Capacitor charging graph - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: HP Calculators (and very old HP Computers) ( /forum-3.html)+--- Forum: HP Prime ( /forum-5.html)+--- Thread: Capacitor charging graph ( /thread-2491.html) |

Capacitor charging graph - Alan72 - 11-25-2014 07:23 AM
I'm trying to plot an exponential graph to show what happens when a capacitor is charged in a series resistor capacitor circuit. I can do it on paper but would like to show it on my Prime. Can anyone help? The formula I'm using on paper is Vc=Vs(1-e^(-t/RC)). This is documented throughout the Internet. I'm trying to produce the curved graph produced as the capacitor charges and as a new user to graphing calls, am unsure of where to start. Can anyone give me a friendly nudge? RE: Capacitor charging graph - CR Haeger - 11-25-2014 01:16 PM
(11-25-2014 07:23 AM)Alan72 Wrote: I'm trying to plot an exponential graph to show what happens when a capacitor is charged in a series resistor capacitor circuit. If all you are interested in doing is creating a graph based on values of Vs, R, C and t (x), then I think you have two options: 1. In Home, store values for Vs, R and C using say 1 Sto>Vs etc... This creates user variable for Vs, while R and C are existing Home/Real variables. or 2. In CAS**, store value for Vs, Vs:=1 , which creates a CAS variable. If you want R and C to be CAS variables as well, edit the names, like R1, C1 and use := . Then, in Function app, enter the function Vs*(1-e^(-X/R*C)) into F1: . This should yield the plot. Editing the variables Vs, R and/or C should adjust the plot. ** If you wanted to work symbolically with the formula for Vc, then setting up the variables and perhaps the Vc formula in CAS is helpful. There are other ways to do this. RE: Capacitor charging graph - Alan72 - 11-26-2014 06:53 AM
Thanks for the assist. I started with an exponential graph and gradually added the variables, checking each time before adding the next. It was the time t variable that caught me out until I took your suggestion and renamed it x. Makes sense really as the x value increases to give me my curve. The rest of the formula didnt change, for instance, the size of resistor was constant so I had figures for all of these. Thanks again. Not sure how to mark this as solved but SOLVED!! RE: Capacitor charging graph - DrD - 11-26-2014 01:41 PM
Advanced Graph (another way of viewing) the exponential charge : V1:Y=V*(1-e^(-X/(R*C))) V:=(your supply voltage here) R:=(your resistor) C:=(your capacitor) R*C = the time constant [shift][Plot]: x-range from 0 to five RC time constants. y-range (your supply voltage) -or- use a supply voltage of 100 for 0-100% x-ticks to RC time constant/10 y-ticks to 1 [Plot] The graph of voltage accumulated on the capacitor versus time with convenient chart values. -Dale- |