A coil of inductance L = 5/8 H and of resistanceR=62.8Ωis connected to the mains alternating voltage of frequency 50Hz. What can be the capacitance of the capacitor (in μF) connected in series with the coil if the power dissipated has to remain unchanged? Take π2=10
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answer is 8.
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Detailed Solution
Pav=EvIvcosϕ=EvEvZR/ZZ has to be same in both cases.Z1=Z2⇒R2+XL2=R2+(XC-XL)2⇒XL=XC-XL⇒2XL=XC⇒2ωL=1ωC⇒C=12ω2L⇒C=12×(2πf)2L=12×4π2×(50)2×5/8=8×10-6F