Combining R and Xc — in a series RC with R = 100 Ω and capacitive reactance Xc = 100 Ω, what is the magnitude of the total impedance Z?

Introduction / Context:
Calculating impedance magnitude in series RC and RL circuits requires vector addition, not simple arithmetic. This question verifies that you can combine resistance and reactance correctly using the Pythagorean relationship that arises from orthogonal phasor components.

Given Data / Assumptions:

• Series RC circuit, sinusoidal steady state.
• R = 100 Ω, Xc = 100 Ω.
• Ideal components and no additional series elements.

Concept / Approach:
Impedance is complex: Z = R - jXc for series RC. The real part is R and the imaginary part is -Xc. The magnitude of a complex number a + jb is sqrt(a^2 + b^2). Therefore, the impedance magnitude is |Z| = sqrt(R^2 + Xc^2). This is analogous to the right triangle formed by resistive and reactive voltage drops or currents in phasor diagrams.

Step-by-Step Solution:

Verification / Alternative check:
If R equals Xc in magnitude, the impedance magnitude is R * sqrt(2). Here, 100 * 1.414 ≈ 141.4 Ω, matching the calculation. The phase angle is phi = arctan(Xc / R) = arctan(1) = 45 degrees (current leads), which is consistent but not required to get |Z|.

Why Other Options Are Wrong:

• Adding magnitudes (200 Ω) ignores orthogonality; reactance is imaginary and must be combined vectorially.
• “100 Ω because reactance cancels” is false; cancellation does not occur between real and imaginary parts.
• “Undetermined” is incorrect; |Z| is computable without knowing the angle.

Common Pitfalls:
Adding R and Xc directly or trying to subtract them numerically, both of which disregard the complex nature of impedance.

Final Answer:
Correct — the impedance magnitude is approximately 141.4 Ω.