalls:
Mixing series and parallel formulas; for parallel circuits, always compute with admittance (G and B). Also note sign convention: capacitive susceptance is positive, giving a leading current.

Final Answer:
34.8°