First slide

Inductance

Question

Two pure inductors, each of self-inductance L are connected in parallel but are well separated from each other, then the total inductance is

Easy

A

L
B

2 L
C

L/2
D

L/4

Solution

short cut: $\frac{1}{L_{e f f}} = \frac{1}{L_{1}} + \frac{1}{L_{2}}$
When the coils are connected in parallel, let the currents in the two coils be i₁ and i₂ respectively. Total induced current
$i = i_{1} + i_{2} or \frac{di}{dt} = \frac{{di}_{1}}{dt} + \frac{{di}_{2}}{dt} . . . . . . . . (1) Now e = - L_{1} (\frac{{di}_{1}}{dt}) = - L_{2} (\frac{{di}_{2}}{dt})$
[ $∵$ In parallel, induced e.m.f across each coil will be same)
$Hence \frac{{di}_{1}}{dt} = - \frac{e}{L_{1}} and \frac{{di}_{2}}{dt} = - \frac{e}{L_{2}} . . . . . . . . . (2)$
Let L' be the equivalent inductance. Then
$e = - L^{'} \frac{di}{dt} or \frac{di}{dt} = - \frac{e}{L^{'}} . . . . . . . . (3)$
From eqs. (1), (2) and (3), we get
$- \frac{e}{L^{'}} = - \frac{e}{L_{1}} - \frac{e}{L_{2}} or \frac{1}{L^{'}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} \begin{array}{ll} ∴ L^{'} = \frac{L_{1} I_{2}}{L_{1} + L_{2}} \\ Here L_{1} = L_{2} = L \\ ∴ L^{'} = \frac{L \times L}{L + L} = \frac{L}{2} \end{array}$

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