Question

Two identical coils of 1200 turns each, are placed side by side such that 60% of the
flux produced by one coil links the other. A current of 10A in the first coil, sets up a
flux of 0.12 mWb. If the current in the first coil changes from +10A to -10A in 20
mS, find (a) the self - inductances of the coils (b) the emf induced in both the coils.

Answer 1

Answer:

Using basic idea and necessary formulas this problem can be solved.

Follow the attachment and excuse me if I have made a mistake.

Answer 2

(a) 0.0144 Wb$A^{-1}$ and 0.0864 Wb$A^{-1}$

(b) 14.4 and 8.64

Given:

Two identical coils of turns = 1200

The flux produced by one coil links the other = 60%

A current in the first coil = 10A

Flux = 0.12 mWb

Current in the first coil changes from +10A to -10A

Time = 20mS

To find:

(a) the self - inductances of the coils

(b) the emf induced in both the coils.

Solution:

For the first coil:

NQ = LI

=> Self inductance (L) = 1200 x 0.12 x 10^3/10

=> L = 0.0144 Wb$A^{-1}$

Emf induced = -Ldi/dt

=> Emf1 = 14.4

For the second coil:

=> Self inductance(M) = 1200 x 0.6 x 0.12 x 10^-3/10

=> M = 0.00864 Wb$A^{-1}$

Emf induced = -Mdi/dt

=> Emf2 = 8.64

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