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In a coil of self-inuctance 0.5 henry, t...

In a coil of self-inuctance `0.5` henry, the current varies at a constant rate from zero to `10` amperes in `2` seconds. The e.m.f. generated in the coil is

A

10 V

B

5 V

C

2.5 V

D

1.25 V

Text Solution

Verified by Experts

The correct Answer is:
C
We have, `(Deltai)/(Deltat)=(10)/(2)=5V`
emf, e `=L(Deltai)/(Deltat)`
`=0.5xx5=2.5V`
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