Power dissipated in an LCR series circuit connected to an A.C. source of emf ε is (a) (ε²[R² + (Lω - (1)/(Cω))²])/(R) (b) (ε²√(R² + (Lω - (1)/(Cω))²))/(R) (c) (ε² R)/(√(R² + (Lω - (1)/(Cω))²)) (d) (ε² R)/([R² + (Lω - (1)/(Cω))²])

Answer: (d). P = I²R with I = (ε)/(Z), Z = √(R² + (Lω - (1)/(Cω))²) P = (ε²R)/(Z²) = (ε²R)/(R² + (Lω - (1)/(Cω))²) So the answer is (d).

Power dissipated in an LCR series circuit connected to an A.C. source… | NEET 2009 (AIPMT)

Q74NEET 2009 · AIPMTEasy

Power, Power Factor & Wattless Current

Power dissipated in an LCR series circuit connected to an A.C. source of emf $\varepsilon$ is

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Answer: (d)

$P = I^2R$ with $I = \frac{\varepsilon}{Z}$ , $Z = \sqrt{R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2}$
$P = \frac{\varepsilon^2R}{Z^2} = \frac{\varepsilon^2R}{R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2}$
So the answer is (d).