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When 100 V d.c is applied across a coil ...

When 100 V d.c is applied across a coil , a current of I A flows through it . When 100 V a.c . Of 50 Hz is applied to the same coil only 0.5 A flows . The inductance of the coil is

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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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