Electromagnetic Induction and Alternating Current 3 Question 9

10. The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A and is decreasing at a rate of 103A/s, then VBVA=V

(1997, 1M)

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Solution:

  1. didt=103A/s

Induced emf across inductance, |e|=Ldidt

|e|=(5×103)(103)V=5V

Since, the current is decreasing, the polarity of this emf would be so as to increase the existing current. The circuit can be redrawn as

Now,

VA5+15+5=VB

VAVB=15V or VBVA=15V



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