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An alternating voltage of $220 \mathrm{~V}, 50 \mathrm{~Hz}$ frequency is applied across a capacitor of capacitance 2 $\mu \mathrm{F}$. The impedence of the circuit is
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Verified Answer
The correct answer is:
$\frac{5000}{\pi}$
Impedence of a capacitor is $\mathrm{X}_{\mathrm{C}}=1 / \omega \mathrm{C}$
$$
\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 2 \times 10^{-6}}=\frac{5000}{\pi}
$$
$$
\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 2 \times 10^{-6}}=\frac{5000}{\pi}
$$
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