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Asked by ashutosharnold1998 19th February 2020, 9:20 PM
Answered by Expert
Answer:
If bulbs are identical their resistances are same
 
Impedence of capacitor and bulb z1 = [ R2 + 1 / ( C ω )2 ]1/2
 
reactance of Capacitor = 106 / ( 500 × 50 ) = 40Ω
 
Impedence of inductor and bulb z2 = [ R2 + ( L ω )2 ]1/2
 
reactance of inductor = 10×10-3 × 50 = 0.5Ω
 
it can be seen that z2 < z1 . Hence current passing through inductor is greater than current passing through capacitor
 
Hence bulb B2 is brighter than bulb B1



Answered by Expert 20th February 2020, 7:57 AM
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