Coils in series
When two coils of inductance L1 and L2 are connected in series and the coefficient of coupling K = 0, then as in series current through each coil is same and potential divides, therefore,
E = e1 + e2
L dI / dt = L dI / dt + L2 dI / dt
L = L1 + L2
When two coils of inductance L1 and L2 are connected in parallel and the coefficient of coupling K = 0 then as in parallel potential remains the same and current divides therefore, I = I1 + I2
dI / dt = dI1 / dt + dI2 / dt
∴ e /Lp = e1 / L1 + e2 / L2 ( using e = L dI / dt)
As e1 = e2 = e
∴ 1 / Lp = 1 / L1 + 1 / L2
Grouping of inductances is governed by formulae similar to that of grouping of resistances
R s = R1 + R2
And 1 / Rp = 1 / R1 + 1 / R2
Note an inductance is said to be ideal if it has no resistance in actual practice due to finite resistance of conductors every inductor has some resistance. The reverse may or may not be true we can have a resistance with or without having inductance. A resistance without inductance is called non-inductive resistance. It is obtained by double folding the wire on itself and coiling it the inductive effects cancel out as same current is flowing in two opposite directions. In a resistance box all resistance coils are made non- inductive.
What is the self inductance of a solenoid of length 40 area of cross section 20 and total number of turns = 800?
Here, L = ? = 40 cm = 0.4 m
A = 20 cm = 20 x 10-4 m2
N = 800
L = μ0 N2 A / I = 4n X 10-7 (800)2 X 20 X10-4 / 0.4
L = 4.02 X 10 - 3 H.
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