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An inductor of 5H inductance carries a steady current of 2A. How can a 50V self-induced emf to 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)s= 0.2s` So, the desired emf can be produced by reducing the given current to zero in 0.2 second. |
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