Finding Voltage graph from current graph of capacitor

In summary, the current graph shows a constant 4 mA from 0 to 4 microseconds, followed by -1 mA from 4 to 7 microseconds. There is a smaller negative slope for the next 3 microseconds.
  • #1
david12445
11
1

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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  • #2
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?
 
  • #3
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)?
 
  • #4
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?
 
  • #5
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
 
  • #6
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.
 
  • #7
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
 
  • #8
david12445 said:
V(t) = (1/C) * .004t,
Yes. You can plot it accordingly. It will be a triangular wave (not symmetric).
 
  • #9
Great, I understand now. Thank you very much for your help
 
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What is the relationship between voltage and current in a capacitor?

In a capacitor, the voltage and current have a direct relationship. This means that as the voltage across the capacitor increases, the current through it also increases.

How can I find the voltage graph from a current graph of a capacitor?

To find the voltage graph from a current graph of a capacitor, you can use the formula V = Q/C, where V is the voltage, Q is the charge stored on the capacitor, and C is the capacitance. You can plot this equation to create the voltage graph.

What does the slope of a current graph of a capacitor represent?

The slope of a current graph of a capacitor represents the capacitance of the capacitor. This is because the slope is equal to the change in current divided by the change in voltage, which is equivalent to the capacitance value.

How does the voltage graph of a capacitor differ from that of a resistor?

The voltage graph of a capacitor differs from that of a resistor in that it is not a straight line. Instead, it has a curved shape that approaches the maximum voltage as the capacitor charges. This is because capacitors store energy, while resistors dissipate energy.

Why is it important to analyze the voltage and current graphs of a capacitor?

Analyzing the voltage and current graphs of a capacitor is important because it allows us to understand the behavior of the capacitor in a circuit. This information can help us design and optimize circuits for various applications.

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