Let f = 50, Hz and C = 100, mu F in an AC cir | vedclass

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(A) Given: $f = 50\, Hz$,$C = 100\, \mu F = 10^{-4}\, F$,$I_m = 1.57\, A$.Angular frequency $\omega = 2\pi f = 2 	imes \pi 	imes 50 = 100\pi\, rad/s

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$E = 50\, \sin(100\, \pi t - \pi / 2)$

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(A) Given: $f = 50\, Hz$,$C = 100\, \mu F = 10^{-4}\, F$,$I_m = 1.57\, A$.Angular frequency $\omega = 2\pi f = 2 	imes \pi 	imes 50 = 100\pi\, rad/s$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi 	imes 10^{-4}} = \frac{100}{\pi} \approx \frac{100}{3.14} \approx 31.85\, \Omega$.Peak voltage $E_m = I_m 	imes X_C = 1.57 	imes 31.85 \approx 50\, V$.In a purely capacitive circuit,the voltage lags behind the current by $\pi / 2$ radians. If current $I = I_m \sin(\omega t)$,then voltage $E = E_m \sin(\omega t - \pi / 2)$.Substituting the values: $E = 50\, \sin(100\pi t - \pi / 2)$.

Let $f = 50\, Hz$ and $C = 100\, \mu F$ in an $AC$ circuit containing a capacitor only. If the peak value of the current in the circuit is $1.57\, A$,the expression for the instantaneous voltage across the capacitor will be (assuming current is $I = I_m \sin(\omega t)$):

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