# Finding Voltage graph from current graph of capacitor

## Homework Statement

I am given the current flowing through a 2 micro-farad capacitor in the form of a graph, and I need to create a voltage graph from this.

I = C dv/dt
Q = VC

## The Attempt at a Solution

The current graph is basic with a constant 4 mA from 0 to 4 microseconds and then -1 mA from 4 to 7 microseconds. I am a bit confused on how to proceed. Is the right way to go finding the total charge at each time interval, and then plugging into the Q = VC formula to find the voltage? So then the total charge would be the area under the current vs time graph? With this I am getting a voltage graph that has a constant positive slope until 4 microseconds, and then a smaller negative slope for the next 3 microseconds, Does this sound about right? Thanks for any help

cnh1995
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The current graph is basic with a constant 4 mA from 0 to 4 microseconds and then -1 mA from 4 to 7 microseconds
You are given the i-t relationship of the capacitor. You'll need to plot it on paper. What is the i-v relationship i.e. how would you write voltage across capacitor as a function of its current?

You are given the i-t relationship of the capacitor. You'll need to plot it on paper. What is the i-v relationship i.e. how would you write voltage across capacitor as a function of its current?
Would the formula for the i-v relationship be v = c/(it)?

cnh1995
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Would the formula for the i-v relationship be v = c/(it)?
No. You have i=Cdv/dt. This equation is in differential form. What should be done so that you'll get it as v as a function of i?

No. You have i=Cdv/dt. This equation is in differential form. What should be done so that you'll get it as v as a function of i?
Would you try to integrate both sides? With that wouldn't you end up with the equation Q = VC? This is the part where I'm confused

cnh1995
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Would you try to integrate both sides?
Right.
V=(1/C)∫i dt.
You have i(t), don't you? Integrate it between the given time intervals and find V as a function of time.
You do not need to take ∫i dt as Q.

Right.
V=(1/C)∫i dt.
You have i(t), don't you? Integrate it between the given time intervals and find V as a function of time.
You do not need to take ∫i dt as Q.

Okay, so with the numbers I gave above, i(t) would just be equal to .004A from 0 to 4 microseconds, making V(t) = (1/C) * .004t, and then the same for the next time interval, does that look right? Thank you for your help

cnh1995
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V(t) = (1/C) * .004t,
Yes. You can plot it accordingly. It will be a triangular wave (not symmetric).

Great, I understand now. Thank you very much for your help

cnh1995