Admittance–impedance conversion: If the admittance of a branch is Y = 20 mS (millisiemens), what is the magnitude of the equivalent impedance |Z|?

Introduction / Context:
Admittance (Y) is the reciprocal of impedance (Z). In many AC problems it is convenient to switch between these domains—especially for parallel networks where admittances add directly. This question checks the unit handling and reciprocal operation using millisiemens.

Given Data / Assumptions:

• Admittance Y = 20 mS = 20 * 10^-3 S = 0.02 S.
• We are asked for the magnitude of impedance |Z| = 1 / |Y|.
• Assume purely resistive magnitude for the numeric conversion (reactive angle not specified).

Concept / Approach:
The reciprocal relationship is Z = 1 / Y. Be careful with milli- prefix: 1 mS = 10^-3 S. Thus 20 mS = 0.02 S, and its reciprocal gives an impedance magnitude in ohms. This is a straightforward unit inversion often tripped up by prefix mistakes.

Step-by-Step Solution:

Verification / Alternative check:
Check dimensional consistency: siemens is 1/ohm; the reciprocal returns ohms. 1 / (20 * 10^-3) = 1 / 0.02 = 50 confirms the arithmetic.

Why Other Options Are Wrong:
20 Ω corresponds to 50 mS, not 20 mS. 5 Ω would require 200 mS. 0.02 Ω is the inverse of 50 S, not 20 mS.

Common Pitfalls:
Forgetting to convert milli- to base SI units; mixing ohms and siemens; applying the reciprocal to each component separately in non-parallel cases (reciprocals add only for admittances in parallel).

Final Answer:
50 Ω