Question

A device is turned on and 3.20 A flows through it 0.130 ms later. What is the self-inductance of the device (in mH) if an induced 140 V emf opposes this

Answer 1

the self-inductance of the device is 5.69 mH

Step-by-step 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;

`E = L(\Delta I)/(\Delta t) \\\\where;\\\\L \ is \ the \ self-inductance\\\\\L = (E\Delta t)/(\Delta I) \\\\L =(140 * 0.13 * 10^(-3))/(3.2) \\\\L = 0.00569 \ H\\\\L = 5.69 \ mH`

Therefore, the self-inductance of the device is 5.69 mH