In a series LCR circuit,R = 18 Omega and the | vedclass

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(D) The true power $P$ consumed in an $AC$ circuit is given by the formula: $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$,where $\cos \phi = \frac{R}{Z}

Vedclass · Accepted Answer

$900$

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(D) The true power $P$ consumed in an $AC$ circuit is given by the formula: $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$,where $\cos \phi = \frac{R}{Z}$.First,calculate the $rms$ current $I_{rms}$ using Ohm's law for $AC$ circuits: $I_{rms} = \frac{V_{rms}}{Z} = \frac{210 \ V}{30 \ \Omega} = 7 \ A$.Next,calculate the power factor $\cos \phi = \frac{R}{Z} = \frac{18 \ \Omega}{30 \ \Omega} = 0.6$.Now,substitute these values into the power formula: $P = 210 \ V 	imes 7 \ A 	imes 0.6$.$P = 1470 	imes 0.6 = 882 \ W$.Rounding to the nearest given option,the true power consumed is approximately $900 \ W$.

In a series $LCR$ circuit,$R = 18 \ \Omega$ and the impedance $Z = 30 \ \Omega$. An $rms$ voltage of $210 \ V$ is applied across the circuit. The true power consumed in the $AC$ circuit is nearly: (in $W$)

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