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 For a current-carrying inductor, emf associated is 20m V. Now, the current through it changes from  6A to 2A in 2 s. The coefficient of mutual  inductance is 

Answer:

Explanation:

$|e|=L\frac{dI}{dt}$

 Here, e= 20mV= 20x 10^-3 V

 $\frac{dI}{dt}=\frac{6-2}{2}=2A/s$

  L= ?

 $\Rightarrow  20\times 10^{-3}=L \times 2$

 $\therefore$     L= 10x 10^-3= 10mH