Question

calculate the mutual inductance between two coils, when a current of 4 A changes to 8 A in 0.5 seconds and induces an e.m.f. of 50 mV in secondary coil.

Answer 1

e = - MdI/dt

So, 50 × 20^-3 = M ( 8 - 4) / 0.5

50 × 10^-3 = M × 8

Mutual Inductance, M = 6.25 × 10^-3 Henry

Answer 2

Answer:

6.25 mH is the required mutual inductance between two coils .

Explanation:

Given ,  a current 4A changes to 8A.

Let $I_1$ = 4A and $I_2$ = 8A

and induces e. m .f (e) = 40mV

Change in current (dI) = $I_ 2- I_1$ = 8 - 4 = 4A

Time  (dt ) = 0.5 sec

As we know that  ,

e = M× $\frac{dI}{dt}$ ⇒ M =  $\frac{e dt}{dI}$

Now , put the given value in the above  formula  we get,

M = 50 × $\frac{0.5}{4}$ =  50 ×$\frac{1}{8}$ =   6.25mH

Hence , the mutual induction  between given two coil is 6.25 mH

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