A frac{2.5}{pi} mu F capacitor and 3000 , Ome | vedclass

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(C) Given: $C = \frac{2.5}{\pi} 	imes 10^{-6} \, F$,$R = 3000 \, \Omega$,$V_{rms} = 200 \, V$,$
u = 50 \, Hz$.First,calculate the capacitive reactan

Vedclass · Accepted Answer

$0.6, 4.8 \, W$

Vedclass · Answer

(C) Given: $C = \frac{2.5}{\pi} 	imes 10^{-6} \, F$,$R = 3000 \, \Omega$,$V_{rms} = 200 \, V$,$
u = 50 \, Hz$.First,calculate the capacitive reactance $X_C = \frac{1}{2\pi 
u C} = \frac{1}{2\pi 	imes 50 	imes (\frac{2.5}{\pi} 	imes 10^{-6})} = \frac{1}{250 	imes 10^{-6}} = 4000 \, \Omega$.The impedance of the circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{3000^2 + 4000^2} = 5000 \, \Omega$.The power factor is $\cos \phi = \frac{R}{Z} = \frac{3000}{5000} = 0.6$.The power dissipated is $P = \frac{V_{rms}^2 \cos \phi}{Z} = \frac{200^2 	imes 0.6}{5000} = \frac{40000 	imes 0.6}{5000} = 8 	imes 0.6 = 4.8 \, W$.

$A$ $\frac{2.5}{\pi} \mu F$ capacitor and $3000 \, \Omega$ resistor are joined in series to an $AC$ source of $200 \, V$ and $50 \, Hz$ frequency. The power factor of the circuit and the power dissipated in it will respectively be:

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