Compute the inductive reactance X_L of a 2 H inductor operating at 60 Hz. Provide the nearest whole-ohm value.

Difficulty: Easy

Correct Answer: 754 Ohm

Explanation:


Introduction / Context:
Inductive reactance quantifies how an inductor opposes AC. It increases with both inductance and frequency. Quick calculations of X_L are essential in designing filters, chokes, and setting impedance levels in power and signal circuits.


Given Data / Assumptions:

  • L = 2 H (henries).
  • f = 60 Hz (mains frequency in many regions).
  • Ideal inductor; standard formula for X_L applies.


Concept / Approach:
The standard relationship is X_L = 2 * pi * f * L. Units: X_L in ohms when f is in hertz and L in henries. The result typically rounds to the nearest ohm for component selection and approximate analysis.


Step-by-Step Solution:

1) Write formula: X_L = 2 * pi * f * L.2) Substitute values: X_L = 2 * pi * 60 * 2.3) Compute: 2 * pi ≈ 6.283; 6.283 * 60 * 2 ≈ 753.98 Ω.4) Round to nearest whole ohm: ≈ 754 Ω.


Verification / Alternative check:
Use approximate pi = 3.14: X_L ≈ 2 * 3.14 * 60 * 2 = 753.6 Ω, which rounds to 754 Ω, confirming the calculation.


Why Other Options Are Wrong:

30 Ω / 60 Ω / 120 Ω: Far below the correct computed value at 60 Hz with 2 H.None of the above: Invalid because 754 Ω matches the computed answer.


Common Pitfalls:
Forgetting the factor 2 * pi, mixing up inductive with capacitive reactance, or rounding too early leading to large errors.


Final Answer:
754 Ohm.

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