If inductive reactance X_L is calculated by X_L = 2 pi f L, what happens to X_L when frequency doubles?

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Multiple Choice

If inductive reactance X_L is calculated by X_L = 2 pi f L, what happens to X_L when frequency doubles?

Explanation:
Inductive reactance scales with frequency: X_L = 2π f L. For a fixed inductor, doubling the frequency makes X_L increase in direct proportion. If f becomes 2f, the new reactance is X_L' = 2π (2f) L = 2 × (2π f L) = 2 X_L. So it doubles. Physically, a higher frequency means current changes more rapidly, and the inductor resists those rapid changes more, increasing its opposition. The other possibilities would require halving the frequency, keeping it the same, or having zero frequency, none of which applies here.

Inductive reactance scales with frequency: X_L = 2π f L. For a fixed inductor, doubling the frequency makes X_L increase in direct proportion. If f becomes 2f, the new reactance is X_L' = 2π (2f) L = 2 × (2π f L) = 2 X_L. So it doubles. Physically, a higher frequency means current changes more rapidly, and the inductor resists those rapid changes more, increasing its opposition. The other possibilities would require halving the frequency, keeping it the same, or having zero frequency, none of which applies here.

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