kf53250 kf53250

17-12-2021
Engineering

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obinnafavour66 obinnafavour66

17-12-2021

Explanation:

Energy stored in an inductor = ½ × L × I²

L is inductance and I is current

I= 3A

L = 500mH = 500 × 10^-3H

=0.5H

Energy= ½ × 0.5 × 3²

= 9/4

= 2.25joules

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