Timeline for Calculating pi network impedance
Current License: CC BY-SA 3.0
14 events
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| Jun 11, 2020 at 15:10 | history | edited | CommunityBot |
Commonmark migration
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| Oct 11, 2017 at 23:37 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 2, 2017 at 6:14 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 1, 2017 at 5:38 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 24, 2017 at 23:12 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 23, 2017 at 20:42 | comment | added | user17592 | @Chu would you perhaps like to write an answer for that comment? Then I can accept it. The other answers may be correct (I cannot check), but weren't really tailored for my needs. | |
| May 23, 2017 at 18:07 | answer | added | Verbal Kint | timeline score: 1 | |
| May 23, 2017 at 15:17 | comment | added | Chu | Slight typo in first comment... I don't think a radio amateur exam will be looking for complex arithmetic. The reactance of C1 is \$\small 93\Omega\$ so the total impedance must be less than this (L/C2/R are in parallel with C1, so this lowers the overall impedance). Hence 1K and 10k are too high. L has a reactance of \$\small 58\Omega\$, and this will add to the series impedance of R/C2. Hence \$\small 93\Omega\$ in parallel with something greater the \$\small 58\Omega\$ will give about \$\small 50\Omega\$. 73! | |
| May 23, 2017 at 15:16 | comment | added | user17592 | @Chu that makes sense, and indeed, the current answers seem more complicated than what I saw in the material so far. But, the answer model says it should be 1kΩ... I guess the correction model is wrong. Thanks & 73! | |
| May 23, 2017 at 15:07 | comment | added | Chu | I don't think a radio amateur exam will be looking for complex arithmetic. The reactance of C1 is \$\small 93\Omega\$ so the total impedance must be less than this (L/C2/R are in parallel with C1, so this lowers the overall impedance). Hence 1K and 10k are too high. L has a reactance of \$\small 80\Omega\$, and this will add to the series impedance of R/C2. Hence \$\small 93\Omega\$ in parallel with something greater the \$\small 80\Omega\$ will give about \$\small 50\Omega\$. 73! | |
| May 23, 2017 at 14:46 | answer | added | DWD | timeline score: 0 | |
| May 23, 2017 at 14:29 | comment | added | user17592 | @glen_geek OK, but how? Could you give me an example? | |
| May 23, 2017 at 14:23 | comment | added | glen_geek | Try using complex math rather than using simple magnitudes. | |
| May 23, 2017 at 14:17 | history | asked | user17592 | CC BY-SA 3.0 |