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NEET 2009
Exam
NEET
Year
2009
Subject
Physics
Chapter
Alternating Current

NEET 2009 · Physics · Alternating Current

Power dissipated in an LCR series circuit connected to an a.c....

Power dissipated in an LCR series circuit connected to an a.c. source of emf ε\varepsilon is :

Options

  1. A\varepsilon^2 R / \sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}
  2. Bε2R/[R2+(Lω−1Cω)2]\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]
  3. Cε2[R2+(Lω−1Cω)2]/R\varepsilon^2 \sqrt{\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]} / \mathrm{R}
  4. Dε2[R2+(Lω−1Cω)2]R\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}

Correct answer

B. ε2R/[R2+(Lω−1Cω)2]\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]

Step-by-step solution

Pav=Erms⋅Irmscos⁡ϕ=ε⋅εz⋅RZ=ε2Rz2=ε2RR2+(ωL−1ωC)2\begin{aligned}\mathrm{P}_{\mathrm{av}}& =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi\\ & =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2}\\ & =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2} \end{aligned}
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