V0=10V,ω=314rad/s
P=VrmsIrmscosϕ
P=VrmsZVrmsZR
=Z2(Vrms)2R
XL=ωL=(314)(20×10−3)=6.280
XC=ωC1=314×100×10−61=31.84Ω
R=50Ω
Z=(XC−XL)2+R2
=(31.84−6.28)2+(50)2=56Ω
⇒P=(56)2(210)2×50=0.79 W
NEET UG 2018 — Physics Electromagnetism
An inductor 20mH, capacitor 100μF and a resistor 50Ω are connected in series across a source of emf, V=10sin(314t). The power loss in the circuit is
Held on 30 Apr 2018 · Verified 9 Jul 2026.
2.74W
0.43W
0.79W
1.13W
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