Hello
I need help, I don't understand how to find out voltage of capacitor when voltage increase or decrease? I see the below formula. I plotted current time graph please help me

 

Is this homework?

Your setup looks OK. What's the problem? What is the analytical expression for i(t)?

 

You are asking to find the capacitor voltage when the voltage increases or decreases, which makes no sense.

The capacitor voltage change is the integral of the current (which is the total charge) into (or out of) the capacitor over time.
What don't you understand about that?

 

Is this homework?

Your setup looks OK. What's the problem? What is the analytical expression for i(t)?

No I plotted graph just for my understanding. Any you asking the value of t1 and t2

 

You are asking to find the capacitor voltage when the voltage increases or decreases, which makes no sense.

The capacitor voltage change is the integral of the current (which is the total charge) into (or out of) the capacitor over time.
What don't you understand about that?

Please look at formula, I want to find out voltage V

 

Please look at formula, I want to find out voltage V

Do you not know how to do the integration in the formula?
That's how you calculate the capacitor voltage.
If not then you need to start studying calculus.

 

Do you not know how to do the integration in the formula?
That's how you calculate the capacitor voltage.
If not then you need to start studying calculus.

Look at handy calculation

 

The handy calculation omitted the 1/C term. Otherwise it looks good to me.

 

So what happened to C and Vo?

 

How did your i(t) suddenly go to just being t?

On what basis, given the plot of i(t) you showed, are you claiming that i(t) = (1 A/s)·t between t = 2 s and t = 4 s?

What happened to the capacitance?

If you would bother to track your units (which I have warned you about repeatedly) you would have seen that the units on your answer don't work out to volts. The fact that you refuse to track your units only proves that you don't really care about getting the answer correct.