0
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Why do you think that the fact that the LED flashed means that it is not necessarily 18 kV? $\endgroup$ Mar 13, 2014 at 19:12

1 Answer 1

1
$\begingroup$

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:

enter image description here

Which means, that the current would behave like:

enter image description here

The current behaves like:

enter image description here

The current does not equal the charge on the plate but they are dependent on each other by the following equation:

enter image description here

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.

$\endgroup$
2
  • $\begingroup$ But what about voltage? If the discharge wasn't from a capacitor , in a hypothetical case, lightning , will we treat it the same way? $\endgroup$
    – user28324
    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. $\endgroup$
    – counter
    Mar 14, 2014 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.