# Finding the total charge from the graph

1. Oct 19, 2015

### red1312

Hi
I hope this the right place to put my question

I didn't understand how it became like this

Thank you

2. Oct 19, 2015

### BvU

A current of 1 A during one second means that 1 Coulomb has passed by.
If the current is constant, all you have to do is multiply current and time to get charge.
For example in the period 1-2 ms the current is constant, so the charge is 1 ms x 10 mA = 10 $\mu$C
Which is also the area in the graph under the red line from 1-2 ms.

For 0-1 you have a non-constant current and the area under the red line is 10 x 1 /2 in mA x ms .

--

Does this help ?

3. Oct 19, 2015

### red1312

Thank you very much
I understood the first part
But sorry I didn't get it the last part which is the 1/2 ,,,

Thank you again

4. Oct 19, 2015

### BvU

In the first msec the current increases from 0 to 10 mA. It isn't 10 mA all through this msec, so 10 x 1 would be too much by a factor of about 2 as you can see from geometry.
In this case the average current during the first msec is 5 mA and the factor of 2 is exact.

Could it be that you aren't all that familiar with integrals and that integrals represent an area ?

Basically all that's done is that we add up all the little bits of charge = $\ \$ i x $\Delta$t $\ \$ for the period 0 - 1 ms and do so for very small time steps. In the limit $\Delta$t $\downarrow$0, this becomes the integral $\int i \ dt$