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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? |
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :L= 5H, |e| = 50V <br/> Let <a href="https://interviewquestions.tuteehub.com/tag/us-718298" style="font-weight:bold;" target="_blank" title="Click to know more about US">US</a> produce the required emf by reducing current to zero. Now `|e|= L (dI)/(dt) or dt= (LdI)/(|e|)= (<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> xx 2)/(50)s` <br/>`=(10)/(50)s= (1)/(5)s= 0.2s` <br/> So, the desired emf can be produced by reducing the given current to zero in 0.2 second.</body></html> | |