Admittance of a parallel RC network: A 470 Ω resistor is in parallel with a 0.2 µF capacitor across a 2.5 kHz AC source. What is the total admittance Y in rectangular form?

Introduction / Context:
Parallel combinations are most conveniently handled in admittance (siemens). For a resistor, G = 1/R is the conductance, and for a capacitor, susceptance is B_C = ωC (positive imaginary in admittance form). Adding these directly yields the network admittance in rectangular form, which is valuable for current-sum phasor analysis.

Given Data / Assumptions:

• R = 470 Ω → G = 1/R.
• C = 0.2 µF = 2 × 10^-7 F.
• f = 2.5 kHz → ω = 2πf.
• Ideal components; sinusoidal steady state.

Concept / Approach:
Compute G = 1/R and B_C = ωC. The total admittance is Y = G + jB_C. Convert to millisiemens (mS) for neat numeric values that match standard option formats.

Step-by-Step Solution:

Verification / Alternative check:
Impedance magnitude check: |Y| ≈ √(2.12^2 + 3.14^2) mS ≈ 3.79 mS → |Z| ≈ 1/|Y| ≈ 264 Ω. This is between 470 Ω and the low capacitive reactance at 2.5 kHz, matching expectations.

Why Other Options Are Wrong:

• 3.14 mS + j2.12 mS: Swaps real and imaginary parts.
• 212 Ω or 318.3 Ω: These are impedances (Ω), not admittance (S), and do not match the computed Y.

Common Pitfalls:

• Using −j for capacitive susceptance in admittance; note Y_C = +jωC (the negative sign applies to impedance, not admittance).

Final Answer:
2.12 mS + j3.14 mS