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Vi = E.e ^-t/cr

 

if having been fully charged to voltage e=10v c=250uf it is allowed to discharge through a 3.3k resistor determine the values of capacitor voltage after time; t=0, t=1, up to t=0

 

b) plot the voltage values against time values

 

c) draw a tangenrt to the urve at t=2s. calculate from the slope the value of rate of change of voltage in V/s

 

d) use differential Calculus to find the gradient of the curve at t=2s

 

e)  compare your values from c) and d) on the accuracy.

 Mar 31, 2020
edited by Guest  Mar 31, 2020
 #1
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Vi = E.e-t/cr     (I believe you left out the negative in the exponent !)

                            You are given  E  = 10 v      c = 250 x 10-6  FARADS        r = 3.3 ohms (?  dimensions left off in your Q)

                             Plugging into the given equation

 

V0 = 10 e-0/(250 x10-6 * 3.3)       this is the value of  V  at t= 0

V1 = 10 e-1 /(250 x 10-6 * 3.3)      this is the value of V  at t=

etc    etc  for other  't' values

 

Start with the graph and show us what you find.....we'l go from there !  cheeky

 Mar 31, 2020
 #2
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not sure how to do the graph 

Guest Mar 31, 2020
 #3
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Calculate the v at  points    t = 0   1    2   3   4 ...... to whatever edpoint is given to you (your question did not have an end point)

                                       

 

t   will be on the x axis     the calculated   v   will be the y axis

 

at t= 0   v = 10      first point   (0,10)        at t = 2   v = ?     second point would be    (2, ?)      etc etc

 

Here is my example......it may or may not be correct...I used 3.3 kOhm   instead of 3.3 ohm  ...I just graphed the equation....your question wants you to plot the points and draw a curve......

 

https://www.desmos.com/calculator/79b8hknsju

ElectricPavlov  Mar 31, 2020
 #4
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say if we used your graph what would the gradient of the curve be at t=2 s

Guest Apr 1, 2020

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