Answer:
the self-inductance of the device is 5.69 mH
Explanation:
Given;
change in current, ΔI = 3.2 A
change in time, Δt = 0.13 ms = 0.13 x 10⁻³ s
induced emf, E = 140
The self-inductance is calculated as;
[tex]E = L\frac{\Delta I}{\Delta t} \\\\where;\\\\L \ is \ the \ self-inductance\\\\\L = \frac{E\Delta t}{\Delta I} \\\\L =\frac{140 \times 0.13 \times 10^{-3}}{3.2} \\\\L = 0.00569 \ H\\\\L = 5.69 \ mH[/tex]
Therefore, the self-inductance of the device is 5.69 mH
