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Standard XII

Physics

Transformer

The coefficie...

Question

The coefficient of mutual induction between two coils is $1.25 H .$ If the rate of fall of current in the primary is $80$ As $^{- 1}$ , then the induced emf in the secondary coil will be

100 V

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64 V

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12.5 V

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0.016 V

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Solution

The correct option is A $100 V$
emf = $- M \frac{d I}{d t}$
$= - 1.25 \times - 80 = 100 V$

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The coefficient of mutual induction between two coils is 1.25H. If the rate of fall of current in the primary is 80 As−1, then the induced emf in the secondary coil will be

The correct option is A 100Vemf = −MdIdt=−1.25×−80=100V

The coefficient of mutual induction between two coils is $1.25 H .$ If the rate of fall of current in the primary is $80$ As $^{- 1}$ , then the induced emf in the secondary coil will be

The correct option is A $100 V$
emf = $- M \frac{d I}{d t}$
$= - 1.25 \times - 80 = 100 V$