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For a inductor coil L = 0.04 H, then workdone by source to establish a current of 5A in it is -A. 0.5 JB. 1.00 JC. 100 JD. 20 J

Answer» Correct Answer - B
Work done by source
`Exxq=E((Deltaphi)/(R))=E(LI_0)/(R)`
`(E/R)LI_0=(I_0)LI_0=LI_0^2`
`0.04xx(5)^2=1.0J`


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