Question

An inductor 20mH a capacitor 50μF and a resistor 40ohm are connected in series across a source of emf V = 10sin 340 t.The power loss in AC circuit is​

Answer 1

Explanation:

Alternating Current. and a resistor of 40 ohm are connected in series across a source of emf V = 10 sin 340 t. the power loss in the ac circuit is, 0.67 W.

Answer 2

☆ANSWER☆

XC=ωC1=340×50×10−61=58.8 Ω

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 Ω

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2    =402+(58.8−6.8)2

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2    =402+(58.8−6.8)2    =4304 Ω

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2    =402+(58.8−6.8)2    =4304 ΩPower, P=irms2R=(ZVrms)2R

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2    =402+(58.8−6.8)2    =4304 ΩPower, P=irms2R=(ZVrms)2R                              =(430410/2)2×40

XC=ωC1=340×50×10−61=58.8 ΩXL=ωL=340×20×10−3=6.8 ΩZ=R2+(XC−XL)2    =402+(58.8−6.8)2    =4304 ΩPower, P=irms2R=(ZVrms)2R                              =(430410/2)2×40                              =430450×40≈0.51 W

HOPE ITS HELP U :)