CR Circuit Question (Calculus)

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The discussion focuses on analyzing the voltage across a capacitor in an RC circuit connected to a 10V d.c. supply, using the equation v = V - Vet/CR. The capacitor is specified as 100µF and the resistor as 47kΩ, resulting in a time constant (CR) of 4.7 seconds. Participants verify the gradient of the voltage graph at t = 5 seconds, with initial calculations yielding -10, which is corrected to -0.1. The importance of correctly applying calculus, particularly the chain rule, is emphasized for determining the gradient.

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A capacitor and resistor are connected as shown across a 10V d.c. supply. The voltage across the capacitor will increase at an exponential rate given by the equation
v = V - Vet/CR
a) Plot the graph of the capacitor voltage from the equation for values of t between 0 and 24 seconds.
vc = V (1-et/CR) , V = 10v, Capacitor = 100µF, Resistor = 47kΩ
CR = 100µF times 47kΩ = 4.7
done part a. but I need someone to check it for me please?

(b)From the graph determine its gradient at t = 5 seconds.
I have done the graph and after calculating it; it gives me a gradient of -10? I don't know if that's the right answer.

(c) Using calculus determine the gradient of the equation and from this the gradient when t = 5 seconds.
I have attempted it my way and I am not getting the right answer. I have been told by the teacher that the answer should be in minus and both gradient answer should be close.
So, any help would be appreciated.
thanks
 

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will.i.am2 said:
A capacitor and resistor are connected as shown across a 10V d.c. supply. The voltage across the capacitor will increase at an exponential rate given by the equation
v = V - Vet/CR
a) Plot the graph of the capacitor voltage from the equation for values of t between 0 and 24 seconds.
vc = V (1-et/CR) , V = 10v, Capacitor = 100µF, Resistor = 47kΩ
CR = 100µF times 47kΩ = 4.7
done part a. but I need someone to check it for me please?(b)From the graph determine its gradient at t = 5 seconds.
I have done the graph and after calculating it; it gives me a gradient of -10? I don't know if that's the right answer.
That's not what you show in the attached image, which is -0.1. Is the slope -10 or -.1? The slope is the vertical rise (or drop) divided by the horizontal run.
will.i.am2 said:
(c) Using calculus determine the gradient of the equation and from this the gradient when t = 5 seconds.
I have attempted it my way and I am not getting the right answer.
Your derivative is incorrect. You are not using the chain rule correctly.

Also, be consistent with the letters you use. You are using both V and v, which is confusing, since these could represent different quantities.
will.i.am2 said:
I have been told by the teacher that the answer should be in minus and both gradient answer should be close.
So, any help would be appreciated.
thanks
One more thing: since this is a calculus problem, it should NOT be posted in the Precalculus section. I am moving it to the Calculus & Beyond section.
 

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