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Leno4ka [110]
2 years ago
15

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
Physics
1 answer:
jeka57 [31]2 years ago
8 0

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;

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

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

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