Parallel Opposing Assignment Help

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Parallel Opposing:

Figure illustrates two coils coupled in parallel where the mutually induced e.m.f in a coil because of change in current in the other coil is negative.

1496_Parallel Opposing.png

Figure: Inductive Coupling in Parallel (Flux Opposing)

Using Kirchoff's voltage law, we can write

V = L1 (di1/dt)  - M (di2/ dt)

Also,

V = L2 (di2/dt)  - M (di1/ dt)

From Eqs. (53) and (54), we get

L ( di1 /dt) - M di2  /dt = L2 (di2/dt)  - M (di1/ dt) ---- (60)

Now

i = i1  + i2

i2  = i - i1

Substituting i2 from Eq. (56) in Eq. (55), we obtain

L  di1/dt  - M d (i - i1 )/dt = L2 + d (i - i1 ) - M(di1/dt)

L  di1/dt  - M d i /dt  + M d i /dt  = L2 + di /dt  - M(di1/dt)

(L1  + L2  + 2M )  di1/dt  = (L2  + M )  di1/dt

∴          di1 /dt  = ( (L2  + M )/(L1 + L2  + 2M)  (di /dt  )

Likewise,         di2 /dt      = (L1+M/ L1 + L2 + 2M  ) di/dt

Using Eqs. (57) and (58) in Eq. (53), we obtain

 V = L1 (L2  + M/ L1 + L2  +2M) di/dt + M(L1 + M )  /  (L1 + L2  + 2M ) (di/dt)

or, V =   (L1 L2  + L 1M + L1 M - M 2)   / (L1 + L2  +2M)            di/dt

or V =   (L1 L2  - M2 )/ (L1 + L2  + 2M)  di/dt

Let L be the equivalent inductance of the parallel combination, then we can write

V = L (di /dt)

From Eqs. (61) and (62), we obtain

 L (di/dt ) =   (L1 L2  - M2 )/ (L1 + L2  + 2M) di/dt                    

L =   (L1 L2 - M2)/ (L1 + L2 +2M)

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