Answer:
(a) The resistance R of the inductor is 2480.62 Ω
(b) The inductance L of the inductor is 1.67 H
Explanation:
Given;
emf of the battery, V = 16.0 V
current at 0.940 ms = 4.86 mA
after a long time, the current becomes 6.45 mA = maximum current
Part (a) The resistance R of the inductor
[tex]R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms[/tex]
Part (b) the inductance L of the inductor
[tex]\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}[/tex]
where;
L is the inductance
R is the resistance of the inductor
t is time
[tex]L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H[/tex]
Therefore, the inductance is 1.67 H
