An ac source of fixed rms voltage is connected across a pure inductor. If the frequency of the source is doubled, the current becomes
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → half
The inductive reactance \( X_L=\omega L \) limits the current in a pure inductor. It is directly proportional to the frequency. For a fixed rms voltage \( V \), the current is \( I=V/X_L \), so it is inversely proportional to the frequency.
When the frequency is doubled, the reactance becomes:
\( \begin{aligned} X_L' &= (2\omega)\,L \\ &= 2X_L \end{aligned} \)
The new current is then:
\( \begin{aligned} I' &= \frac{V}{X_L'} \\ &= \frac{V}{2X_L} \\ &= \frac{I}{2} \end{aligned} \)
Therefore, the current becomes half.
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 amplitude of the current is, then
\( \displaystyle i_m=\frac{v_m}{X_L} \qquad (7.14) \)
… The inductive reactance is directly proportional to the inductance and to the frequency of the current.”