A 0.4 µF capacitor is in series with a resistor and driven by a 4 kHz, 40 V_ac (RMS) source. If the series current is 40 mA (RMS), what is the total series impedance magnitude of the circuit?

Introduction:
Finding series impedance from known RMS voltage and current is a direct application of Ohm law in phasor form. This problem reinforces the relationship Z = V / I for AC magnitudes in a series RC network.

Given Data / Assumptions:

• Capacitance C = 0.4 µF.
• Frequency f = 4 kHz.
• Source voltage V = 40 V RMS.
• Series current I = 40 mA RMS.

Concept / Approach:
For a series circuit, the same current flows through all elements. The magnitude of total impedance is Z = V / I irrespective of the resistor-capacitor split. Phase affects power but not this magnitude calculation.

Step-by-Step Solution:
1) Convert current: I = 40 mA = 0.04 A.2) Use Z = V / I.3) Compute Z = 40 / 0.04 = 1000 Ω.4) Express as 1 kΩ.

Verification / Alternative check:
The capacitive reactance can be estimated: Xc = 1 / (2 * pi * f * C) ≈ 1 / (2 * 3.1416 * 4000 * 0.4e-6) ≈ 99.5 Ω. A reasonable resistor magnitude would then bring the total to roughly 1000 Ω when combined vectorially, consistent with the calculation from V/I.

Why Other Options Are Wrong:
10 Ω and 100 Ω: too small relative to 40 V at 40 mA.
1 MΩ: would limit current to microamp range at 40 V, not 40 mA.
2.2 kΩ: more than the computed value and inconsistent with V/I.

Common Pitfalls:
Mixing peak and RMS values, or trying to sum R and Xc arithmetically. For magnitude from measured V and I, simply use Z = V / I.

Final Answer:
1 kΩ