Use differentiation to find the rate of change of voltage at 6 ms

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devdev

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Nov 8, 2021
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Hello!

I've got a task and got stuck. Here is the content:

A capacitor was connected in series with a resistor and then charged up from a power supply. Use differentiation to find the rate of change of voltage at 6 ms.

Circuit information:

Capacitor = 80 nF (80x10-9 F) Resistor = 40 kΩ (40x103 Ω) Supply voltage, = 5

I know I have to use the charging voltage equation v = V(1-e^-t/T) and differentiate it, my problem is I don't know which differentiation function I should use. I know the basics, like ax^n -> axn^n-1 etc but don't know how to apply these rules in this task. I have been stuck for 2 hours on it.

Thank you in advance.
 
You are right that you need to differentiate your formula [imath]v = V\left(1-e^{\frac{t}{T}}\right)[/imath], but do you know which variable?
 
Thank you for your response. I am not sure and that's my problem. I guess if the task requires finding the rate of change of the voltage then V and t are variables, am I right?
 
devdev said:
Yes, so using the formula v = V(1-e^-t/T) ---> v=4.233225166, my circuit utilizes 5V supply voltage, t = 0.006 and T = 0.0032
Click to expand...
So calculate \(\displaystyle \frac{dv}{dt}\) and evaluate at t = 0.006
 
Thanks, I thought it is more complicated. I have seen some working out like: ?̇(?) = dv/dt=V/T * e ^-t/T and I couldn't understand how it is derivated.
 
devdev said:
Thanks, I thought it is more complicated. I have seen some working out like: ?̇(?) = dv/dt=V/T * e ^-t/T and I couldn't understand how it is derivated.
Click to expand...
Can you calculate:

\(\displaystyle \frac{d}{dx}\left[e^{ax}\right ] \) ....................... you have to use chain rule
 
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