Find the source phasor: A 12 kΩ resistor in series with a 90 mH inductor is connected to an 8 kHz AC source. If the circuit current is I = 0.3∠0° mA (reference), what is the source voltage in polar form?
Electronics
RL Circuits
Difficulty: Medium
Choose an option
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A3.84∠20.6° V
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B12.8∠20.6° V
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C0.3∠20.6° V
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D13.84∠69.4° V
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E3.84∠–20.6° V
Answer
Correct Answer: 3.84∠20.6° V
Explanation
Introduction / Context:This problem reinforces core AC circuit analysis: compute series RL impedance, then use Ohm’s law in phasor form to find the source voltage given a reference current phasor.
Given Data / Assumptions:
- R = 12 kΩ, L = 90 mH.
- Frequency f = 8 kHz ⇒ ω = 2 * pi * 8000 rad/s.
- I = 0.3∠0° mA (phasor reference).
- Ideal components, steady state sinusoidal.
Concept / Approach:Compute XL, then the series impedance Z = R + jXL. The source voltage phasor is V = I * Z. The impedance angle is θ = arctan(XL / R).
Step-by-Step Solution:
ω = 2 * pi * 8000 ≈ 50265 rad/s.XL = ω * L = 50265 * 0.09 ≈ 4524 Ω.Z = 12000 + j4524 Ω ⇒ |Z| = sqrt(12000^2 + 4524^2) ≈ 12829 Ω, θ ≈ arctan(4524/12000) ≈ 20.6°.|V| = |I| * |Z| = 0.0003 * 12829 ≈ 3.84 V, ∠V = 20.6°.Verification / Alternative check:Component drops: VR = I * R ≈ 0.0003 * 12000 = 3.6 V; VL = I * XL ≈ 0.0003 * 4524 ≈ 1.36 V. Vector sum of 3.6 V (real) and 1.36 V (positive j) gives ≈ 3.84∠20.6°, consistent.
Why Other Options Are Wrong:
- 12.8∠20.6° V and 13.84∠69.4° V: Magnitude errors; do not match I * Z.
- 0.3∠20.6° V: Units confusion (current magnitude).
- 3.84∠–20.6° V: Incorrect sign for inductive lead angle.
Common Pitfalls:
- Forgetting to convert milliamps to amps when multiplying by ohms.
- Mixing rectangular and polar without accounting for the phase angle.
Final Answer:3.84∠20.6° V