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Asked by ashutosharnold1998 | 19 Feb, 2020, 09:20: PM
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 Thiyagarajan K | 20 Feb, 2020, 07:57: AM
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