(A) $1.891 \mu F$ (B) $2.649 \mu F$

(C) $3.296 \mu F$

(D) $5.198 \mu F$

$\text{Power} \;P = V \times I \rightarrow I = \large\frac{P}{V} $$ = \large\frac{20W}{50V} $$ = 0.4A$

$\Rightarrow \text{ Resistance}\; R = \large\frac{V}{I} $$ = \large\frac{50V}{0.4A} $$ = 125 \Omega$

$\Rightarrow \text{Impedence} Z = \large\frac{V_{\text{power supply}}}{I}$$ = \large\frac{250V}{0.4A} $$ = 625 \Omega$

$\text{Impedence} \; Z = \large\sqrt ( R^2 + ( \large\frac{1}{\omega C})^2)$

$\text{Impedence} \; Z = \large\sqrt ( R^2 + ( \large\frac{1}{2 \pi V C})^2)$

$\Rightarrow Z^2 = R^2 + ( \large\frac{1}{2\pi V C})^2$

Re-arranging and solving for C, we get: $C = \large\frac{1}{2\pi\sqrt (Z^2 - R^2)}$$ = \large\frac{1}{2\pi \times 50 \sqrt(625^2 - 125^2)}$

$\Rightarrow C = \large\frac{1}{2\pi \times 50 \times 612.87}$$ \approx 5.198 \mu F$

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