If frequency increase, IL will

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Decrease

Explanation: In an AC circuit containing an inductor, the current $I_L$ is inversely proportional to the frequency $f$ of the source. As the frequency increases, the inductive reactance $X_L$ increases, which results in a decrease in the current flowing through the inductor. Working Principle: The inductive reactance is defined as $X_L = 2\pi f L$. According to Ohm's Law for AC circuits, the current is $I_L = \frac{V}{X_L} = \frac{V}{2\pi f L}$. Since the applied voltage $V$ and the inductance $L$ remain constant, an increase in frequency $f$ leads to a higher reactance, thereby reducing the current $I_L$. Key Formulas: $X_L = 2\pi f L$ — Inductive Reactance formula; $I_L = \frac{V}{2\pi f L}$ — Current in an inductive circuit Key Points: Inductive reactance is directly proportional to frequency ($X_L \propto f$).; Current in a purely inductive circuit is inversely proportional to frequency ($I_L \propto \frac{1}{f}$).; At DC conditions ($f=0$), an ideal inductor acts as a short circuit. Advantages: Used in frequency-selective filtering (Low-pass filters).; Essential for current limiting in ballasts and reactors. Disadvantages: Increases reactive power demand as frequency fluctuates.; Potential for resonance issues in power systems. Applications: AC motor starting circuits.; Radio frequency tuning circuits.; Power electronic filtering. Additional Info: The inverse relationship $I_L \propto 1/f$ only holds for an ideal inductor where $R=0$.. Option A is incorrect because current increases when frequency decreases, not increases. Answer: B is correct — The current $I_L$ decreases because the inductive reactance $X_L$ increases linearly with the frequency. Exam Tip: Remember that for capacitors, the relationship is reversed ($I_C \propto f$) because capacitive reactance $X_C = \frac{1}{2\pi f C}$ decreases as frequency increases.

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