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).
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AZ = 150 ∠–90°Ω
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BZ = 150∠90°Ω
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CZ = 150
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DZ = 150∠180°Ω
Answer
Correct Answer: Z = 150 ∠–90°Ω
Explanation
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:
Write impedance: Zc = −j * 150 Ω.Magnitude: |Zc| = 150 Ω.Angle: ∠Zc = −90° (purely capacitive).Polar form: Zc = 150 ∠ −90° Ω.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