Question

An Inductor is made from 20 turns of wire (of negligible resistance), wrapped into a cylinder of diameter 25 mm and length 68 mm, around a material of relative permeability 500. 5 more identical Inductors are then constructed, and all 6 of them are connected together in parallel. This network of Inductors is then connected to a 9.0 V cell, of internal resistance 2.0 Ω. Find the magnitude of the initial rate of change of the current flowing through the cell.

Answer 1

Given,

Number of turns, N = 20

Length of cylinder, l = 68 mm = 68*10^-3 m

Diameter of the cylinder, D = 25 mm

Radius, R = D/2 = 25/2 = 12.5 mm

                = 12.5*10^-3 m

Now,

relative permeability, k = 500

Now,

We know,

Inductance, L = (*N²*A)/l

                      = (500*4*3.14*10^-7*20*20*3.14*12.5*10^-3*12.5*10^-3)/(68*10^-3)

                     = 1.81*10^-3 H

Now,

5 are added in series,

So, total inductance, L' = 5*L = 5*1.81*10^-3

                                      = 9.1*10^-3 H

Now,

Resistance, R = 2 ohms

Voltage, V = 9V

In L-R circuit, we know current is given by,

=> i(t) = (V/R)*{1 - e^-Rt/L}

Now, rate of change of current is

=> d{i(t)}/dt = (V/L)*e^-Rt/L

At t=0,

=> d{i(t)}/dt = V/L = 9/(9.1*10^-3)

                      = 989.01 A/sec