Timeline for Voltage - Energy drop
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2016 at 5:12 | answer | added | Patel Rishi | timeline score: 0 | |
| Aug 7, 2015 at 2:49 | answer | added | Alfred Centauri | timeline score: 1 | |
| Aug 7, 2015 at 2:21 | answer | added | Bill N | timeline score: 0 | |
| Aug 6, 2015 at 22:31 | answer | added | codex1962 | timeline score: 0 | |
| Aug 6, 2015 at 22:26 | comment | added | Bill N | Your first energy calculation needs clarification. Either use parentheses or MathJax/LaTeX notation. What quantities are you dividing? Why don't you multiply $q\Delta V$? q = $1.602\times10^{-19} C$ and $\Delta V$ = 5 V. | |
| Aug 6, 2015 at 22:18 | answer | added | user87299 | timeline score: 0 | |
| Aug 6, 2015 at 21:17 | comment | added | docscience | Taking the resistance to an extreme (infinite resistance) there are no electrons moving (through the wires in the loop) but you still have the same drop, the same potential, the same voltage (of course assuming an ideal battery) | |
| Aug 6, 2015 at 21:15 | comment | added | docscience | Less total resistance in circuit 2, therefore less current going through both resistors. Therefore less energy dissipation in either resistor in circuit 2. | |
| Aug 6, 2015 at 21:12 | comment | added | pastillman | woops - I initially had a 9v battery :) (corrected) | |
| Aug 6, 2015 at 21:11 | history | edited | pastillman | CC BY-SA 3.0 |
deleted 2 characters in body
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| Aug 6, 2015 at 21:08 | comment | added | DJohnM | Any reason the 5 V resulted in 9 Joules of energy? | |
| Aug 6, 2015 at 21:07 | review | First posts | |||
| Aug 6, 2015 at 21:18 | |||||
| Aug 6, 2015 at 21:02 | history | asked | pastillman | CC BY-SA 3.0 |