Series RLC numeric check — inductor voltage from current and reactance: Assume the same series RLC as in the earlier item with loop current 5 A (rms). If the inductor’s reactance is X_L = 15.56 Ω at the operating frequency, what is the magnitude of the inductor’s RMS voltage V_L?

Introduction / Context:Voltage drops across reactive elements in a series AC circuit follow the same IX rule as resistive drops follow IR, but with a 90-degree phase shift. This question links a previously established series current to the inductor’s voltage magnitude using its reactance.

Given Data / Assumptions:

• Series loop current I = 5 A (rms).
• Inductive reactance X_L = 15.56 Ω at the test frequency.
• Ideal inductor (ignore winding resistance for this calculation).

Concept / Approach:The magnitude of an inductor’s RMS voltage is V_L = I * X_L. Phase considerations matter for vector addition in KVL, but magnitude computation is straightforward with the given current and reactance.

Step-by-Step Solution:

Verification / Alternative check:Phasor diagram: V_L is orthogonal to V_R and opposite in direction to V_C. Its magnitude still equals I * X_L, independent of R and C values.

Why Other Options Are Wrong:

Common Pitfalls:Adding magnitudes instead of phasors when forming the source voltage; here we only compute the single-element magnitude.

Final Answer:77.8 V