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

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;

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

the assestment will probley apele with cause and effect

connect the batteries in series and hope