Difficulty: Easy
Correct Answer: 5 A
Explanation:
Introduction / Context:In any single-loop series circuit (R, L, and/or C in series), the same instantaneous current flows through every element because there is only one current path. This question tests the core idea of series topology and how it applies to an inductor’s branch current reading.
Given Data / Assumptions:
Concept / Approach:Kirchhoff’s Current Law applied to a series loop implies a single branch; therefore the current is identical through R, L, and C at every instant. Phasor relationships change how voltage distributes (phase shifts), but not the series current equality. Thus the inductor current must equal the source current.
Step-by-Step Solution:
Recognize a single series path → same current in all components.Given I_source(rms) = 5 A → I_L(rms) = 5 A by series rule.Voltage magnitudes will differ (V_R = IR, V_L = IX_L, V_C = I*X_C), but the current is common.Verification / Alternative check:Measure with an ammeter inserted in series before the inductor and again between L and C; both readings match the source current (within instrument tolerance).
Why Other Options Are Wrong:
“Depends only on L” or “greater than 5 A”: violates series current continuity.“Zero at any frequency”: only true if the loop is open; here the loop is closed.“Cannot be determined”: the series rule determines it directly.Common Pitfalls:Confusing series (same current) with parallel (same voltage). Also, mixing up current magnitude with reactive voltage magnitudes, which can be large at resonance without changing current equality.
Final Answer:5 A
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