Timeline for Voltage in a shortcircuit connected in parallel to a voltage source and two resistors
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2019 at 10:07 | comment | added | Álvaro Franz | Dear Ofek, I unmarked the green tick because I am still not happy with my understanding of this case. It does not mean at all that I don't appreciate your help. | |
| Dec 25, 2019 at 18:55 | comment | added | Ofek Gillon | Thanks! You too! | |
| Dec 25, 2019 at 18:54 | comment | added | Álvaro Franz | There should be some sort of mix-accepted-answer because really... it happens a lot, that a few answers together are the most complete answer. Anyways... merry christmas. | |
| Dec 25, 2019 at 18:53 | comment | added | Ofek Gillon | You're welcome, but I should thank uriyaba for his explanation also because he made most of the effort :) | |
| Dec 25, 2019 at 18:52 | comment | added | Álvaro Franz | A million thanks. | |
| Dec 25, 2019 at 18:51 | comment | added | Ofek Gillon | I used the fact that the current is the same everywhere using the current law (in every node the current doesn't go elsewhere) | |
| Dec 25, 2019 at 18:51 | vote | accept | Álvaro Franz | ||
| Dec 26, 2019 at 10:06 | |||||
| Dec 25, 2019 at 18:50 | comment | added | Ofek Gillon | The voltage across the loop will be $$10V - 13\Omega \cdot I - 7 \Omega \cdot I = 0 \Rightarrow 10 V = 20 \Omega \cdot I $$ $$ I = \frac{10 V}{20\Omega} = 0.5 \frac{V}{\Omega} = 0.5 A$$ | |
| Dec 25, 2019 at 18:49 | comment | added | Álvaro Franz | How does the Kirchhoff's law result in I being half an Ampere? (Sorry I am not making fun or anything, I really don't see it) I know Kirchhoff's law 1 and 2 but I do not see them applying here and resulting in half an Ampere current. | |
| Dec 25, 2019 at 18:47 | comment | added | Ofek Gillon | I just applied Kirchhoff's law for the right circuit and got that the current needs to be half an ampere. So by defining the down-right corner to be "0 volts" I just continued using Ohm's law to figure out the voltage in each place in the circuit. | |
| Dec 25, 2019 at 18:45 | comment | added | Álvaro Franz | For each resistor you drew a voltage of half it's resistance value because I = 0,5 A. Could you please explain where those values come from? Those number make it be this way but why is it not I = 0,2 and then there is some voltage in the blank cable too? Thanks a lot for taking your time. | |
| Dec 25, 2019 at 18:37 | history | answered | Ofek Gillon | CC BY-SA 4.0 |