Power dissipated in an L-C-R series circuit connected to an AC source of emf ε is
NEET UG 2009 — Physics Electromagnetism
2009mcqmedium
Power dissipated in an L−C−R series circuit connected to an AC source of emf ε is
Official previous-year question
Held on 30 Apr 2009 · Verified 9 Jul 2026.
Options
A
[R2+(Lω−Cω1)2]ε2R
B
Rε2R2+(Lω−Cω1)2
C
Rε2[R2+(Lω−Cω1)2]
D
R2+(Lω−Cω1)2ε2R
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Solution
The emf of an LCR circuit is ε. The impedance of a series LCR circuit is given as, Z=[R2+(ωL−ωC1)2] The power factor in a series LCR circuit is given as, cosϕ=∣Z∣R The power dissipated in the circuit is given as, P=VrmsIrmscosϕ=ε×∣Z∣ε×∣Z∣R=[R2+(ωL−ωC1)2]ε2R Thus, the power dissipated in the series LCR circuit is [R2+(ωL−ωC1)2]ε2R.
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