If the frequency of the signal is doubled, how will the reactance of a circuit containing only an inductor be affected?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → doubled
A pure inductor limits the current through its inductive reactance \( X_L=\omega L \), where \( \omega=2\pi\nu \). For a given inductor, the reactance is directly proportional to the frequency. So doubling the frequency doubles \( X_L \).
\( \begin{aligned} \frac{X_L'}{X_L} &= \frac{2\pi\,(2\nu)\,L}{2\pi\nu L} \\ &= 2 \end{aligned} \)
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… The quantity ωL is analogous to the resistance and is called inductive reactance, denoted by XL:
\( X_L=\omega L \qquad (7.13) \)
… The inductive reactance is directly proportional to the inductance and to the frequency of the current.”