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An inductor of 5 H inductance carries a E steady current of 2A. How can a 50 V self-induced emf be made to appear in the inductor ? |
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Answer» Solution : L =5H, |e| = 50V Let us produce the required EMF by REDUCING current to zero. Now , `|e| = L (dI)/(DT) " or " dt = (LdI)/(|e|) = (5 xx 2)/(50) s ` ` = 10/50 s = 1/5= 0.2 s` ? So, the desired emf can be produced by reducing the given current to zero in 0.2 second. |
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