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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 |
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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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