Wouldn't it just be RC=ln(V/V0)/tkuruman said:The discharge equation is V = V0e-t/RC from which you get -t/(RC) = ln(V/V0). So what is RC (not R/C) ?
I have no idea. I've never used ln before and I don't know how to rearrange using it. However, what I got this time is RC=t/ln(V/V0)kuruman said:It would not be that. Please try again doing the algebra steps very carefully.
but if the gradient of the graph gives -t/RC would it just be C=t/R (but it gives me the wrong answer)Daniel2244 said:I have no idea. I've never used ln before and I don't know how to rearrange using it. However, what I got this time is RC=t/ln(V/V0)
The gradient of which graph? What did you plot against what?Daniel2244 said:but if the gradient of the graph gives -t/RC would it just be C=t/R (but it gives me the wrong answer)
Ln(V/V0) against timekuruman said:The gradient of which graph? What did you plot against what?
Yes, sorry, I was meant to put -1/RC. The gradient I got was 0.015 so would I then do e0.015 which gives me 1.01 which is 1% dicharge at 10 seconds. but how does this help me find the capacitance?kuruman said:No. The gradient is -1/(RC). There is no t. What gradient do you get from your plot? How will you (did you) use it to find the time constant RC?
No, you would not do that. You missed the point of the exercise. You plotted ln(/V0) against time. That's good. The equation is $$\ln(V/V_0)=-\frac{t}{RC}$$ and the plot gives you a straight line of gradient 0.015. OK so far? Now define ##\ln(V/V_0) = y## and ##\frac{1}{RC}=m##. The equation becomes ##y=-mx.## This is the equation of a straight line of gradient m = 0.015 which, by definition, is also equal to ##1/(RC)##. Given the previous sentence, what is ##RC?##Daniel2244 said:The gradient I got was 0.015 so would I then do e0.015 ...
Ok, so would I do 1/0.015=66.6 then 66.6/99400=6.7069E^-4 (99400 being resistance) then after converting to microF i get 670.69microF.kuruman said:No, you would not do that. You missed the point of the exercise. You plotted ln(/V0) against time. That's good. The equation is $$\ln(V/V_0)=-\frac{t}{RC}$$ and the plot gives you a straight line of gradient 0.015. OK so far? Now define ##\ln(V/V_0) = y## and ##\frac{1}{RC}=m##. The equation becomes ##y=-mx.## This is the equation of a straight line of gradient m = 0.015 which, by definition, is also equal to ##1/(RC)##. Given the previous sentence, what is ##RC?##
Thanks man, your help was much appreciated!kuruman said:Yes, except that I would not carry that many significant figures.
Capacitance is a measure of an object's ability to store electric charge. It is important in the discharge equation because it determines the rate at which a capacitor will discharge.
Capacitance is calculated by dividing the amount of charge on a capacitor by the potential difference between its plates. It is measured in units of Farads (F).
In the discharge equation, capacitance and voltage have an inverse relationship. This means that as capacitance increases, voltage decreases, and vice versa.
The factors that affect capacitance include the distance between the plates, the area of the plates, and the type of material between the plates. These factors impact the discharge equation by altering the amount of charge that can be stored on the plates and how quickly the capacitor will discharge.
The discharge equation is used in practical applications involving capacitors, such as in electronic circuits and power systems. It helps engineers and scientists determine the behavior of capacitors and how they can be used effectively in various devices.