0 \text{ A}$
Final current, $I_2 = 10.0 \text{ A}$
Time interval, $\Delta t = 450 \text{ ms} = 450 \times 10^{-3} \text{ s}$
Step 2:
Calculate the change in current, $\Delta I$:
$\Delta I = I_2 - I_1 = 10.0 \text{ A} - 29.0 \text{ A} = -19.0 \text{ A}$
Step
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