Timeline for instant current in very long transmission line
Current License: CC BY-SA 4.0
10 events
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| Sep 5, 2019 at 17:20 | history | edited | analogsystemsrf | CC BY-SA 4.0 |
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| Sep 5, 2019 at 17:15 | comment | added | analogsystemsrf | you can always model any wire as series inductance and parallel capacitance; yes, the capacitance needs charging but the inductance limits the instantaneous current. Thus Zo (line impedance, sqrt(L / C)) is what the power supply must charge. | |
| Sep 5, 2019 at 10:13 | vote | accept | attle | ||
| Sep 5, 2019 at 10:10 | comment | added | attle | analogsystemsrf-- sorry to add but does this mean if a high voltage( say 1000v) power supply is connected to any transmission line that is impedance matched, it would see an initial very high abit very very short current spike? If it does, does these spike affect power supplies longevity at all? | |
| Sep 5, 2019 at 9:34 | comment | added | analogsystemsrf | cm64 ---- we need to view the wires as combined inductances and capacitances, thus the impedance (of twisted-pair, maybe 100 ohms; or a precise coax of 50 ohms) defines the V/I ratio. | |
| Sep 5, 2019 at 9:32 | comment | added | analogsystemsrf | Jason ---I bet you are correct. Can the voltage become 0.1 amp * 1,000 ohms, which makes V_far_end_at_load be 100 volts? How do we write the boundary conditions? A better query ( to you ) is "What is conserved?" Are the electrons conserved? Thus the line current sets the wave-solution? | |
| Sep 5, 2019 at 9:27 | history | edited | analogsystemsrf | CC BY-SA 4.0 |
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| Sep 5, 2019 at 9:27 | comment | added | cm64 | So based on your logic a simple resistor circuit would get a huge current when turned on because it has a zero like resistor in series(wires) before the resistor and that doesn't happen. | |
| Sep 5, 2019 at 9:26 | comment | added | Jasen Слава Україні | I think you get a higher voltage at the resistor when the reflection starts. | |
| Sep 5, 2019 at 9:22 | history | answered | analogsystemsrf | CC BY-SA 4.0 |