L=L1 + L2 + 2M
AND
L=(L1L2)/(L1+L2)
Prove both the equations.......
Equations are from chapter ELECTROMAGNETIC INDUCTION....
Answers
We know that, in a parallel network the voltage remains constant and the current divides at each parallel inductor. If IL1, IL2, IL3 and so on ILn are the individual currents flowing in the parallel connected inductors L1, L2 and so on Ln, respectively, then the total current in the parallel inductors is given by
ITotal = IL1 + IL2 + IL3 . . . . + In
If the individual voltage drops in the parallel connection are VL1, VL2, VL3 and so on VLn, then the total voltage drop between the two terminals VT is
VTotal = VL1 = VL2 = VL3 . . . . = Vn
The voltage drop in terms of self inductance can be expressed as V = L di/ dt. This implies total voltage drop,
VT = LT di/dt
⇒ LT d/dt (IL1 + IL2 + IL3 . . . . + In)
⇒ LT ( (di1)/dt + (di2)/dt + (di3)/dt . . . .)
Substituting V / L in place of di/dt, the above equation becomes
VT = LT (V/L1+ V/L2 + V/L3 . . . .)
As the voltage drop is constant across the circuit, then v = VT. So we can write
1/LT = 1/L1 + 1/L2 + 1/L3 . . . . .
This means that the reciprocal of total inductance of the parallel connection is the sum of reciprocals of individual inductances of all inductors. The above equation is true when there is no mutual inductance is affect between the parallel connected coils.
For avoiding complexity in dealing with fractions, we can use product over sum method to calculate the total inductance. If two inductors are connected in parallel, and if there is no mutual inductance between them, then the total inductance is given as
LT = (L1× L2)/(L1+ L2)
We know that, in a parallel network the voltage remains constant and the current divides at each parallel inductor. If IL1, IL2, IL3 and so on ILn are the individual currents flowing in the parallel connected inductors L1, L2 and so on Ln, respectively, then the total current in the parallel inductors is given by
ITotal = IL1 + IL2 + IL3 . . . . + In
If the individual voltage drops in the parallel connection are VL1, VL2, VL3 and so on VLn, then the total voltage drop between the two terminals VT is
VTotal = VL1 = VL2 = VL3 . . . . = Vn
The voltage drop in terms of self inductance can be expressed as V = L di/ dt. This implies total voltage drop,
VT = LT di/dt
⇒ LT d/dt (IL1 + IL2 + IL3 . . . . + In)
⇒ LT ( (di1)/dt + (di2)/dt + (di3)/dt . . . .)
Substituting V / L in place of di/dt, the above equation becomes
VT = LT (V/L1+ V/L2 + V/L3 . . . .)
As the voltage drop is constant across the circuit, then v = VT. So we can write
1/LT = 1/L1 + 1/L2 + 1/L3 . . . . .
This means that the reciprocal of total inductance of the parallel connection is the sum of reciprocals of individual inductances of all inductors. The above equation is true when there is no mutual inductance is affect between the parallel connected coils.
For avoiding complexity in dealing with fractions, we can use product over sum method to calculate the total inductance. If two inductors are connected in parallel, and if there is no mutual inductance between them, then the total inductance is given as
LT = (L1× L2)/(L1+ L2)
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