Growth and Decay: The voltage (vc) across a charging capacitor at time T is given by

Hi,

I could really use some help. I have hit a wall with my studies.

3. The voltage (vc) across a charging capacitor at time T is given by:

V_c = V_s [ 1 - e^{ -T/(CR)}]



Where the supply voltage is Vs = 10v, the capacitance C = 10μF and resistance R = 20KΩ. Determine:

3.1. The time for the voltage to reach 5v

3.2. The voltage at 1ms after charging starts

3.3. The rate of voltage change at 1ms after charging starts

3.4. Show that the rate of voltage change when the capacitor is discharging from the supply voltage value will be the same as when charging from the same value at an any given time


Any help you could provide would be gratefully recieved!

Thanks

Daz

Daz said:

Hi,

I could really use some help. I have hit a wall with my studies.

3. The voltage (vc) across a charging capacitor at time T is given by:

V_c = V_s [ 1 - e^{ -T/(CR)}]

Where the supply voltage is Vs = 10v, the capacitance C = 10μF and resistance R = 20KΩ. Determine:

3.1. The time for the voltage to reach 5v

3.2. The voltage at 1ms after charging starts

3.3. The rate of voltage change at 1ms after charging starts

3.4. Show that the rate of voltage change when the capacitor is discharging from the supply voltage value will be the same as when charging from the same value at an any given time


Any help you could provide would be gratefully recieved!

Thanks

Daz

Click to expand...

You are given an equation - and you need to evaluate a parameter when the value of other parameters are given.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

As I explained in my other question. I'm stuck, I have tried Kahn academy but I'm getting stuck in my own work. I have cut outs from a John bird book. But I'm missing stuff. If you could help point me in the right direction I will work on it and complete it. I'm purely using this forum for help.

thank you

Daz said:

3. The voltage (vc) across a charging capacitor at time T is given by:

V_c = V_s [ 1 - e^{ -T/(CR)}]

Where the supply voltage is Vs = 10v, the capacitance C = 10μF and resistance R = 20KΩ. Determine:

3.1. The time for the voltage to reach 5v

Click to expand...

They gave you a formula with variables V_c, V_s, T, C, and R. They have given you values for V_c, V_s, C, and R. They have asked you to find the value of T.

You have posted this question to "Calculus", but I see nothing there other than simple ("Beginning") algebra; namely; they've asked you to evaluate a given equation for given values of the given variables. (here) So try using what you learned back in algebra: plug the given numbers in for the given variables in the given equation, and solve for the one value that's left.

Daz said:

3.2. The voltage at 1ms after charging starts

Click to expand...

Unless I'm missing something, this time they gave you V_s, T, C, and R, and are asking you for V_s. So "plug-n-chug"!

Daz said:

3.3. The rate of voltage change at 1ms after charging starts

Click to expand...

How does the rate of change in f(t) relate to the derivative, with respect to time t, of f(t)? How might one apply that knowledge to this question? (Or are you not familiar with derivatives?)

Please be complete. Thank you!

For 3.1 i have an answer of 13.86 seconds

For 3.2 i have an answer of 9.995

For 3.3 i have an answer of 0.0005 V/s

However i am struggling with the rate of change question could you provide more guidance? I have forgotten most of my maths information from college and trying my best to piece this all together for an Open Universtity course.

Thanks