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An inductor of self inductance 12 H carries a steady current of 2A. How can a 60V self-induced e.m.f. be made to appear in the inductor? |
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Answer» SOLUTION :`L=12H,""I=2A,VAREPSILON=60V` Let `(dI)/(DT)` be the rate of variation of current required, so that, `(dI)/(dt)=(varepsilon)/(L)=(60)/(12)=5A//s""therefore (dI)/(dt)xxt=2andt=(2)/((dI)/(dt))=(2)/(5)=0.4s` The current should be uniformly reduced from 2 to zero in 0.4 seconds. |
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