Inductors in parallel and series with AC source: L1 = 25 µH in parallel with L2 = 50 µH, then in series with L3 = 20 µH, driven by 60 V rms at 300 kHz. What is the total rms current drawn?

Introduction / Context:
Analyzing combinations of inductors with an AC source requires computing the equivalent inductance and then using inductive reactance to find the current. This problem reinforces parallel–series reduction for inductors and conversion from inductance to reactance at a specified frequency.

Given Data / Assumptions:

• L1 = 25 µH, L2 = 50 µH in parallel.
• L3 = 20 µH in series with the parallel pair.
• Source: 60 V rms, f = 300 kHz (sinusoidal).
• Ideal inductors; no resistance; steady-state AC.

Concept / Approach:

For two inductors in parallel: L_p = (L1 * L2) / (L1 + L2). The series combination is L_total = L_p + L3. Inductive reactance is X_L = 2 * π * f * L_total. The current magnitude is I_rms = V_rms / X_L.

Step-by-Step Solution:

Verification / Alternative check:

Order-of-magnitude check: tens of microhenries at hundreds of kilohertz yields reactance on the order of tens of ohms; 60 V over ~70 Ω gives about 0.85 A, consistent.

Why Other Options Are Wrong:

87 mA and 337 mA underestimate the current (would require much higher reactance). 87 A is unrealistically high for these values.

Common Pitfalls:

Forgetting that inductors in parallel combine like resistors in parallel; mixing microhenry units; using f instead of 2πf for reactance.

Final Answer:

870 mA