rates of change (1 Viewer)

yellowhighlighterr · Jul 22, 2023

part b
maybe its the late night
maybe im just losing brain cells

jimmysmith560 · Jul 23, 2023

Would the following working help?

$I=\frac{V}{R}+Ae^{-\frac{R}{L}t} \\ \\ I=0, t=0 \\ \\ 0=\frac{V}{R}+Ae^{-\frac{R}{L}t\times 0} \rightarrow \frac{V}{R}= -A\rightarrow A=-\frac{V}{R}$

yolo tengo · Jul 23, 2023

jimmysmith560 said:
Would the following working help?

$I=\frac{V}{R}+Ae^{-\frac{R}{L}t} \\ \\ I=0, t=0 \\ \\ 0=\frac{V}{R}+Ae^{-\frac{R}{L}t\times 0} \rightarrow \frac{V}{R}= -A\rightarrow A=-\frac{V}{R}$

jimmy where r u getting these worked out solutions

yellowhighlighterr · Jul 23, 2023

yolo tengo said:
jimmy where r u getting these worked out solutions

my question exactly bruh-

rates of change (1 Viewer)

yellowhighlighterr

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jimmysmith560

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yolo tengo

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yellowhighlighterr

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