Grouping of inductors

GROUPING OF INDUCTORS

Inductors can be connected either in series or parallel or combination of both. Here it is assumed that the inductors are connected electrically and isolated magnetically i.e., there is no magnetic coupling between the elements.

Case 1: Introductions in series

Let L₁ and L₂ be the two linear time variant inductors in series. Let V be the voltage applied and be the current flowing.

The self induced emf in L₁ = e₁ = -L₁ dt / di

The self induced emf in = L₂ = e₂ = -L₂  di / dt

The total self induced emf = e₁ + e₂

According to Lenz's law, this self induced emf is to oppose the applied voltage i,e.,

- (L₁ + L₂) di / dt = –V

⇒ (L₁+L₂)di/dt = – V....(i)

If L be the equivalent of the above series combination, the induced emf in it

= –L di / dt

Therefore, L di/dt = V .....(ii)

From equation (i) and (ii) we write that

L di/dt = (L₁ + L₂) di/dt

L = (L₁ + L₂)

Thus, the inductors in series can be added to obtain the equivalent single inductor.

Case 2: Inductors in parallel

From the parallel combination, the voltage across L₁ = the across L₂ = the applied voltage.

i,e.

L₁ di₁ / dt = L₂ d_i2/dt = V ... (i)

d_i1/dt = V/L1 and di₂/dt = V/ L₂ ......(ii)

But by KCL, i = i₁ + i₂ ...(iii)

Differentiating on both sides of equation (iii), we get

Note: If 'n' inductors are in parallel,

1/L =1/L₁ +1/L₂ +1/L₃+.....1/L_n