Equivalent Inductance of Parallel Connected Inductors

When the potential difference between the terminals of two or more inductors are same, it is said that they are connected in parallel. As potential difference or voltage across the terminals are same therefore current flowing through the inductors may or may not be same. If the parallel connected inductors have the same value of inductance then equal current will flow through them. If the value of inductance are not same then current through the inductors will be different.

Equivalent Inductance without Magnetic Coupling between Inductors

Consider the figure below. In the figure n inductors having inductance of L₁, L₂,�..,L_n are connected in parallel and a voltage V is applied. Let current I₁, I₂,�.I_n is flowing through the inductors L₁, L₂,��L_n.

The emf generated in the inductance will oppose the cause i.e. current flowing through the inductor as per Lenz�s Law. Therefore,

V = L₁di1/dt = L₂di2/dt = ���=L_ndin/dt

But we know that,

I = i₁ + i₂ +�..+i_n

 Differentiating both side w.r.t time,

dI/dt = di₁/dt + di₂/dt +�..+di_n/dt

        = V/L₁ + V/L₂ + ��..+ V/L_n   �����.(1)

If the parallel combination of inductors is to be replaced by a single equivalent inductor then,

V = LdI/dt   (assuming current I is flowing the equivalent inductor)

dI/dt = V/L �������(2)

Comparing equation (1) and (2) we get,

V/L = V/L₁ + V/L₂ + ��..+ V/L_n  

1/L = 1/L₁ + 1/L₂ + �..+ 1/L_n

Thus from the above equation we can find the value of equivalent inductance L. It shall be noted that the above formula is valid if there is no mutual coupling between the inductors. If there exists mutual coupling between the inductors the value of equivalent inductance will depend upon the type of coupling i.e. additive coupling or subtractive coupling. Well, we will consider the two types of coupling and will find the equivalent inductance.

Equivalent Inductance for Additive Mutual Coupling

Suppose two inductors with inductance L₁ and L₂ are connected in parallel as shown in figure below. The coupling between the inductors is assumed to be additive. This additive coupling is shown by two dots on the same side of the inductor in the figure.

Suppose the current flowing through inductor L₁ and L₂ being i₁ and i₂ and the mutual inductance is M.

Voltage across L₁ and L₂ = V

But voltage across L₁ = L₁di₁/dt + Mdi₂/dt = V

and across L₂ = L₂di₂/dt + Mdi₁/dt = V 

Therefore we can write,

L₁di₁/dt + Mdi₂/dt = L₂di₂/dt + Mdi₁/dt

di₁/dt (L₁ - M) = di₂/dt (L₂ - M)

di₁/dt = [(L₂ - M) / (L₁ - M)] di₂/dt  ������.(1)

Now, the total current flowing i = i₁ + i₂

Differentiating the above equation w.r.t time we get,

di/dt = di₁/dt + di₂/dt

        = [(L₂ - M) / (L₁ - M)] di₂/dt + di₂/dt

        = di₂/dt [{(L₂ - M) / (L₁ - M)} + 1]

        = di₂/dt [(L₁+L₂-2M) / (L₁ - M)]  �������..(2)

Now, if the above combination of inductor is replaced by an equivalent inductor L then

V = Ldi/dt  (total current i will flow through the equivalent inductor)

    = L x di₂/dt [(L₁+L₂ - 2M) / (L₁ - M)] ��.. from equation (2)

But V = L₂di₂/dt + Mdi₁/dt

Therefore,

L₂di₂/dt + Mdi₁/dt = L x di₂/dt [(L1+L2-2M) / (L₁ - M)]

Putting the value of di₁/dt from equation (1),

? L₂di₂/dt + M[(L₂ - M) / (L₁ - M)] di₂/dt = L x di₂/dt [(L₁+L₂-2M) / (L₁ - M)]

?L₂ + M[(L₂ - M) / (L₁ - M)] = L [(L₁+L₂-2M) / (L₁ - M)]

?[L₂(L₁ - M)  + M(L₂ - M)] / (L₁ - M) = L [(L₁+L₂-2M) / (L₁ - M)]

?[L₂(L₁ - M)  + M(L₂ - M)]  = L (L₁+L₂-2M)

?L = [L₂(L₁ - M)  + M(L₂ - M)] / (L₁+L₂-2M)

      = (L₂L₁ � M²) / (L₁+L₂-2M)

?Equivalent Inductance L = (L₂L₁ � M²) / (L₁+L₂-2M)

Equivalent Inductance for Subtractive Mutual Coupling

Suppose two inductors with inductance L₁ and L₂ are connected in parallel as shown in figure below. The coupling between the inductors is assumed to be subtractive. This coupling is shown by two dots on the opposite side of the inductor in the figure.

Suppose the current flowing through inductor L₁ and L₂ being i₁ and i₂ and the mutual inductance is M.

Voltage across L₁ and L₂ = V

But voltage across L₁ = L₁di₁/dt - Mdi₂/dt = V

Notice that while finding the emf developed across L₁, minus sign is used between L₁di₁/dt and Mdi₂/dt as the coupling between the inductors is subtractive.

and across L₂ = L₂di₂/dt - Mdi₁/dt = V 

Therefore we can write,

? L₁di₁/dt - Mdi₂/dt = L₂di₂/dt - Mdi₁/dt

?di₁/dt (L₁ + M) = di₂/dt (L₂ + M)

?di₁/dt = [(L₂ + M) / (L₁ + M)] di₂/dt  ������.(1)

Now, the total current flowing i = i₁ + i₂

Differentiating the above equation w.r.t time we get,

di/dt = di₁/dt + di₂/dt

        = [(L₂ + M) / (L₁ + M)] di₂/dt + di₂/dt

        = di₂/dt [{(L₂ + M) / (L₁ + M)} + 1]

        = di₂/dt [(L1+L₂ + 2M) / (L₁ + M)]  �������..(2)

Now, if the above combination of inductor is replaced by an equivalent inductor L then

V = Ldi/dt  (total current i will flow through the equivalent inductor)

    = L x di₂/dt [(L₁+L₂ + 2M) / (L₁ + M)] ��.. from equation (2)

But V = L₂di₂/dt + Mdi₁/dt

Therefore,

L₂di₂/dt - Mdi₁/dt = L x di₂/dt [(L₁+L₂ + 2M) / (L₁ + M)]

Putting the value of di₁/dt from equation (1),

L₂di₂/dt - M[(L₂ + M) / (L₁ + M)] di₂/dt = L x di₂/dt [(L₁+L₂ + 2M) / (L₁ + M)]

?L₂ - M[(L₂ + M) / (L₁ + M)] = L [(L₁+L₂ + 2M) / (L₁ + M)]

?[L₂(L₁ + M)  - M(L₂ + M)] / (L₁ + M) = L [(L₁+L₂ + 2M) / (L₁ + M)]

?[L₂(L₁ + M)  - M(L₂ + M)]  = L (L₁+L₂ + 2M)

?L = [L₂(L₁ + M)  - M(L₂ + M)] / (L₁+L₂ + 2M) 

   = (L₂L₁ � M²) / (L₁+L₂ + 2M)

? Equivalent Inductance L = (L₂L₁ � M²) / (L₁+L₂ + 2M)

To summarize,

• Equivalent Inductance L of parallel connected inductors L₁, L₂, �.,L_n having no mutual coupling among them can be find by using

         1/L = 1/L₁ + 1/L₂ +��..+1/L_n

• If two inductors having inductance L₁ and L₂ are connected in parallel and the coupling between them is additive, then equivalent inductance L is given as

        L = (L₂L₁ � M²) / (L₁+L₂ - 2M)

• If two inductors having inductance L₁ and L₂ are connected in parallel and the coupling between them is subtractive, then equivalent inductance L is given as

        L = (L₂L₁ � M²) / (L₁+L₂ + 2M)