An AC circuit contains only an inductor. By what percent (%) will its inductive reactance be affected when inductance is halved and frequency is tripled?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 50\% \)
The inductive reactance is \( X_L = \omega L = 2\pi fL \), so it is proportional to the product fL.
\( \displaystyle \frac{X_L'}{X_L} = \frac{f'}{f}\times\frac{L'}{L} = 3\times\frac{1}{2} = 1.5 \)
\( \displaystyle \text{Percentage change} = \frac{X_L'-X_L}{X_L}\times100\% = (1.5-1)\times100\% = 50\% \)
The reactance increases by 50%.