Physics · Alternating Current · NEET
It increases. XL = 2πfL, so XL is directly proportional to frequency f. If you double the frequency, XL doubles. A coil opposes any change in current, and fast AC changes current more often, so the coil fights harder at high frequency. This is why an inductor is called a 'high-frequency blocker'.
XC = 1/(2πfC), so XC is inversely proportional to frequency. Higher f means smaller XC. A capacitor charges and discharges each half cycle; at high frequency it charges and discharges faster, so more charge flows per second, which means more current and less opposition. That is why XC falls as f rises.
At very high frequency, XL = 2πfL tends to infinity, so the inductor acts like an open circuit (blocks current, like a break in the wire). XC = 1/(2πfC) tends to zero, so the capacitor acts like a short circuit (a plain wire, no opposition). NEET loves this limit case - it lets you simplify a messy circuit into just resistors.
It depends on the element. Inductive reactance is DIRECTLY proportional (XL rises with f). Capacitive reactance is INVERSELY proportional (XC falls with f). They move in opposite directions, which is exactly why they cancel each other at one special frequency called resonance, where XL = XC.
They become equal at the resonant frequency. Setting XL = XC gives 2πfL = 1/(2πfC), which solves to f = 1/(2π√(LC)). Below this frequency XC is larger (circuit is capacitive); above it XL is larger (circuit is inductive). At exactly this frequency the reactances cancel and impedance is minimum.
An ac source is connected to a capacitor C. Due to decrease in its operating frequency:
For very high frequencies, the effective impedance of the circuit (shown in the figure) will be (network reduces to 3 Ω resistive value at the high-frequency limit):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Inductive reactance: XL = ωL = 2πfL. Capacitive reactance: XC = 1/(ωC) = 1/(2πfC). Here f is the frequency of the AC source, L is inductance in henry, and C is capacitance in farad. Both have the unit ohm (Ω).
If frequency is doubled, XL doubles (since XL is proportional to f) and XC becomes half (since XC is inversely proportional to f). NCERT Example 7.4 shows this directly: doubling the frequency halves XC and doubles the current in a capacitor.
For DC, frequency f = 0, so XC = 1/(2πfC) = infinity - the capacitor blocks it fully. For AC, f is not zero, so XC is finite and current flows. The higher the AC frequency, the smaller XC becomes and the easier current passes.
An inductor blocks high-frequency AC more, because XL = 2πfL rises with frequency. For DC (f = 0) the inductor's reactance is zero, so DC passes freely through an ideal inductor once steady. So an inductor passes DC but opposes high-frequency AC - the opposite of a capacitor.
Because they oppose each other, they cancel at one frequency where XL = XC, giving resonance (f = 1/(2π√(LC))). This is the basis of tuning circuits, minimum impedance, and maximum current questions. Knowing the direction each reactance moves lets you decide whether a circuit is inductive or capacitive at a given frequency.
