FCC Exam Question: 3-16B6

In rectangular coordinates, what is the impedance of a network composed of a 0.001-microfarad capacitor in series with a 400-ohm resistor, at 500 kHz?

A. 318 - j400
B. 400 + j318
C. 318 + j400
D. 400 - j318
Correct Answer: D

Explanation: The impedance (Z) of a series circuit in rectangular coordinates is expressed as Z = R ± jX, where R is the resistance and X is the reactance. For a series RC circuit, the formula is Z = R - jXc, with the negative sign indicating capacitive reactance. 1. **Calculate the capacitive reactance (Xc):** The formula for capacitive reactance is Xc = 1 / (2πfC). Given: * f = 500 kHz = 500,000 Hz * C = 0.001 microfarads = 0.001 × 10⁻⁶ F = 1 × 10⁻⁹ F Xc = 1 / (2 * π * 500,000 Hz * 1 × 10⁻⁹ F) Xc ≈ 1 / (3.14159) Xc ≈ 318.3 ohms 2. **Form the impedance:** The resistor (R) is 400 ohms. Substituting R and Xc into the impedance formula for an RC circuit: Z = R - jXc Z = 400 - j318.3 ohms Rounding Xc to 318, the impedance is 400 - j318 ohms. Options B and C are incorrect because the positive 'j' would represent inductive reactance. Option A incorrectly swaps the resistance and reactance values.

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Includes Elements 1, 3, 6, 7R, 8, and 9.