Series RL impedance magnitude: A 1.2 kΩ resistor is in series with a 15 mH inductor across a 10 kHz AC source. What is the total impedance magnitude?
Electronics
RL Circuits
Difficulty: Easy
Choose an option
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A152.6 Ω
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B1,526 Ω
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C1,200 Ω
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D942 Ω
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E1,920 Ω
Answer
Correct Answer: 1,526 Ω
Explanation
Introduction / Context:Here we compute the magnitude of series RL impedance by combining resistance and inductive reactance via the root-sum-square relationship, a routine but essential AC task.
Given Data / Assumptions:
- R = 1.2 kΩ = 1200 Ω.
- L = 15 mH; f = 10 kHz.
- Ideal components, sinusoidal steady state.
Concept / Approach:XL = 2 * pi * f * L. Total impedance magnitude is |Z| = sqrt(R^2 + XL^2).
Step-by-Step Solution:
XL = 2 * pi * 10000 * 0.015 = 2 * pi * 150 ≈ 942.48 Ω.|Z| = sqrt(1200^2 + 942.48^2) ≈ sqrt(1,440,000 + 888,276) ≈ sqrt(2,328,276) ≈ 1,526 Ω.Therefore, the impedance magnitude is about 1.53 kΩ.Verification / Alternative check:Angle check: θ = arctan(XL / R) ≈ arctan(942.5 / 1200) ≈ 38.9°, reasonable for RL with R slightly dominant. This supports the computed magnitude.
Why Other Options Are Wrong:
- 152.6 Ω and 942 Ω: Confuse steps (either divide by 10 or use XL alone).
- 1,200 Ω: Ignores inductive reactance contribution.
- 1,920 Ω: Overestimates by summing instead of using root-sum-square.
Common Pitfalls:
- Adding R and XL directly instead of vectorially.
Final Answer:1,526 Ω