Capacitor Charge Drain and Surge Impedance

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When a capacitor charged at 12 volts is connected to a cable with a characteristic impedance of 84 ohms and a significantly higher DC resistance of 840 ohms, it can draw a current of approximately 1/7th of an amp for a very short duration, such as 1 nanosecond. This current draw indicates that charge is indeed removed from the capacitor at that rate. The propagation speed of the signal is about 20 cm/ns, meaning that only the DC resistance of the first 10 cm of the cable affects the current flow. Since this resistance is likely much lower than 84 ohms, it plays a crucial role in determining the current drawn from the capacitor. Therefore, the characteristics of the cable significantly influence the behavior of the capacitor in this scenario.
kmarinas86
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If I have a cable of 84 ohms of characteristic (or surge) impedance (which is independent of length) and whose DC resistance is far greater than 84 ohms (let's say 840 ohms), would I get, by hooking up a capacitor charged at 12 volts across it for a very short period of time (let's say 1 nanosecond, so not enough time for the capacitor to lose any significant charge nor any significant voltage), draw 1/7th of an amp from that capacitor for that period of time? Does this really mean that charge is removed from the capacitor at the rate of 1/7th of an amp?
 
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Yes.
 
Signal propagation speed is approx 2*10^5 km/s, or 20 cm/ns. For the DC resistance in a section of the cable to have any influence on the current from the capacitor, the signal must propagate to that section, and back.

This means that only the DC resistance of the first 10 cm of the cable is important, and it is probably much smaller than 84 Ohm.
 
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