Question

. Consider arrangements of n inductors, inductances L1, L2, ... Ln. Assuming that all inductors are far enough separated that thee mutual inductance between any pair of the inductors can be taken to be zero, find the equivalent inductance Leq of the inductor arrangement when the inductors are(a) all in series and (b) all in parallel. (c) Now consider just two inductors in series, with inductances L1 and L2, but now they are close together on a single circuit board, so their mutual inductance is M. In this case, what is the equivalent inductance Leq of the two inductors together? (For the parallel case, see problem 83 in the textbook.)

Answer 1

Given

the inductors are in parallel combination

let L1,L2 L3 ,...... Ln are the inductances of each inductor connected in parallel  

the potential is same across each conductor

we know that V = L*di/dt

VT = L_t (di1/dt +di2/dt+di3/dt+ .............+di.n/dt)

VT = L_t (v1/L1 +v2/L2+v3/L3+............+vn/Ln)

VT = V1 = V2 = V3 =............. Vn = V

1/L_t = 1/L1+1/L2+1/L3+ ---------+1/Ln

and in series the current through ecah inductor is same  

V = L di/dt

VT = L_T di/dt

VT = V1+V2+V3+..............Vn

LT= di/dt(L1+L2+L3+............+Ln)

if two inductors are in parallel combination  

1/LT = 1/L1+1/L2

LT = L2*L1/(L1+L2)

if two inductors are in series combination  

LT = L1+L2

LT = (L1+L2)