Hi everyone. In this last video of the series, we learn how to reduce inductors connected in series and parallel to one equivalent inductor. So as a reminder, inductor is a dynamic element whose voltage current relationship is given as follow: V, the voltage drop across the inductor, is equal to L di/dt, where L is the inductance, I is the current through that inductor. Integrating this equation, we can get the current through the inductor as the integral of the voltage across the inductor divided over L.

So let's start with a series network. So here I have two inductors in series in the circuit, and the circuit is supplied by one voltage supply V_T here that is sourcing a current i. I'd like to reduce the circuit to an equivalent one that has one equivalent inductor, let's call it L_T, such that the two circuits are equivalent. So here I have V_T and i. Okay, considering the circuit on the left, I have the inductor and the voltage source in parallel, so they share the same voltage. So now I can write down the voltage across the inductor V_T in terms of the current through the inductor through this equation L di/dt, where i is the current through the inductor, which is the same as the current sourced out from our voltage source.

So now let's take the equation and do circuit analysis on the circuit on the right to see how we would represent L_T in terms of L1 and L2. So far I have an expression for V_T in terms of the equivalent inductance now. So let's go to the circuit on the right. So I have my voltage supply V_T. So I wrote a KVL here: we have V_T equal to V1 + V2. From the previous slide, we have V_T represent terms of the equivalent inductor, so this is what I have here. Now I can write down expression for V1, where V1 is the voltage across L1, as equal to the derivative of the current going through inductor 1 times L1. But since the current going through L1 is the same as the current supplied because here they are in series, which is also the same as the current going through L2, so i1 is equal to i2 and are equal to i. And I can get V2, which is the voltage across inductor 2, as L2 di/dt. Now if you look at this right hand side, it can be written as (L1 + L2)*di/dt. So in order for this equation to match, I should have L_T equal to L1 + L2. So I could replace these two series inductors with one equivalent inductor whose value is L1+L2, and I get the same voltage for the supply. So in general, if I have series inductors, I can reduce them to one equivalent inductor whose inductance is equal to summation of the individual inductances.

Now let's go for a parallel combination. So let's assume that I have a current source here feeding a parallel network of two inductors. I'd like to replace these two parallel inductors with one equivalent inductor, let's call it L_T. Okay, note that here all elements are in parallel, so they all must have the same voltage drop V. For the circuit, we can write down an expression for i_T in terms of the equivalent inductance because i_T here is the same as the current going through that inductor, so it would equal to 1/L_T integral of V dt, where V here is the voltage across the inductor, which is the same as the voltage across the current source because they are in parallel. Okay, so now I manage to write down the supply current in terms of the equivalent inductance and the voltage across the supply current. Now let's take this result and apply to the circuit on the right. We do circuit analysis and see how we can get an expression for L_T.

So here I have i_T, and we have this node here: the current in is i_T, the current out is i1 and i2, so we can write down this KCL equation. Now I can replace i_T with the equation that we derived using the equivalent inductor from the previous slide. So here I have i_T, and again remember that they are all in parallel, so they all have the same voltage drop. Okay, now I can use the inductor physics to replace i1. So i1 would be 1/L1 integral of the voltage across L1, which is the same voltage across L2, which is the same as the voltage across the current source. Same for i2 equal to 1/L2 integral of the voltage across the second inductor. Now the right hand side of this equation can be written as (1/L1 + 1/L2) times the integral of V dt. So now this is the right hand side, this is the left hand side. For both sides to match, we must have 1/L_T = 1/L1 + 1/L2, which is equivalently give you L_T = L1*L2/(L1 + L2), which is the same combination as if we have parallel resistors: we multiply them together and we divide over the summation. Okay, so in general, if I have more than two inductors connected in parallel, I can replace them with one equivalent inductor such that 1/L_T = 1/L1 + 1/L2 + ... + 1/L_n.

So let's practice that using the simple problem. So here I have two nodes A and B, and I'd like to get the equivalent inductance between the two nodes. We do circuit reduction. So first of all, here I have a series combination, so let's replace that by L_S. So now I have A, I have B, and here I have L1. I replace the series combination between L2 and L3 with one equivalent inductance, let's call it L_S. Okay, so since L2 and L3 are in series, then L_S would be simply their summation, so we get 40 mH. Now we have L1 in parallel to L_S, I can combine them using a parallel equation. So for two parallel inductors, the equivalent inductance would be the multiplication divided over the summation, so this gives us 24 mH.

So in conclusion, we learned in this video about series and parallel connections of inductors, and we learn how to get the equivalent inductance. So for series connection, the equivalent inductance is simply the summation of the individual inductances. For parallel combination, one over the equivalent inductance is equal to the summation of the inverse of all inductances. Specifically, if we have two inductors in parallel, the equivalent inductor would equal to L1*L2/(L1 + L2). Note here that the way we combine the inductance is similar to the way that we combine resistors: if they are in series, we add them together; if they are in parallel, we use this inverse equation. Thank you so much for listening.