Introduction to Electric Circuits

32 Electric circuit elements

Example 2.17

Calculate the mutual inductance between two coils having self-inductances of
2.5 mH and 40 mH if

(1) they are so placed that the coefficient of coupling is 0.8;

(2) one of the coils is wound closely on top of the other;

(3) the coils are places as shown in Fig. 2.23.

L1

L2

Figure 2.23

Solution

1 From Equation (2.41) we have that

M = kV/(L~L2) = 0.8V'(2.5 x 40) = 8 mH

2 Since the coils are wound one on top of the other, then virtually all the flux

produced will link with both coils and so k = 1. Thus

M = k~/(LIL2) = V/(2.5 x 40) = 10 mH

3 In this case the magnetic axes of the two coils are at right angles so that

there is no magnetic coupling and so k = 0 and M = 0.

Inductance M series

The diagram of Fig. 2.24 shows two coils connected in series electrically and

coupled magnetically. The total emf induced in coil 1 is the sum of the self-

induced emf due to the current changing in itself and the mutually induced emf

due to the current changing in coil 2. Thus

E 1 -- Ell -k- E12-- L~di/dt + Mdi/dt- (L~ + M)di/dt (2.42)

Similarly

E2 = E22 + El2 = L2di/dt + Mdi/dt- (L2 + M)di/dt

The total emf induced in the series combination is therefore given by

(2.43)

E 1 -~ E 2 = (L 1 + L 2 + 2M)di/dt (2.44)