In AC circuit analysis, a pure capacitor with capacitive reactance Xc = 150 Ω is connected across a sinusoidal source. Express the capacitor’s impedance Z in polar form (magnitude ∠ angle, in ohms).

Introduction / Context:
A capacitor in steady-state AC is represented by a purely imaginary impedance that is negative on the imaginary axis. Converting this to polar form reinforces phasor skills used in RC/RLC analysis and filter design.

Given Data / Assumptions:

• Pure capacitor with capacitive reactance Xc = 150 Ω.
• No series resistance or parasitics; ideal sinusoidal steady state.
• Polar form is written as |Z| ∠ angle, in ohms.

Concept / Approach:

The impedance of a capacitor is Zc = −jXc. On the complex plane, −j corresponds to an angle of −90° (pointing straight down the negative imaginary axis). The magnitude is |Z| = Xc for a pure capacitor.

Step-by-Step Solution:

Verification / Alternative check:

Rectangular-to-polar check: a point at (0, −150) has magnitude √(0^2 + 150^2) = 150 and angle arctan(−∞) = −90°, confirming the polar form.

Why Other Options Are Wrong:

150∠90° Ω describes +j150 Ω (inductor). 'Z = 150' omits the essential angle and would imply a resistor. 150∠180° Ω would be a negative real resistor, which is not a capacitor.

Common Pitfalls:

Confusing sign of the imaginary part; swapping inductor and capacitor angles; leaving off the angle in polar form.

Final Answer:

Z = 150 ∠–90°Ω