Series and Parallel Combination of Inductors

 Now we'll talk about inductor combinations in series and parallel, And in series inductors, two or more inductors are linked one after another in the same wire, and because they're connected in the same wire, the current through all of the inductors remains constant. As an example,

Series and Parallel Combination of Inductors

We have two inductors linked one after the other on the same wire in this circuit, hence the current across both inductors will remain constant. Let's imagine the current in this circuit is equal to I, and we know that the voltage across an inductor is equal to the inductor's inductance multiplied by the current rate of change through the inductor, which is defined by DT.

Now we can calculate the voltage across our first inductor, which is equal to the inductance, L sub 1, multiplied by the time rate of change of current through the inductor, di by DT. The voltage across the second inductor will be equal to L 2 multiplied by di by DT. Now we'll use KVL to calculate the equivalent inductance in this loop, and then we'll use KVL to determine the equivalent inductance after that.

So we have plus V minus L 1 di by DT minus l2 di by DT equal to zero, or plus V-helvin di by DT minus l2 di by DT equal to zero. We want to compute the equivalent inductance, thus instead of these two inductors, we should use one equivalent inductor. Let's imagine that in place of these two inductors, we use the equivalent inductor. Let's suppose L sub D Q is the equivalent inductance. When these two inductors are replaced with equivalents,

We'll use a circuit similar to this one, where the inductance is l eq, the voltage across it is V, and the current through the inductor is I. By dt, we can see that the voltage V equals l eq di. As a result, we get l eq di by dt. From these two variables, we may derive di by dt common. As a result, we have L 1 + L 2 multiplied by DT inside the bracket. 

We can see that L equivalent is equal to L 1 + L 2 when we compare the left and right sides, therefore we have our equivalent inductance. It's the sum of the first inductor's inductance and the second inductor's inductance. So, anytime we have inductors linked in series, we simply sum their inductance values to get the equivalent inductance.

When we link n inductors in series, and Alvin is the inductance of the first inductor to the inductance of the second inductor, and Helen is the inductance of the NH inductor, the equivalent inductance is equal to L 1 + L 2 all the way to L and an oven.

all the inductance values are same that is L 1 is equal to L 2 is equal to L 3 all the way to Helen and they are equal to L then the equivalent inductance is equal to n times L and from this result it is clear that when inductors are connected in series the net inductance or the equivalent inductance increases for example let's say there are 3 inductors connected in series and the inductance of each inductor is equal to 2 milli Henry then the equivalent inductance is equal to 3 multiplied to 2 milli Henry this is equal to 6 million REE now compared to 2 milli Henry 6 million reads more and therefore when three inductors are connected in series the inductance increases now we will have discussion on the parallel combination of inductors and for this I have taken this particular network in which 2 inductors are connected such that they have the same potential difference across them and the potential difference is equal to V so V volts is the potential difference or the voltage across the two inductors we are having and if I is the current leaving the positive terminal of these souls then it will get divided at this node and let's say i1 is the current through the first inductor and I to is the current through the second inductor now applying KCl at this node we have high equal to i1 plus i2 

we know for an inductor the current through it is equal to one over L integration V DT now when we focus on the first inductor we can see that current is equal to i1 so i1 is equal to 1 over L 1 Alvin his conductance and the voltage across the inductor is equal to V so we have integration V DT similarly for the second inductor current i2 will be equal to 1 over L 2 integration V DT and when we have the equivalent inductor then current I will be equal to 1 over L equivalent integration V DT now we will put I 1 I 2 an I in this equation and we will have 1 over L equivalent integration VD t equal to 1 over Alvin integration VD t plus 1 over L 2 integration V DT so from here it is clear that 1 over L equivalent 1 over L equivalent is equal to 1 over L 1 plus 1 over L 2 so remember this result and from this we can have he'll equivalent it will be equal 12 1 multiplied 12 2 divided by L 1 plus L 2 so the result we are having for parallel combination of inductors is same as the result

we had in case of parallel combination of resistors and the series combination of capacitor now when we have an inductors connected in parallel then the equivalent inductor can be calculated from one over L equivalent equal to one over L 1 plus 1 over L 2 all the way to 1 over L N and when all the inductances are equal and they're equal to L then we have 1 over L equivalent equal to n times 1 over L or we can say hell equivalent is equal to L over N so from this result it is clear that when inductors are connected in parallel the overall inductance or the net inductance or the equivalent inductance reduces for example let's say 3 inductors are connected in parallel and the value of inductance of each inductor is equal to 2 milli Henry then the equivalent inductance is equal to 2 milli Henry divided by 3 now compared to 2 milli Henry 2 over 3 million reias less and therefore the equivalent inductance reduces in parallel combination