Timeline for Parallel voltage sources (diode on one branch)
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2022 at 19:46 | comment | added | Hearth | The ideal diode and voltage source look like what you're actually working with is a slightly less ideal model of a diode, where the diode is modelled as a voltage source (of somewhere between 0.55 and 0.7 V) when forward-biased and an open circuit when reverse-biased, rather than the ultra-simplified short-circuit/open-circuit model you may have been using previously. | |
| Nov 27, 2022 at 19:27 | answer | added | Circuit fantasist | timeline score: 0 | |
| Nov 27, 2022 at 10:06 | answer | added | SUNITA GUPTA | timeline score: 1 | |
| Nov 27, 2022 at 9:00 | vote | accept | dboko | ||
| Nov 27, 2022 at 9:00 | comment | added | dboko | Got it, thanks. In my head I was imagining that this circuit would be connected to something else. | |
| Nov 27, 2022 at 8:56 | comment | added | Jonathan_the_seagull | @dboko How will current enter from V2? Think of the + and - as the two probes of a multimeter that is used to measure the voltage in the arm containing the diode. | |
| Nov 27, 2022 at 8:52 | comment | added | dboko | when V1 ≤ 0.6 V, I think that current going through the resistor would be equal to the current coming from the plus side of V2...from V2 the voltage would drop by IR, and then by V1 to reach 0. | |
| Nov 27, 2022 at 8:51 | answer | added | Jonathan_the_seagull | timeline score: 3 | |
| Nov 27, 2022 at 8:18 | comment | added | BeB00 | Think about what the voltage drop across the resistor would be: V=IR, what is I in this case? | |
| Nov 27, 2022 at 7:46 | history | edited | ocrdu | CC BY-SA 4.0 |
deleted 44 characters in body; edited tags; edited title
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| S Nov 27, 2022 at 7:37 | review | First questions | |||
| Nov 27, 2022 at 7:46 | |||||
| S Nov 27, 2022 at 7:37 | history | asked | dboko | CC BY-SA 4.0 |