Alternating Current

JEE Main level, numerical answer, medium

5 of 190

Question 5 of 190

An inductor of reactance 100Ω100 \Omega, a capacitor of reactance 50Ω50 \Omega, and a resistor of resistance 50Ω50 \Omega are connected in series with an AC source of 10 V , 50 Hz. Average power dissipated by the circuit is ________ W.

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Answer: 1

P=VrmsIrmscos⁡ϕP=Vrms×Vrmsz×RzP=Vrms2×Rz2z=R2+(xL−xC)2Z=502ΩP=100×502500×2=1 W\begin{aligned} & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \times \frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}} \times \frac{\mathrm{R}}{\mathrm{z}} \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}}^2 \times \frac{\mathrm{R}}{\mathrm{z}^2} \\ & \mathrm{z}=\sqrt{\mathrm{R}^2+\left(\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{C}}\right)^2} \\ & \mathrm{Z}=50 \sqrt{2} \Omega \\ & \mathrm{P}=100 \times \frac{50}{2500 \times 2}=1 \mathrm{~W}\end{aligned}

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