RC circuit and discharging a capacitor

AI Thread Summary
In an RC circuit with a resistor of 4700 ohms and a capacitor of 39 microfarads arranged in series, the voltage across the capacitor can be graphed using the equation V(t) = Vo * e^(-t/RC). The initial voltage (Vo) is determined by the peak voltage set on the function generator, which can be adjusted, for example, to 20V peak-to-peak. When using a square wave function generator, the capacitor charges during the horizontal portion of the wave and discharges during the downward transition. The time constant (RC) is calculated to be approximately 0.183 seconds, suggesting a frequency setting of around 2 Hz for the function generator to effectively observe the charging and discharging behavior. Understanding these parameters allows for accurate plotting of Vcap versus time on paper.
-EquinoX-
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Homework Statement



RC circuits and discharging a capacitor

ok, say I have the following:

R = 4700 ohm
C = 39 micro farads
they are arranged in series and the battery is replaced using a function generator, which we set to produce a square wave (on/off). With any frequencies set.

How can I graph the Vcap vs time?

Homework Equations


The Attempt at a Solution



I know the equation V(t) = Vo * e^(-t/RC) can be used.. however in this case what is Vo (initial voltage) here?

if this is so then what is the purpose of the square wave function generator? Doesn't it matter on how big the frequencies are set?
 
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Are the R and C in series or parallel? :smile:
 
aniketp said:
Are the R and C in series or parallel? :smile:

oops forgot to mention that, it's in series
 
If you are familiar with Laplace transformation, this problem can simply be solved by making a laplace transformation of the circuit and calculating the necessary value.
 
no.. I am not familiar with Laplace transformation
 
anyone?
 
When you swith on the function generator, a peak voltage will be across RC circuit. During horizontal protion of the square wave, capacity will start charging. During vertically down ward portion and next horizontal portion of the square wave the capacity will discahrge.
 
so I want to get this numerically... is there a way to compute it
 
If you know the peak voltage and the period of the square wave, it is possible to find V(t).
 
  • #10
well the period of the square wave I know, but how do I get the peak voltage?

I am just using a normal function generator.. so maybe the peak voltage is known...
 
  • #11
You can select the peak voltage of the function generator, say 20 V peak-to-peak and period 500 ms. (greater then time constant RC). Find the voltage at every 50 ms. Plot the graph V(t) vs time.
 
  • #12
I don't want to do this experimentally.. I want to do this on paper actually... so if the peak voltage is 20V and period was 500 ms what is the Vo
 
  • #13
How can I graph the Vcap vs time?
To draw graph use V(t) = 20*e^-t/RC formula.
Put the values of t as 50ms. 100ms, ,,and so on. Substitute the values of RC. Find the values of V(t)
 
  • #14
ok.. gotcha.. is there actually a setting in the function generator to set the peak voltage... or is that hard coded in every function generator
 
  • #15
Yes. It is there. It can be varied, say from 0 to 30V.
 
  • #16
What does the amplitude take account for?
 
  • #17
-EquinoX- said:
What does the amplitude take account for?

Which amplitude? Amplitude of square wave it self is the Vo.
 
  • #18
What would be a good frequency to set on the function generator if the amplitude was around 2-5V and the RC is as I specified in my first post... I need some range of the frequency
 
  • #19
In the given circuit the time constant RC = 0.183 S. So to draw the graph, if possible set the function generator to 2 Hz.
 
  • #20
what is S there? 2 Hz.. isn't that really2 small? What should I set the time/div to be
 
  • #21
-EquinoX- said:
what is S there? 2 Hz.. isn't that really2 small? What should I set the time/div to be

Set to 500 ms.
 
  • #22
and what is S?
 
  • #23
RC is the time constant and it is measured in seconds.
 
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