Timeline for When use Voltage divider in circuit analysis
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 25, 2016 at 15:22 | history | tweeted | twitter.com/StackElectronix/status/813042204893704196 | ||
| Nov 28, 2016 at 15:20 | answer | added | StartJ | timeline score: 1 | |
| Apr 30, 2013 at 17:51 | vote | accept | Cesar | ||
| Apr 27, 2013 at 16:47 | comment | added | Kaz | This is not a voltage divider question. Forget about voltage dividers. You need to know how to reduce networks of resistors and voltage sources to equivalent circuits with just one voltage source in series with a resistor: Thevenin's theorem. A voltage divider is just a special case that comes up often. | |
| Apr 27, 2013 at 6:50 | comment | added | jippie | In your top circuit with the given values R4 does nothing, hence you can remove it from the equation. The potential at [4] and [6] is the same because \$\frac{R2}{R5} = \frac{R3}{R6}\$ | |
| Apr 27, 2013 at 5:32 | answer | added | Sebastian Valencia | timeline score: 1 | |
| Apr 27, 2013 at 4:41 | comment | added | placeholder | For reference the \$V_{34}=2.5V\$ this can be obtained (as Oli hinted) purely by inspection and with almost no math. | |
| Apr 27, 2013 at 1:38 | answer | added | Kurt E. Clothier | timeline score: 2 | |
| Apr 27, 2013 at 0:38 | answer | added | Oli Glaser | timeline score: 3 | |
| Apr 27, 2013 at 0:35 | review | Close votes | |||
| Apr 27, 2013 at 3:13 | |||||
| Apr 27, 2013 at 0:15 | review | First posts | |||
| Apr 27, 2013 at 0:22 | |||||
| Apr 26, 2013 at 23:56 | history | asked | Cesar | CC BY-SA 3.0 |