JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2025
- Easy
- Numerical
An inductor of reactance \(100 \Omega\), a capacitor of reactance \(50 \Omega\), and a resistor of resistance \(50 \Omega\) are connected in series with an AC source of 10 V , 50 Hz. Average power dissipated by the circuit is ________ W.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \times \frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}} \times \frac{\mathrm{R}}{\mathrm{z}} \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}}^2 \times \frac{\mathrm{R}}{\mathrm{z}^2} \\ & \mathrm{z}=\sqrt{\mathrm{R}^2+\left(\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{C}}\right)^2} \\ & \mathrm{Z}=50 \sqrt{2} \Omega \\ & \mathrm{P}=100 \times \frac{50}{2500 \times 2}=1 \mathrm{~W}\end{aligned}\)
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