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KVL Exercise - Basic |
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| Feb6-13, 09:30 AM | #1 |
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KVL Exercise - Basic
1. The problem statement, all variables and given/known data
Click on the link https://www.diigo.com/item/image/2sb3i/yjim 2. Relevant equations KVL Law Similar Example : https://www.diigo.com/item/image/2sb3i/orwa 3. The attempt at a solution (a) In the left loop : -8-12+VR2 = 0 VR2 = -20 v (b) In the big loop : -8-12+7-9-V2-3+VR1 = 0 -8-12+7-9-V2-3+1 = 0 V2 = -24 Why is my solution wrong? If I take the right loop for (b) using the right answer for (a) : -4+7-9-V2-3+1 = 0 V2 = -8 which is true. Is the wrong with the negative voltage (-8) ?? thanks. |
| Feb6-13, 09:48 AM | #2 |
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Bottom line: assume that the "-8V" is really "8V" oriented according to the "+ -". |
| Feb6-13, 10:03 AM | #3 |
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thank you
but I see that there is no difference if it was "+8" instead of "-8" ??? |
| Feb6-13, 10:15 AM | #4 |
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KVL Exercise - BasicFor part (a) your equation summing the potential drops would become: +8-12+VR2 = 0 |
| Feb6-13, 10:27 AM | #5 |
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ok, how the solution of (a) will be if :
https://www.diigo.com/item/image/2sb3i/y0us |
| Feb6-13, 10:35 AM | #6 |
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| Feb6-13, 10:44 AM | #7 |
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I'm sorry
but can you tell me the equation of (a) if the potential change was "-8"? I thought that the equation +8-12+VR2 = 0 is used when the potential change = -8. |
| Feb6-13, 11:41 AM | #8 |
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| Feb6-13, 11:44 AM | #9 |
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aha
thank you Mr. gneill |
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