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?
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A77.8 V
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B31.1 V
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C3.11 V
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D15.56 V
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ECannot be determined without R and C
Answer
Correct Answer: 77.8 V
Explanation
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:
Use V_L = I * X_L.Substitute I = 5 A and X_L = 15.56 Ω → V_L = 5 * 15.56 = 77.8 V.Therefore the correct magnitude is 77.8 V (rms).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:
31.1 V and 15.56 V: correspond to using a wrong current or confusing R with X_L.3.11 V: off by an order of magnitude.“Cannot be determined”: incorrect; I and X_L suffice for |V_L|.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