RC Circuit: Effects of Increasing R on E and V

[M. next]

Supposing we have an RC circuit (see figure)
As we increase R, what will happen to the visualized signal of E and V(capacitor) and V(resistor)?


And a general question about measuring the (Tao) [the one that equals to RC] BUT FROM THE OSCILLOSCOPE..

They mentioned it is the time for which V(resistor) decreases to half its initial value (but I still don't get it)

Thanks in advance

 

[truesearch]

Hi M.next... I cannot see your diagram ! (I can picture what you mean but...)
One thing to note... RC is called the time constant (tau) but this is NOT the time taken for the voltage across R to 1/2.
The time for it to 1/2 is 0.693RC (ln2 x RC) so measuring the time to become 1/2 from the oscilloscope (not difficult !) enables you to calculate RC
Hope this helps... your diagram would be useful !

 

[M. next]

I donno why it is not uploading :/!
Concerning the tau, please elaborate supposing you can imagine an oscilloscope signal..

 

[truesearch]

If you have a simple R in series with a C connected to a battery of voltage E...then when the switch is closed the current will be a max (= E/R) and will decrease exponentially. The voltage across R will also decrease exponentially from a value of E.
When the voltage across R has decreased to E/2 then the time taken = 0.693RC.
If you can display the V across R on a CRO you should be able to measure the time for V to become V/2.
I am not certain what you mean by...'as we increase R...!
Hope this helps.
As an example: if you have 1 microfarad in series with 1 megohm then time constant = 1 second so it will take 0.693s for the voltage across R to become 1/2

 

[M. next]

Thank you very much truesearch