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 induce d 140 V emf opposes this
1 answer:
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;
Therefore, the self-inductance of the device is 5.69 mH
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