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?

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:

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