RC circuit and discharging a capacitor

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SUMMARY

The discussion focuses on analyzing an RC circuit with a resistor of 4700 ohms and a capacitor of 39 microfarads, arranged in series, using a function generator to produce a square wave. The equation V(t) = Vo * e^(-t/RC) is used to calculate the voltage across the capacitor over time, where Vo is the peak voltage set on the function generator. Participants clarify that the peak voltage can be adjusted, with a suggested value of 20V and a frequency of 2 Hz, allowing for the calculation of V(t) at intervals of 50 ms to graph Vcap vs time.

PREREQUISITES
  • Understanding of RC circuits and time constants
  • Familiarity with the exponential decay function V(t) = Vo * e^(-t/RC)
  • Basic knowledge of function generators and their settings
  • Ability to perform calculations involving microfarads and ohms
NEXT STEPS
  • Learn how to use a function generator to set peak voltage and frequency
  • Explore Laplace transformations for circuit analysis
  • Study the effects of different frequencies on RC circuit behavior
  • Practice graphing exponential decay functions based on RC time constants
USEFUL FOR

Students studying electrical engineering, hobbyists working with RC circuits, and educators teaching circuit analysis concepts.

-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?
 
Last edited:
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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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