Identify the graph which correctly represents the variation of capaci…

The correct answer is Option (3)

The capacitive reactance is inversely proportional to the frequency:

    \( \displaystyle X_C = \frac{1}{2\pi f C} \;\Rightarrow\; X_C \propto \frac{1}{f} \)

So the graph must be a rectangular hyperbola: \( X_C \) is very large at low frequency, falls steadily as \( f \) increases, approaches zero for very large \( f \) without ever touching the axes, and is a smooth curve with no oscillation.

Why the other options are incorrect

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  Option (1) — Incorrect. \( X_C \) would rise with \( f \) and be zero at \( f = 0 \); this is the graph of the inductive reactance \( X_L = 2\pi f L \).

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  Option (2) — Incorrect. A straight line with a negative slope would mean \( X_C \) decreases uniformly and reaches zero at a finite frequency; but \( X_C \propto 1/f \) is not a straight line.

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  Option (4) — Incorrect. \( X_C \) does not oscillate with frequency; a wave-like curve that takes negative values has no relation to \( 1/(2\pi f C) \).

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Comparing it to i_m = v_m/R for a purely resistive circuit, we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and is denoted by X_c,

    \( \displaystyle X_c = 1/\omega C \qquad (7.17) \)

… But it is inversely proportional to the frequency and the capacitance.”