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In a series resonant circuit, having L, C and R as its elements, the resonant current is i. The power dissipated in the circuit at resonance is

1. i²R
2. ωL − 1 / ωC
3. zero
4. i²R where ω is the angular resonance frequency.

Correct answer: i²R

Solution

At resonance in an LCR series circuit, the impedance is purely resistive, and the power dissipated is given by P = I²R, where I is the current and R is the resistance. The other options are either incorrect or irrelevant to power dissipation.

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