Measuring Capacitance given a voltmeter + ammeter

AI Thread Summary
To measure the capacitance of an industrial capacitor using a voltmeter and ammeter, the relationship C=Q/V is essential. The ammeter records current as a function of time, allowing for the calculation of charge (Q) through integration of current over time, as I = dQ/dt. The confusion arises in connecting the ammeter readings to the charge, with clarification that Q equals the integral of current over time, not simply I multiplied by time. The discussion emphasizes the need to utilize the time-dependent aspects of the measurements to accurately determine capacitance. Understanding these principles is crucial for solving the problem effectively.
mikel2009
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Homework Statement


Given the following:
• a battery,
• a digital voltmeter (which records voltage as a function of time),
• a digital ammeter (which records current as a function of time),
• a battered industrial capacitor,
• two copper plates,
• and several bits of copper wire.

measure the capacitance of the industrial
capacitor

(there's a second segment of the question, but i understand it)

Homework Equations


C=Q/V. etc.
I=dQ/dt.


The Attempt at a Solution



I have attempted to write a solution to this, but am only confusing myself.
The voltage is a given, however I'm uncertain how I'm supposed to obtain Q from the dQ/dt graph, in order to determine capacitance.
someone please shed some light -.-
 
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This is just an assumption that I thought: Hope it helps
The ammeter is recording current as a function of time, so the reading is actually I/t.
From here we can calculate Q.
Q = I from ammeter readings . t
the I is the reading from the ammeter, so it becomes Q = I/t . t
the t cancel out so Q = I
Does it makes sense?
 
the usual way to measure the capacitance works with an ac voltage source!

Your way is to vague!
 
mikel2009 said:
The voltage is a given, however I'm uncertain how I'm supposed to obtain Q from the dQ/dt graph, in order to determine capacitance.
someone please shed some light -.-
Have you had integrals in calculus class?

jk0921 said:
This is just an assumption that I thought: Hope it helps
The ammeter is recording current as a function of time, so the reading is actually I/t.
Well, not really. The ammeter reading is I, period.

saunderson said:
the usual way to measure the capacitance works with an ac voltage source!

Your way is to vague!
It's not "his way", it's the way given in the problem statement and he is stuck with it. :smile:
 
You answered this yourself with the provided equations.

Differentiate C = Q/V
Bearing in mind that I = dQ/dt as well as the nature of a capacitance of a capacitor

Then look at what you are provided and it's very simple. Note the function of time parts of the question.

P.S. aren't you leaving this a bit late ;)
 
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