# Total Charge from integration

Discussion in 'Homework Help' started by Tera-Scale, Jan 14, 2011.

1. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
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5
I have an equation for the instantaneous current, and i want to find the total charge present in a capacitor after the first 7 seconds. And Q=It. Do I have to find the definite integral of the instantaneous current and multiply it with the time (7sec)? (Attached)

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Last edited: Jan 14, 2011
2. ### Georacer Moderator

Nov 25, 2009
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If you have an RC circuit in your hands, no. Unless if the exercise asks you do do it that way.

Of course you can integrate the current and get to the charge, but to get the percentage you need to know the total achievable charge too. Do you have any info on the total charge or the voltage applied on the capacitor and its capacity?

Also, you got something wrong. The initial equation is $dQ=i \cdot dt$. If the current is constant that turns into the formula you know, $Q=i \cdot t$. But for variable current you have $Q=\int i \cdot dt +Q_0$, so your equation is wrong.

Anyway, posting the original question will help alot.

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3. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
I dont have any information about the voltage applied, max capacity etc accept the instantaneous current equation attached and the time. (0-7sec). Yes the current is varying according to the graph. I try it that way thanks.

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4. ### Georacer Moderator

Nov 25, 2009
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Oh, my bad. I read "present" as "percent" in the first line and thought what would the total current be.
So just go ahead and calculate the charge through integration.

5. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Which integration? The one you mention, or the integral of product of the instantaneous current and time?

6. ### Georacer Moderator

Nov 25, 2009
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1,266
The integral on the last LaTeX image on post #2.

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