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An inductor of 5 H inductance carries a 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)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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