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Question

A coil having a resistance of 5Ω and an inductance of 0.02H is arranged in parallel with another coil having a resistance of 1Ω and an inductance of 0.08H. Calculate the power absorbed (in W) when a voltage of 100V at 50Hz is applied.
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Solution

XL=ωL1=(2π×50)(0.02)=6.28Ω
Z1=R21+X2L1
=(5)2+(6.28)2=8.0 Ω
P1=(Irms)1Vrmscosϕ1 =(1008) (100) (58)
=781.25 W
|XL2=ωL2=(2π×50)(0.08)=25.13Ω
Z2=R22+X2L2
25.15 Ω
P2=(irms)2Vrmscosϕ2
=(10025.15)(100)(125.15)
=15.8 W
PTotal=P1+P2=797 W

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