Convert to polar impedance: A 47 Ω resistor in series with an inductive reactance of 120 Ω is connected to an AC source. What is the total impedance in polar form?

Introduction / Context:
Converting a series R and jXL into polar form is a staple AC analysis skill, linking rectangular and polar representations of impedance for phasor calculations.

Given Data / Assumptions:

• R = 47 Ω, XL = 120 Ω (inductive; positive imaginary part).
• Series connection; Z = R + jXL.

Concept / Approach:
Compute magnitude |Z| by root-sum-square and phase angle θ = arctan(XL / R). For an inductor in series with a resistor, the impedance angle is positive (current lags voltage).

Step-by-Step Solution:

Verification / Alternative check:
Rectangular check: 129∠68.6° ≈ 47 + j120 when converted back, confirming consistency.

Why Other Options Are Wrong:

• 47∠68.6° Ω and 120∠68.6° Ω: Use component magnitudes rather than the combined magnitude.
• 129∠31.4° Ω: Complementary angle; wrong quadrant for inductive series.
• 127∠60° Ω: Rounded magnitude/angle do not match computed values.

Common Pitfalls:

• Using θ = arctan(R/XL) instead of XL/R.
• Forgetting the positive angle sign for inductive reactance in series.

Final Answer:
129∠68.6° Ω