RL impedance — is the input impedance set by combining resistance R and inductive reactance Xl together?

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
Impedance generalizes resistance to AC by including reactive effects. In RL circuits, both resistance and inductive reactance determine current magnitude and phase angle. The question asks whether both contributions combine to set the impedance seen by the source.


Given Data / Assumptions:

  • Single frequency sinusoidal operation.
  • Ideal inductor of inductance L with reactance Xl = ω * L.
  • Ideal resistor R with no frequency dependence.


Concept / Approach:
For a series RL, total impedance is Z = R + j * Xl, where Xl = ω * L. In parallel RL, total admittance is Y = 1/R + 1/(j * Xl) = 1/R − j/(ω * L). In either case, both R and Xl contribute to the net complex quantity that defines the circuit’s response. Therefore, the impedance or admittance must be computed by combining these parts using complex arithmetic, not simple addition of magnitudes.


Step-by-Step Solution:

Series: Z = R + j * (ω * L).Magnitude: |Z| = sqrt(R^2 + (ω * L)^2).Angle: θ = arctan((ω * L)/R), indicating current lag.Parallel: Y = 1/R − j/(ω * L); then Z = 1/Y, still determined by both R and L.


Verification / Alternative check:
Use an LCR meter to measure impedance at multiple frequencies. As frequency increases, measured |Z| and angle change in a manner consistent with ω * L growth, confirming that both R and Xl define the result.


Why Other Options Are Wrong:

  • “False” would ignore the inductor’s contribution and predict a frequency independent response, which is inaccurate for any RL network that includes a nonzero inductance.


Common Pitfalls:
Adding magnitudes directly rather than vectorially. Always treat the reactive term as imaginary to avoid errors in magnitude and phase calculations.


Final Answer:
True

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