Timeline for voltage divider with 2 power sources, Vout?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2018 at 23:34 | comment | added | Trevor_G | @DontAskTheEye it's all the same just written different ways. As long as you have your solution, all is well. | |
| Jan 14, 2018 at 22:25 | comment | added | DontAskTheEye | I have accepted somebody's else response and i just noticed @G36 comment right here. Thus, if i combine G36's comment and my 'accepted answer' i found a 'good' detailed solution: Vout - I* R2=Vb => Vout = I* R2 + Vb. Where I = (Va-Vb)/(R1+R2) | |
| Jan 14, 2018 at 22:03 | comment | added | DontAskTheEye | i would accept your response but you magically popped out vout equation without showing how to do it. Question A) was my most important of all. Anyway, overly it is an almost comprehensive and good answer, thanks.. | |
| Jan 14, 2018 at 21:59 | vote | accept | DontAskTheEye | ||
| Jan 14, 2018 at 21:59 | |||||
| Jan 14, 2018 at 16:24 | comment | added | Trevor_G | @G36 indeed.. though let's not confuse the OP more ;D | |
| Jan 14, 2018 at 16:13 | comment | added | G36 | Or Vout = Va - I*R1 = Va - (Va - Vb)*R1/(R1+R2) = 5V - (5V - 3.3V)*200/(200+667) = 4.60784314V | |
| Jan 14, 2018 at 16:06 | history | edited | Trevor_G | CC BY-SA 3.0 |
added 113 characters in body
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| Jan 14, 2018 at 15:58 | history | answered | Trevor_G | CC BY-SA 3.0 |