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An inductor of 5 H inductance carries a steady current of 2 A. How can a 50 V self induced e.m.f. be made to appear in the inductor ? |
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Answer» `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)/(8)s = 0.2 s` So, the desireed emf can be produced by reducing the given current to zero in `0.2` second. |
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