Finding Voltage from capacitance, current and time

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SUMMARY

The discussion focuses on calculating the voltage across a 2µF capacitor given a current-time graph. The user understands the distinction between current (Ic) and voltage (Vc) in capacitor circuits but struggles to convert current values into voltage values. The key equations discussed include Q=CV for charge and voltage relationships, and the differentiation of this equation to relate current to voltage over time. The user is advised to utilize the relationship between charge and voltage to derive the necessary voltage values from the current data.

PREREQUISITES
  • Understanding of capacitor behavior in electrical circuits
  • Familiarity with the equations Q=CV and its differentiation
  • Basic knowledge of current (I) and charge (Q) relationships
  • Ability to interpret current-time graphs
NEXT STEPS
  • Study the differentiation of the equation Q=CV to relate current and voltage
  • Learn about the charging and discharging equations for capacitors, specifically Vc=Vo(E^-t/RC)
  • Explore the concept of current as the rate of charge flow, ΔQ/ΔT
  • Investigate practical examples of capacitor voltage calculations using current data
USEFUL FOR

Students studying electrical engineering, educators teaching capacitor theory, and anyone involved in analyzing capacitor circuits and their voltage-current relationships.

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Homework Statement


The question asks
the current through a 2uF capacitor is show. sketch the capacitor voltage Vc between t = 0ms give that the voltage at t=0ms is 0V. Your graph must be correctly scaled with numerical values.

There is a graph included showing a current rise for 0-2 milliseconds. 0 current for 2-4 milliseconds negative current for 3 milliseconds.

The Attempt at a Solution


I understand this question is asking for the difference between graphing Ic and Vc. I have the graph drawn and understand the difference between Ic and Vc in a charging and discharging circuit.

Now, I must convert my current values into voltage values and then add it to my graph. This is where I am stuck.

I know that:

Vc=Vo(E^-t/rc) and ic =-Vo/R(E^-t/rc) are used when a capacitor is charging and discharging.
both of those have voltages in the formula, so I can't use them.

I know Q=CV
I also know coulomb is 1s/A
can I change that to be A= Δ Q/Δ T ??
and rearrange it to Q= Δ A x Δ t ??

I can't figure out how to get the current to voltage!
 
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If you are given the current into or out of a capacitor you know the amount of charge moving into or out of the capacitor (since current is defined to be charge/time). Look up the relationship between charge and voltage for a capacitor. If you know the capacitance and charge at any instant then you can determine the voltage on the capacitor.
 
You are on the right lines by starting with Q=CV. Then differentiate both sides wrt time.
 

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