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An inductor and a resistor are connected in series to an AC source. The current in circuit is 500 mA, if the applied AC voltage is 8sqrt2 V at a frequency of 175/(pi) Hz and the current in the circuit is 400 mA, if the same AC voltage at a frequency of (225)/(pi)Hz is applied . The values of the inductance and the resistance are respectively |
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Answer» 60 m H ,71 `Omega` `implies""R^(2)+L^(2)omega^(2)=(V/I)^(2)` Here,`""I_(1)=500xx10^(-3)A` `""omega_(1)=175/(pi)xx2pi(rad)/s=350(rad)/s` `""V_(1)=8sqrt2` `implies""R^(2)+L^(2)(350)^(2)=((8sqrt2)/(500xx10^(-3)))^(2)` `implies""R^(2)L^(2)(350)^(2)=512""...(i)` `implies""R^(2)L^(2)(550)^(2)=800""...(ii)` Solving , we GET `R=sqrt71Omega" and "L=60 mH` |
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