## Parallel Aiding Assignment Help

Assignment Help: >> Inductive Coupling in Parallel - Parallel Aiding

Parallel Aiding:

Figure illustrates two coils coupled in parallel where the fluxes are additive as per dot convention.

Figure: Inductive Coupling in Parallel (Flux Aiding)

Using Kirchoff's voltage law, we may write

V = L1 (di1/ dt ) + M (di2/ dt)

Also,

V = L 2(di2/ dt ) + M (di1/ dt)

From Eqs. (2.42) and (2.43), we get

L  (di1/ dt ) + M( di2/ dt)  = L (di2/ dt ) + M( di1/ dt )

Now

∴ i = i1  + i2

i2  = i - i1

Substituting i2 from Eq. (45) in Eq. (44), we obtain

L1  (di1/ dt)  + M d (i - i1 )/ dt = L2 d (i - i1 ) / dt + M (di1/ dt)

Or, L1  (di1/ dt)  + M di / dt  - M di1 / dt = L2 (di / dt) - L2 (di1 / dt )+ M (di1/ dt)

or,  (L 1 + L2  - 2M )  (di1 /dt) = (L2  - M )  di1/dt

∴          di2/dt  = ((L2  - M )/ (L1 + L2  - 2M )) di1/ dt

Similarly,

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

Using Eqs. (46) and (47) in Eq. (42), we get

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 may write

V = L (di /dt)

From Eqs. (50) and (51), we obtain

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

L =   (L1 L2 - M2)/ (L1 + L2 - 2M)