When `100" volt" dc` is applied across a coil, a current of `1 amp` flows through it, when `100 V` ac of `50 Hz` is applied to the same coil, only 0.5 amp flows. Calculate the resistance and inductance of the coil.
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In case of a coil i.e. L-R circuit . `l=(V)/(Z)withZ=sqrt(R^(2)+(omegaL)^(2))` So when dc is applied`omega=0,Z=R` And hence`I=(V)/(R)i.e.R=100Omega` And when ac of 50 Hz isapplied `I=(V)/(Z)i.e.Z=(V)/(I)=(100)/(0.5)=200Omega` But`Z=sqrt(R^(2)+(omegaL))` `2pifL^(2)=200^(2)-100^(2)=3xx10^(4)` L=(sqrt3xx10^(2))/(2pixx50)=(sqrt3)/(pi)=0.55H`
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