In a series LCR circuit, the resistance is 18 | vedclass

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Question

(D) The formula for true power $(P)$ consumed in an $LCR$ circuit is given by $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$.
We know that $I_{rms} = \fr

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(D) The formula for true power $(P)$ consumed in an $LCR$ circuit is given by $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$.
We know that $I_{rms} = \frac{V_{rms}}{Z}$, where $Z$ is the impedance.
Also, the power factor $\cos \phi = \frac{R}{Z}$, where $R$ is the resistance.
Substituting these into the power formula: $P = V_{rms} \cdot \left( \frac{V_{rms}}{Z} ight) \cdot \left( \frac{R}{Z} ight) = \frac{V_{rms}^2 \cdot R}{Z^2}$.
Given values: $V_{rms} = 220 	ext{ V}$, $R = 18 	ext{ } \Omega$, $Z = 33 	ext{ } \Omega$.
$P = \frac{(220)^2 \cdot 18}{(33)^2} = \frac{48400 \cdot 18}{1089}$.
$P = \frac{871200}{1089} = 800 	ext{ W}$.
Therefore, the true power consumed is $800 	ext{ W}$.

In a series $LCR$ circuit, the resistance is $18 \text{ } \Omega$ and the impedance is $33 \text{ } \Omega$. An r.m.s. voltage of $220 \text{ V}$ is applied across the circuit. The true power consumed in the circuit is: (in $\text{ W}$)

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