Capacitive Reactance

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A 50 $\mu$F capacitor is connected across an AC source whose angular frequency is 200 rad/s. Estimate the capacitive reactance that limits the alternating current through the capacitor.

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Capacitive reactance represents the opposition a capacitor offers to alternating current and decreases with frequency because faster charging and discharging lets more charge flow per second. It is expressed as $X_C = \dfrac{1}{\omega C}$. Substituting $\omega = 200$ rad/s and $C = 50\times10^{-6}$ F gives $X_C = 1/(200 \times 50\times10^{-6}) = 1/(0.01) = 100\;\Omega$. The option 10 $\Omega$ results from a tenfold arithmetic slip in the product $\omega C$. The option 1000 $\Omega$ overestimates by ignoring the factor of frequency correctly. The option 50 $\Omega$ simply echoes the capacitance number without computation. A capacitor passes high frequencies readily and blocks DC, the opposite trend to an inductor, because at zero frequency no charge oscillation occurs and the reactance becomes infinite. Physically, at higher frequency the plates reverse charge so rapidly that a larger current is sustained for the same voltage amplitude, which is exactly why the opposition falls. This is the NCERT relation that lets capacitors act as frequency-selective elements in filters and coupling networks. A unit check confirms $1/[(rad/s)(F)]$ reduces to ohms, and a hundred ohms is a sensible value at low frequency for a microfarad-scale capacitor; halving the supply frequency would double this reactance, consistent with the inverse dependence on $\omega$.

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