Connecting numerous inductor in parallel in a circuit connects their terminals to the same two locations. This combination has pros and cons over a single inductor. Designing and optimizing electrical circuits requires understanding the purpose, pros, and cons of parallel inductors. Parallel inductors enhance circuit inductance. Connecting many inductor in parallel increases the circuit’s inductance. When the inductance needed surpasses a single inductor, this is advantageous. Parallel inductors increase current handling and distribution, making them appropriate for electronics and power electronics applications.

**Benefits of Parallel Inductors**

Parallel inductors have many benefits. First, it provides more circuit control than a single inductor due to its higher overall inductance. Distributing current across many inductors allows parallel inductors to manage greater currents. This reduces saturation and boosts performance. Using several inductors allows for a reduced footprint while retaining inductance. This helps in space-constrained situations.

**Parallel Inductor Drawbacks**

Inductor in parallel have pros and cons. The circuit design and construction complexity is a big negative. Properly connecting and distributing current among inductors complicates circuit planning and may increase impedance and resonance. Multiple inductors may raise cost and production difficulties. This requires considerable consideration and analysis to improve parallel inductor use while minimizing their downsides.

Parallel inductors have their ends connected to other inductors. Like resistors, inductors coupled in parallel have a total inductance slightly less than their smallest inductance.The current across parallel inductors is not the same as the total current. However, summing the currents via all parallel inductors yields the total current.

When the total current is less than the current through each inductor, the magnetic field each inductor creates is also less.The smallest resistor in parallel blocks less current than the largest, therefore most of the current goes through it.When a circuit’s current changes, it picks the inductor with the least resistance if the inductors are connected in parallel. Because each inductor opposes the current change.

**Inductor paralleled **

Connecting two inductors in parallel maintains the same voltage, and if the total current changes, the voltage loss is less than in series. Low voltage lowers inductance for the same current change.

These are the most crucial considerations when paralleling inductors. We shall now discuss the parallel link of inductors with and without considering their interactions.

**Make plans**

Parallel-connected inductors without magnetic coupling.As mentioned, one end of an inductor is connected to a node and all of its other ends are parallel to another node. The figure below shows n parallel inductors.

Assume no magnetic connection between inductors. The total inductance equals the sum of the reciprocals of each inductance. Discuss how to acquire this comment.In parallel networks, the voltage keeps the same while the current splits at each parallel inductor. If the parallelly connected inductors L1, L2, and so on have currents IL1, IL2, IL3, and so on, their total current is

- Total IL1+IL2+IL3+ In

Parallel links have voltage dips VL1, VL2, VL3, etc. Total voltage drop between terminals is VT.

**How VTotal works: VL1=VL2=VL3=Vn**

- The self-inductance voltage drop is V = L di/dt. The power drops completely.
- VT=LTd/dt
- IL1+IL2+IL3+In d/dt
- LT (di1, di2, di3, etc.)

With di/dt replaced by V/L, the equation becomes

LT = (V/L1+V/L2+V/L3)

Since voltage loss across the circuit is constant, v = VT. To write

“1/LT”=”1/L1+1/L2+1/L3.”

The parallel link’s total inductance reciprocal equals the sum of each inductor’s reciprocal. When parallel coils have no mutual inductance, the equation above is true.

The product over sum approach can calculate total inductance without fractions. The total inductance of two parallel inductors without mutual inductance is

LT = (L1-L2)/(L1+L2).

**A parallel-connected inductor.**

One 20 Henry and one 30 Henry inductor are in the circuit. What is their total inductance when linked in parallel?

We know that 1/LT = 1/L1 + 1/L2 to calculate string inductance.

If L1=20,

30 Henry L2

This indicates LT = (L1*L2)/(L11+L2) = (20*30)/((20+30)) = 600 / 50 = 12.

LTotal = 12 Henry is the inductance.

**Rows of mutually linked inductors**

When inductors are magnetically coupled, the formula for total inductance must be altered since each inductor’s magnetic field moves differently. Parallel inductors’ magnetic fluxes will connect.

The mutual inductance increases when the fluxes formed match the magnetic flux. These coils are “Aiding” coils. Mutual inductance decreases when flux is opposite magnetic flux. These are “opposing” coils. The distance between the coils affects this shared inductance.

See the image below, which displays two inductors connected in parallel with self-inductances L1 and L2 and a mutual inductance M.

**Parallel Inductors Help**

See Figure (a). Magnetic fields assist connect parallel inductors L1 and L2. To find the circuit’s total current:

i = i1 + i2

Since di/dt = (di1)+(di2),

How to find inductor or parallel branch voltage:

L1 or L2 (di1)/dt + M (di2)/dt

This equals L1 (di1)/dt plus M (di2)/dt.

Di1/dt (L1–M) = di2/dt.

= di2/dt ((L2 – M)/(L1 – M)) …………. (2)

Putting equation 2 into equation 1 yields

Di2/dt ((L2–M))/((L1–M)) +

(di2)/dt = (L2–M)/((L1–M)) + 1. (3)

In a parallel inductor circuit, LT is total inductance. Voltage is given by

V=LTd/dt

L-1 + M-2 = LT di/dt

LT = 1/di/dt ^ L1 (di1) + M (di2)

Equation 2 into the first equation yields

D/dt = 1/LT ^ L1 (di2)/dt (L2–M)/((L1–M)) + M (di2)/dt

1/LT = L1 (L2–M)/((L1–M)) + M(di2)/dt. (4)

**Combining equations 3 and 4 yields**

1/1 = LT^L1 (L2–M)/((L1–M)) + M

Simplifying the equation yields

The formula is LT = (L1 L2–M2)/(L1+L1)-2M.

The magnetic flux of L1 on L2 or L2 on L1 is 2M. If their magnitudes are equal and they are properly magnetically connected, two inductors have the same equivalent inductance (L). Because LT = L1 = L2 = M. If the joint inductance is 0, the total inductance is L 2.

**Parallel-Helping Inductors Example**

If 25mH and 45mH inductor are connected in parallel, find their total inductance. A 20mH common inductance is reported.

Therefore, L1 = 25 mH.

L2 = 45 mH

M = 20 mH

The formula for assisting inductor total inductance is LT = (L1 L2– M2)/(L1+ L1)-2M.

LT = (25*45- 202)/(25+45-2*20)

- = (1125-400)/(70-40)
- = 725/30
- = 24.166mH

#### Total inductance is 24.166 milli Henry.

Different parallel inductors

In Figure (b), inductors L1 and L2 are connected in parallel and have opposite magnetic fields. The total inductance is

L1–L2–M2 = L1+L2+2M

Two inductors with equal strengths and perfect magnetic coupling cancel each other out, hence their equivalent inductance is zero. If two inductors allow current, their total inductance is (L ± M) ÷ 2.

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