- #1

- 1

- 0

L=13H

R=160 ohms

I=0.4A

W = 1/2Li^2 = 1/2(13)(0.4)^2 = 1.04 J

I am unsure where to go from here. I can calculate the stored energy, but what do I do for the rate of thermal energy?

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- Thread starter cmalle09
- Start date

- #1

- 1

- 0

L=13H

R=160 ohms

I=0.4A

W = 1/2Li^2 = 1/2(13)(0.4)^2 = 1.04 J

I am unsure where to go from here. I can calculate the stored energy, but what do I do for the rate of thermal energy?

- #2

berkeman

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