1.

The flux linked with each turn of coil is 0.1 Wb and total number of turns is 1000. If current passing through coil is 10 amp, then self inductance of coil is ...... mH

Answer»

0.1
10
`10^4`
`10^(-4)`

Solution :`N phi=LI`
`THEREFORE L=(NPHI)/I`
`=(1000xx0.1)/10`
=10H
`=10xx1000x10^(-3)` H
`therefore L=10^4` mH


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