Capacitor Charge Drain and Surge Impedance

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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on Phys.org
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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