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An air cored coil has a self-inductance of 0.1 H. A soft-iron core of relative permeability 100 is introduced and the number of turns is reduced to 1//10^(th). The value of self-inductance now is : |
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Answer» SOLUTION :`L=0.1=(mu_(0)N^(2)A)/(l), L.(mu_(r )mu_(0)N_(1)^(2)A)/(l)`, `N_(1)=(N)/(10)` `THEREFORE(L.)/(0.1)=(1000xx mu_(0)(N//10)^(2)A.l)/(l xx mu_(0) N^(2)A)`or`L. = 1 H` |
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