Timeline for Is my circuit solution correct?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2019 at 19:49 | vote | accept | make2r | ||
| Aug 30, 2019 at 8:06 | answer | added | dalex78 | timeline score: 0 | |
| S Aug 29, 2019 at 18:32 | history | suggested | user199402 | CC BY-SA 4.0 |
MathJax formatting. Toned down the bold. Fixed schematic and added values to it.
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| Aug 29, 2019 at 16:33 | review | Suggested edits | |||
| S Aug 29, 2019 at 18:32 | |||||
| Aug 29, 2019 at 14:43 | history | edited | make2r | CC BY-SA 4.0 |
added 1 character in body
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| Aug 29, 2019 at 14:34 | comment | added | make2r | Yeah I get it only thru the parallel part which is then actually a series the other resistor plays no role in tau | |
| Aug 29, 2019 at 14:31 | history | edited | make2r | CC BY-SA 4.0 |
added 16 characters in body
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| Aug 29, 2019 at 14:30 | review | Close votes | |||
| Sep 4, 2019 at 3:05 | |||||
| Aug 29, 2019 at 14:12 | comment | added | Arsenal | Still not quite there I fear. The inductor will only keep the current active which is running though itself not the one through the whole circuit. | |
| Aug 29, 2019 at 14:02 | comment | added | Bart | "tau=L/Rtotal=0.1/26=0.003" Where did you get a 𝜏 of 3 ms? After the switch opens, the inductor "defluxes" over 10 and 15 ohms in series. Which gives 𝜏=100mH/25 Ω = 4ms. Also, 0.1/26 does not equal 0.003. | |
| Aug 29, 2019 at 13:10 | comment | added | Andy aka | Your diagram should indicate which is R1, R2 etc.. Don't expect folk to keep glancing up and down because it drives me mad doing that and you don't want to get me mad. | |
| Aug 29, 2019 at 13:06 | comment | added | make2r | so I=UL/Rtotal -> 1.92=UL/25 -> UL=48V Tau=L/Rtotal=0.1/25=0.004 , iL=1.92*e^(-0.006/0.004)=0.42A , uL=48*e^(-0.006/0.004)=10.71V , energy of the inductor at t=6ms WL=(L*i^2)/2=0.008J | |
| Aug 29, 2019 at 12:59 | comment | added | Arsenal | Yes that is correct. | |
| Aug 29, 2019 at 12:55 | comment | added | make2r | So the current goes thru R1 and R2 only it won go thru R3=20Ω | |
| Aug 29, 2019 at 12:48 | comment | added | Arsenal | I think it's useful to think of an inductor as a form of current source instead of voltage, because "it tries" to keep the current steady. And if you think about the current which is going to flow, which of the three resistors is/are in the current path when the switch opens? | |
| Aug 29, 2019 at 12:30 | history | asked | make2r | CC BY-SA 4.0 |