# RC Time Constant and Average Slope

1. Jun 1, 2006

### Trista

In this lab, I'm supposed to find the average slope between a charging and discharging capacitor. For the slope of charging and for the slope discharging I used the basic formula y2-y1/x2-x1 choosing arbritary points on both lines. Charging slope = 4.6/-15, and discharging slope = 3.48/-57. Now how do I go about finding an average slope? Add the two together and divide by 2?

Multiplying RC (my capacitor and resistors are as follows: C1 = 1000 microFarads Resistor 1 = 10 kOhms) together and I get 10. My average slope should equal that, I suppose, or at least close. If I add my charging and discharging slopes together and then divide by 2, I get 2.33, not 10, so I'm thinking that I'm really doing something wrong here.

Thanks for any insight there is out there.

2. Jun 1, 2006

### vsage

Try picking two charging slopes or two discharging slopes. What you're doing doesn't make a whole lot of sense, and what I don't quite get about your data is why both slopes are negative, I'm assuming you have some sort of voltage on the plates vs. time graph that you're plucking values off.

Edit 1: What type of graph are you using anyway to determine your slopes? Is it fairly accurate, as if from an oscilloscope? You may want to extend tangent lines off of present curves to better determine a slope if you feel your graph is particularly accurate.

Edit 2: Anyway, there could be a number of reasons why your data is off, but most likely you chose "bad" points, or points not representative of the average slope (ie points where the slope is lower than average or points where the slope is too above average). Choose points closer to the middle ground of your charging and discharging graphs.

Last edited by a moderator: Jun 1, 2006
3. Jun 2, 2006

### andrevdh

It seems to me that the average slope of the charging and discharging graph should be zero if you started the discharging time measurements from the full power supply voltage $V_o$ (a condition that is difficult to achieve in practice):

Discharge:

$$\frac{dV}{dt}=-\frac{V}{RC}$$

Charge:

$$\frac{dV}{dt}=\frac{V}{RC}$$

but I am not sure why anyone would want to check this.

If the discharge measurements started at some other initial voltage over the capacitor, say $V_a$ then the gradient of the discharge graph would be

Discharge:

$$\frac{dV}{dt}=-\frac{V_a}{V_o}\frac{V}{RC}$$

so that the average gradient will be time dependent. I guess that this type of investigation will thus reveal how well you succeeded in discharging your capacitor from the fully charged, $V_o$, condition. Was this a stipulation in the experimental procedures?

Last edited: Jun 2, 2006