I conducted an experiment , i put a silver foil on a a CRT TV , then i open the TV, charging the foil (acting like a capacitor plate), if i approach a grounded rod to it , it will discharge with a spark of about 1.8 cm length , which is equivalent to about 18 kV. What i am really interested in is once that spark goes in the rod and becomes current, i wanted to know the voltage of that current, so i asked a question on SE and the answer that it will be 18 kV, but when i connected a LED 3V in the wire connecting the rod to the ground , it worked. Another thing, will the current be simply the charge on the plate , or a normal relation between V and R?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Why do you think that the fact that the LED flashed means that it is not necessarily 18 kV? $\endgroup$– DumpsterDoofusCommented Mar 13, 2014 at 19:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
I cannot tell you if 18kV is correct but yes - it must be a high voltage to create such a spark. Addressing the current: It is not simply a U=RI behavior since you have a capacity in your circuit which means, that the current get less over time. This is what the circuit looks like:
Which means, that the current would behave like:
The current behaves like:
The current does not equal the charge on the plate but they are dependent on each other by the following equation:
Keep in mind that if the voltage gets too low for a spark on a certain distance the discharge process stops.
I hoped that this answer helped you and contains the information you liked to know.
-
$\begingroup$ But what about voltage? If the discharge wasn't from a capacitor , in a hypothetical case, lightning , will we treat it the same way? $\endgroup$ Commented Mar 14, 2014 at 5:17
-
$\begingroup$ In this case the voltage of the discharge equals the voltage of the capacitor.If the discharge doesn't come from a capacitor you need to know how it behaves. You can e.g. generate a discharge using a high voltage source - in that case the voltage would remain mostly constant. I assume that a natural lightning behaves somehow similar to a capacitor since it actually is a capacitive effect. $\endgroup$– counterCommented Mar 14, 2014 at 11:27