Finding Voltage graph from current graph of capacitor

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Homework Help Overview

The discussion revolves around deriving a voltage graph from a given current graph for a capacitor, specifically a 2 micro-farad capacitor. The current is presented as a piecewise function, with a constant current followed by a negative current over specified time intervals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between current and voltage in capacitors, questioning how to derive voltage from the current graph. There is discussion about using the area under the current-time graph to find charge and applying relevant equations.

Discussion Status

Participants have engaged in a productive dialogue, with some suggesting integration of the current function to find voltage as a function of time. There is acknowledgment of the need to clarify the relationship between current and voltage, and some participants have provided guidance on the integration process.

Contextual Notes

There is a focus on the mathematical relationships and assumptions inherent in capacitor behavior, including the need to integrate the current over time to find voltage. The discussion reflects uncertainty about the correct application of formulas and the interpretation of the current graph.

david12445
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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.

Homework Equations


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
 
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david12445 said:
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?
 
cnh1995 said:
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)?
 
david12445 said:
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?
 
cnh1995 said:
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
 
david12445 said:
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.
 
cnh1995 said:
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
 
david12445 said:
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
 
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