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A coil having a resistance of 5 Omega an...

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

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`X_L_1=omegaL_1=(2pixx50)(0.02)=6.25Omega`
`:. Z_1=sqrt(R_1^2+X_(L_1)^2)`
`=sqrt((5)^2+(6.28)^2)=8.0Omega`
`P_1=(I_("rms"))_(1)V_("rms")cosphi_(1)=(100/8)(100)(5/8)`
`=781.25W`
`X_(L_2)=omegaL_2=(2pixx50)(0.08)`
`=25.13Omega`
`:. Z_2=sqrt(R_2^2+X_(L_2)^2)`
`=25.15Omega`
`=(100/25.15)(100)(1/25.15)`
`=15.8W`
`:. P_("Total")=P_1+P_2=797W`.
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