NEET Physics Alternating Current Questions Solved

NEET - 2009

Power dissipated in an L-C-R series circuit connected to an AC source of emf ε is 

(a) ε2RR2+Lω-1Cω2

(b) ε2R2+Lω-1Cω2R

(c) ε2R2+Lω-1Cω2R

(d) ε2RR2+Lω-1Cω2

Power dissipated in series L-C-R.

 

P=Irms2R=εrms2RZ2=ε2RR2+ωL-1ωC2

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