Electro Magnetic Induction And Alternating Currents Question 489

Question: A varying current in a coil changes from 10A to zero in 0.5 sec. If the average e.m.f induced in the coil is 220V, the self-inductance of the coil is

Options:

A) 5 H

B) 6 H

C) 11 H

D) 12 H

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Answer:

Correct Answer: C

Solution:

  • Initial current

    $ (I_{1})=10A $ ;

    Final current $ (I_{2})=0 $ ;

    Time (t) = 0.5 sec and induced e.m.f. $ (\varepsilon )=220 $ V. Induced e.m.f. $ (\varepsilon ) $

    $ =-L\frac{dI}{dt}=-L\frac{(I_{2}-I_{1})}{t}=-L\frac{(0-10)}{0.5}=20L $ Or,

    $ L=\frac{220}{20}=11H $ [where L = Self inductance of coil]



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