In a series LR circuit,power of 400 W is dis | vedclass

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(D) Given: $P = 400 \ W$,$V_{rms} = 250 \ V$,$f = 50 \ Hz$,$\cos \phi = 0.8$.$1$. Calculate impedance $Z$:$P = V_{rms} I_{rms} \cos \phi = \frac{V_{rm

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$400$

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(D) Given: $P = 400 \ W$,$V_{rms} = 250 \ V$,$f = 50 \ Hz$,$\cos \phi = 0.8$.$1$. Calculate impedance $Z$:$P = V_{rms} I_{rms} \cos \phi = \frac{V_{rms}^2}{Z} \cos \phi$$400 = \frac{250^2}{Z} 	imes 0.8$$Z = \frac{62500 	imes 0.8}{400} = \frac{50000}{400} = 125 \ \Omega$.$2$. Calculate resistance $R$ and inductive reactance $X_L$:$R = Z \cos \phi = 125 	imes 0.8 = 100 \ \Omega$.$X_L = Z \sin \phi = 125 	imes \sqrt{1 - 0.8^2} = 125 	imes 0.6 = 75 \ \Omega$.$3$. For unity power factor,the circuit must be in resonance,so $X_C = X_L = 75 \ \Omega$.$X_C = \frac{1}{2 \pi f C} = 75$$C = \frac{1}{2 \pi 	imes 50 	imes 75} = \frac{1}{7500 \pi} \ F = \frac{10^6}{7500 \pi} \ \mu F = \frac{10000}{75 \pi} \ \mu F = \frac{400}{3 \pi} \ \mu F$.Comparing with $C = (\frac{n}{3 \pi}) \ \mu F$,we get $n = 400$.

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