Series R–L–C with phasor current: A 90 Ω resistor, a coil with reactance 30 Ω, and a capacitor with reactance 50 Ω are in series across a 12 V (rms) AC source. What is the current through the resistor?
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A130 mA
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B9 mA
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C90 mA
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D13 mA
Answer
Correct Answer: 130 mA
Explanation
Introduction / Context:In a series R–L–C circuit, the same current flows through all series elements. To find that current, you compute the total impedance magnitude and then apply Ohm’s law using rms values. This is a foundational AC analysis skill.
Given Data / Assumptions:
- R = 90 Ω; X_L = +30 Ω; X_C = 50 Ω.
- Source voltage V_s(rms) = 12 V.
- Series connection; ideal components.
Concept / Approach:The series impedance is Z = R + j(X_L − X_C). The current magnitude is I = V_s / |Z|. Since the circuit is series, this I is also the current through the resistor (and all components).
Step-by-Step Solution:
Net reactance X = 30 − 50 = −20 Ω (capacitive).|Z| = √(R^2 + X^2) = √(90^2 + 20^2) = √(8100 + 400) = √8500 ≈ 92.2 Ω.I = V_s / |Z| = 12 / 92.2 ≈ 0.130 A = 130 mA.Verification / Alternative check:Because |Z| is only slightly larger than R, the current should be a little below 12/90 ≈ 0.133 A. Our computed 0.130 A is consistent.
Why Other Options Are Wrong:
- 90 mA or 13 mA or 9 mA: These either underestimate or grossly underestimate the current given the relatively small reactive unbalance and the 12 V source.
Common Pitfalls:
- Adding reactances arithmetically without considering their signs (capacitive vs inductive).
Final Answer:130 mA