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Q.
Reactance of a capacitor of capacitance $ C\mu F $ for AC frequency $ \frac{400}{\pi } $ 0 Hz is $ 25\,\Omega . $ The value of C is:
Bihar CECEBihar CECE 2001Alternating Current
Solution:
Capacitive reactance of an AC circuit containing only capacitor with an AC source is
$X_{C}=\frac{1}{\omega C}$
or $X_{C}=\frac{1}{2 \pi f C}$
or $C=\frac{1}{2 \pi f X_{C}}$
Here, $f=\frac{400}{\pi} H z, X_{C}=25\, \Omega$
$\therefore C=\frac{1}{2 \pi \times \frac{400}{\pi} \times 25}=\frac{1}{2 \times 400 \times 25}$
$=5 \times 10^{-5} F$
$=50 \,\mu \,F$