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The IMPEDANCE of a series RL circuit at angular frequency $\omega$ is
A$R + L$
B$R - j\omega L$
C$R + j\omega L$
D$R + 1/(j\omega L)$
Answer & Solution
Correct answer: C. $R + j\omega L$
1. Series impedances ADD: $Z_{RL} = Z_R + Z_L = R + j\omega L$.
2. Magnitude: $|Z| = \sqrt{R^2 + (\omega L)^2}$.
3. Phase: $\arctan(\omega L / R)$ — between 0° (pure R) and 90° (pure L) as $\omega L$ increases.
4. Option A is dimensionally wrong. Option B has wrong sign on $j\omega L$ (would be capacitive!). Option D has the capacitor formula.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — AC", Vol II, Ch 5 (Series RLC impedance)._
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