For a series LCR circuit the power loss at resonance is (a) (V²)/([ω L - (1)/(ω C)]) (b) I² Lω (c) I² R (d) (V²)/(Cω)

Answer: (c). Power is dissipated only in the resistor; L and C take no average power. At resonance Z = R, cos φ = 1, so P = I²R. So the answer is (c) I²R.

For a series LCR circuit the power loss at resonance is | NEET 2002 (AIPMT)

Q79NEET 2002 · AIPMTEasy

Power, Power Factor & Wattless Current

For a series LCR circuit the power loss at resonance is

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Answer: (c)

Power is dissipated only in the resistor; $L$ and $C$ take no average power.
At resonance $Z = R$ , $\cos\phi = 1$ , so $P = I^2R$ .
So the answer is (c) $I^2R$ .

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