The average power consumed by the given circu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) From the given circuit,the voltage is $V = 200\sqrt{2} \sin(300t)$,so the $RMS$ voltage is $V_{rms} = 200 \, V$ and angular frequency $\omega = 30

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(B) From the given circuit,the voltage is $V = 200\sqrt{2} \sin(300t)$,so the $RMS$ voltage is $V_{rms} = 200 \, V$ and angular frequency $\omega = 300 \, rad/s$.Given $R = 3000 \, \Omega$,$L = \frac{50}{3} \, H$,and $C = \frac{10}{3} \, \mu F = \frac{10}{3} 	imes 10^{-6} \, F$.Inductive reactance $X_{L} = \omega L = 300 	imes \frac{50}{3} = 5000 \, \Omega$.Capacitive reactance $X_{C} = \frac{1}{\omega C} = \frac{1}{300 	imes \frac{10}{3} 	imes 10^{-6}} = \frac{1}{10^{-3}} = 1000 \, \Omega$.Impedance $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{3000^{2} + (5000 - 1000)^{2}} = \sqrt{3000^{2} + 4000^{2}} = 5000 \, \Omega$.The average power consumed is $P = I_{rms}^{2} R = \left(\frac{V_{rms}}{Z}ight)^{2} R$.$P = \left(\frac{200}{5000}ight)^{2} 	imes 3000 = \left(\frac{1}{25}ight)^{2} 	imes 3000 = \frac{3000}{625} = 4.8 \, W$.

The average power consumed by the given circuit is $W$.

Similar Questions

$A$ $50 \,W$ bulb connected in series with a heater coil is put to an $AC$ mains. Now the bulb is replaced by a $100 \,W$ bulb. The heater output will ...........

If a resistance of $100\, \Omega$,an inductance of $0.5\, H$,and a capacitance of $10 \times 10^{-6}\, F$ are connected in series to a $50\, Hz$ $AC$ supply,then the impedance is:

An alternating voltage of frequency $\omega$ is induced in an electric circuit consisting of an inductance $L$ and capacitance $C$,connected in parallel. Then across the inductance coil:

Find the impedance of the circuit in $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module