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Self-inductance of a coil is 50mH. A current of 1 A passing through the coil reduces to zero at steady rate in 0.1 s, the self-induced emf isA. 5 VB. 0.05 VC. 50 VD. 0.5 V

Answer» Correct Answer - D
Here, `L=50xx10^(-3)H`
`(dI)/(dt)=((1-0))/(0.1)=10`
emf, `e=(L.dI)/(dt)=50xx10^(-3)xx10`
`=50xx10^(-2)=0.5V`


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