Accessible AC (#3)

Electricity and Magnetism Level pending

Let X L X_L be a positive real number representing the inductive reactance (in ohms) associated with an inductor at an angular frequency ω \omega .

For a 2.5 H 2.5 H inductor, calculate the following quantity:

d X L d ω = ? \large{\frac{d X_L}{d \omega} = ?}


The answer is 2.5.

Steven Chase by Steven Chase , 37, USA

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1 solution

Arjen Vreugdenhil
Dec 12, 2017

The inductive reactance of an inductor is X L = L ω X_L = L\omega .

Thus we must calculate d d ω L ω = L . \frac{d}{d\omega} L\omega = L. Substitute L = 2.5 H L = \SI{2.5}{H} to obtain the answer: 2.5 H = 2.5 Ω / Hz \SI{\boxed{2.5}}{H} = \SI{2.5}{\Omega/Hz} .

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