Question

The inductor arrangement in the figure at the right, with L1 = 40.0 mH, L2 = 30.0 mH, L3 = 15.0 mH, and L4 = 20.0 mH, is to be connected to a varying current source. Determine the equivalent inductance of the arrangement.

These are my teachers requirements if you could please follow them I would really appreciate it thank you :)

In addition to being neat and clear, and actually answering the question, you must:

1) show the original principle (often in equation form)

2) substitute variables as needed?

3) solve it first (before substituting numbers)

4) show every substitution (with units and correct sig figs)

5) present a boxed answer (with units and correct sig figs)

In a Force problem in more than one dimension, you MUST start with a Free Body Diagram.

Answer 1

Inductor in Series Circuit

The current, ( I ) that flows through the first inductor, L1 has no other way to go but pass through the second inductor and the third and so on. Then, series inductors have a Common Current flowing through them, for example:

IL1 = IL2 = IL3 = IAB …etc.

In the example above, the inductors L1, L2 and L3 are all connected together in series between pointsA and B. The sum of the individual voltage drops across each inductor can be found using Kirchoff’s Voltage Law (KVL) where, VT = V1 + V2 + V3 and we know from the previous tutorials on inductance that the self-induced emf across an inductor is given as: V = L di/dt.

So by taking the values of the individual voltage drops across each inductor in our example above, the total inductance for the series combination is given as:

By dividing through the above equation by di/dt we can reduce it to give a final expression for calculating the total inductance of a circuit when connecting inductors together in series and this is given as:

Inductors in Series Equation

Ltotal = L1 + L2 + L3 + ….. + Ln etc.

Inductors in Parallel Circuit

In the previous series inductors tutorial, we saw that the total inductance, LT of the circuit was equal to the sum of all the individual inductors added together. For inductors in parallel the equivalent circuit inductance LT is calculated differently.

The sum of the individual currents flowing through each inductor can be found using Kirchoff’s Current Law (KCL) where, IT = I1 + I2 + I3 and we know from the previous tutorials on inductance that the self-induced emf across an inductor is given as: V = L di/dt

Then by taking the values of the individual currents flowing through each inductor in our circuit above, and substituting the current i for i1 + i2 + i3 the voltage across the parallel combination is given as:

By substituting di/dt in the above equation with v/L gives:

We can reduce it to give a final expression for calculating the total inductance of a circuit when connecting inductors in parallel and this is given as:

Parallel Inductor Equation

so here

L2 and L3 are in parallel so effective of them is

L_eff = L2*L3 / (L2 + L3) = 15 * 30 / (15+30) = 10 mH

this is connected in series with L1 and L4

so L_total (of the circuit) = L1 + L4 + L_eff = 40 + 20 + 10 = 70mH