Timeline for Voltage in a shortcircuit connected in parallel to a voltage source and two resistors [closed]
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 30, 2019 at 11:52 | vote | accept | Álvaro Franz | ||
| Dec 26, 2019 at 12:45 | history | closed |
BioPhysicist GiorgioP-DoomsdayClockIsAt-90 ZeroTheHero Kyle Kanos John Rennie |
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| Dec 26, 2019 at 10:06 | history | edited | Álvaro Franz | CC BY-SA 4.0 |
added 1520 characters in body; edited title
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| Dec 25, 2019 at 18:51 | vote | accept | Álvaro Franz | ||
| Dec 26, 2019 at 10:06 | |||||
| Dec 25, 2019 at 18:37 | answer | added | Ofek Gillon | timeline score: 1 | |
| Dec 25, 2019 at 18:08 | answer | added | uriyabsc | timeline score: 1 | |
| Dec 25, 2019 at 17:47 | comment | added | Álvaro Franz | @OfekGillon Thanks for your kind answer. I cannot understand the fact that the voltage difference between the "right side" will be 0. Would that be true if there was only the loop without the right part of the circuit? I would highly appreciate it if you describe what is going on in that loop. | |
| Dec 25, 2019 at 17:37 | comment | added | Ofek Gillon | There will be current in the loop, but it won't leak out of it. The voltage difference between the "left side" of the resistors is $10V$ but the voltage difference between the "right side" will be $0$, meaning it will behave like a short circuit for your problem | |
| Dec 25, 2019 at 17:28 | history | edited | Álvaro Franz | CC BY-SA 4.0 |
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| Dec 25, 2019 at 14:30 | comment | added | Álvaro Franz | @AlfredCentauri This is a very good approach. It helps me get that the answer is $U_A = -10 V$. But still, I wonder what happens in the loop. A teacher said that the cable with nothing in it, acts as a shortcircuit, so I can just imagine it's not there. That would be very clear if there wasn't a voltage source in the loop. But the fact that there is a voltage source in the loop makes it hard to believe that there won't be any current and voltage on that side of the circuit. | |
| Dec 25, 2019 at 11:58 | comment | added | Alfred Centauri | Alvaro, have you tried adding a resistor of resistance $R$ from $A$ to ground, solving for the voltage across it in terms of $R$, and then taking the limit as $R\rightarrow\infty$? Note: the ground symbol here only means (in this case) that this is the zero volt reference node, i.e., imagine placing the black lead of your voltmeter there and leaving it. | |
| Dec 25, 2019 at 11:35 | review | Close votes | |||
| Dec 26, 2019 at 12:45 | |||||
| Dec 25, 2019 at 11:07 | history | edited | Álvaro Franz | CC BY-SA 4.0 |
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| Dec 25, 2019 at 11:01 | history | asked | Álvaro Franz | CC BY-SA 4.0 |