Now, let us derive an expression for the equivalent inductance of two inductors connected in parallel. Suppose we have two inductors like this connected in parallel between the points, let us say A and B. Let this be the point A and this one is the point B. And between A and B we have provided let us say an em of E total and through the point A I have passed a current I, since A is a junction the current at A will split to I1 and I2, clear. I1 flow through the first inductor and I2 flows through the second inductor.
They will again meet here and flow as I, that is come out of B as I itself. Now let us say L1 be the self inductance of the first inductor and L2 be the self inductance of the second inductor. Now since these two coils are. These two inductors are close to each other, the system has an induct, mutual inductance of let us say m.
Clearly in both of these coils current is flowing from left to right, clear. And one more point we know that since this is a parallel connection, for a parallel connection EMF remains the same and if I, let us now call that as A1. E2.
We know that E total is equal to E1 equal to E2, since for a parallel connection the EMF remains the same. Now, when you consider the current at the junction A, you know the total current I is equal to I1 plus I2, clear. Differentiating with respect to T, you know di by dt is equal to di1 dt plus di2 dt, clear, call that as equation number 1. Now let us consider the emf e1. You know emf will be emf across the inductor. First inductor will be due to two reasons, one due to self induction that is self induction will be L1 times di1 by dt plus the emf due to mutual induction that is m times since the flux due to mutual induction will be the presence of current in the second coil.
So Di2 by dt. We already discussed about this earlier in the series combination of inductors. Clear.
So, even this is the emf due to self-induction and this one is the emf due to mutual induction. And clearly for mutual induction, the reason for the mutual induction is the current due to second coil. So, here you have to take Di2 by dt. Now, the EMF across the second coil let us call that as let us call that as E2 call this equation as equation number 2. You know E2, E2 for the EMF E2 will also be due to two reasons one the self induction of the coil itself that is L2 times di2 by dt since the current flowing through the second coil is I2 plus. Mutual induction times dI1 by dt, clear, similar to this one.
Now, we have the EMF across these two inductors remains the same for parallel connection. So, E1 equal to E2, which means equating these 2 and 3, that is 2 equal to 3, you will get L1 dI1 by dt. plus m di2 by dt equals L2 di2 by dt. plus m di1 by dt.
Now, taking the di1 by dt terms to the one side, to one side you will get, let me erase this, that is taking this m di1 to the left, you will get L1 di1 by dt minus m di1. by dt is equal to take this term to the right. So, L2 di2 by dt minus m di2 by dt. That is taking di1 by dt common di1 by dt times L1 minus m. equal to di2 by dt here di2 by dt is common times L2 minus m.
Therefore, di1 by dt will be you can write this as first write the L2 minus m term first L2 minus m. divided by L1 minus m times dI2 by dt, clear. Call this as equation number 4. Now, we know that dI1 by dt will be Di1 by dt plus Di2 by dt, clear?
Implies, you know, Di by dt is equal to, let me write this one equation for instead of Di1 by dt. So, this will be L2 minus m by L1 minus m. m dI2 by dt plus this term you write that as such, write that as dI1 by dt itself. So, plus dI2 divided by dt. Instead of dI1 by dt, I just put the equation 4 and dI2 by dt, that one itself.
So, this in here the di2 by dt is common. So, you can simplify this as di by dt will be taking this L di2 by dt common you will get L2 minus m divided by. L1 minus m plus 1 times dI2 divided by dt, call this as equation number let us say 5. Now, you know for a for the parallel connection the total emf, e total be equal to E2, clear or E1 equal to E2 equal to E total.
Now let us consider E total equal to E2. So total EMF will be you know total EMF is equal to equivalent EMF L equivalent into di by dt where i is the total current. E2 you know that we already have written that. You can check the equation 3 that is L2 di2 by dt plus m di1 by dt. At the start of this derivation we have written this equation P2.
Now let us put L equal to n instead of di by dt you can write this one. that is L2 minus M by L1 minus m plus 1 is equal to L2 into, sorry, L2 plus m times di2 by dt is equal to L2 di2 by dt plus m into, you know. Di1 by dt is equal to L1, L2 minus m by L1 minus m Di1, sorry Di2 by dt. We already derived this equation before at the start of the equation, clear. Now, here the Di2 by dt term is common.
So, you can write this as L equivalent, you can cross multiply this that is L2 minus m plus L1 minus m divided by L1 minus m, L1 minus m times di2 by dt is equal to L2 plus m times L2 minus m divided by L1 minus m in whole brackets dI2 divided by dt. Clear, dI2 by dt is common. You can cancel this one and this one or upon simplifying this you will get Let me erase this. You know l equivalent times l you just write l1 plus l2 minus m minus m will be minus 2m divided by l1 minus m in equal to you can take this to the left. That is L2 into L1 minus m plus m into L2 minus m whole divided by L1 minus m.
So, L equivalent is equal to L1 plus L2 minus 2m. divided by L1 minus m equal to you can write this as L1 L2 minus L2m plus mL2 minus m square whole divided by L1 minus m, clear. You can cancel this L1 minus m and L1 minus m also this L2m and mL2 both gets get cancelled, clear.
So, L equivalent will be, equivalent inductance will be, you can take this numerator to the right side. So, L1 L2 minus m square divided by L1 plus L2 minus 2m. So, this is the expression. Clear.