# RC circuit and discharging a capacitor

1. Apr 1, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data

RC circuits and discharging a capacitor

ok, say I have the following:

R = 4700 ohm
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?

2. Relevant equations

3. 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: Apr 1, 2009
2. Apr 1, 2009

### aniketp

Are the R and C in series or parallel?

3. Apr 1, 2009

### -EquinoX-

oops forgot to mention that, it's in series

4. Apr 1, 2009

### Kruum

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.

5. Apr 1, 2009

### -EquinoX-

no.. I am not familiar with Laplace transformation

6. Apr 2, 2009

anyone?

7. Apr 2, 2009

### rl.bhat

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.

8. Apr 2, 2009

### -EquinoX-

so I want to get this numerically... is there a way to compute it

9. Apr 2, 2009

### rl.bhat

If you know the peak voltage and the period of the square wave, it is possible to find V(t).

10. Apr 2, 2009

### -EquinoX-

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. Apr 2, 2009

### rl.bhat

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. Apr 2, 2009

### -EquinoX-

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. Apr 2, 2009

### rl.bhat

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. Apr 2, 2009

### -EquinoX-

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. Apr 2, 2009

### rl.bhat

Yes. It is there. It can be varied, say from 0 to 30V.

16. Apr 2, 2009

### -EquinoX-

What does the amplitude take account for?

17. Apr 2, 2009

### rl.bhat

Which amplitude? Amplitude of square wave it self is the Vo.

18. Apr 3, 2009

### -EquinoX-

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. Apr 3, 2009

### rl.bhat

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. Apr 3, 2009

### -EquinoX-

what is S there? 2 Hz.. isn't that really2 small? What should I set the time/div to be