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    A series LCR circuit (R = 30 Ω, XL\text X_\text LXL​​ = 40 Ω, XC\text X_\text CXC​​ = 80 Ω) is connected to an AC source of 200 V and 50 Hz.
    Question

    A series LCR circuit (R = 30 Ω, XL\text X_\text L​ = 40 Ω, XC\text X_\text C​ = 80 Ω) is connected to an AC source of 200 V and 50 Hz. The power dissipated in the circuit is:

    A.

    480 W

    B.

    240 W

    C.

    48 W

    D.

    24 W

    Correct option is A

     Given Parameters: Resistance (R)=30 Ω Inductive reactance (XL)=40 Ω Capacitive reactance (XC)=80 Ω RMS voltage (Vrms)=200 V Frequency =50 HzZ=R2+(XLXC)2=302+(4080)2=900+1600=50 ΩIrms=VrmsZ=20050=4 Acosφ=RZ=3050=0.6(35)P=Vrms×Irms×cosφ=200×4×0.6=480 W\begin{aligned}&\textbf{ Given Parameters:} \\&\quad \circ~\text{Resistance } (R) = 30\,\Omega \\&\quad \circ~\text{Inductive reactance } (X_L) = 40\,\Omega \\&\quad \circ~\text{Capacitive reactance } (X_C) = 80\,\Omega \\&\quad \circ~\text{RMS voltage } (V_{rms}) = 200\,\text{V} \\&\quad \circ~\text{Frequency } = 50\,\text{Hz} \\\\&Z = \sqrt{R^2 + (X_L - X_C)^2} \\&\quad = \sqrt{30^2 + (40 - 80)^2} \\&\quad = \sqrt{900 + 1600} \\&\quad = 50\,\Omega \\&I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{50} = 4\,\text{A} \\\\&\cos\varphi = \frac{R}{Z} = \frac{30}{50} = 0.6 \left( \frac{3}{5} \right) \\\\&P = V_{rms} \times I_{rms} \times \cos\varphi \\&\quad = 200 \times 4 \times 0.6 \\&\quad = 480\,\text{W}\end{aligned}​​

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