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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
where ω is the angular resonance frequency.
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i²R [ωL -(1/ωC)] - zero
- i²ωL
- π²R
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Correct Option: D
| At resonance Lω = | 1 | ,ω |
| Cω |
| = | 1 | |
| √LC |
| Current through circuit i = | E | |
| R |
Power dissipated at Resonance = i²R
