# Calculating total charge

1. May 18, 2010

### kaiser0792

1. The problem statement, all variables and given/known data
The current at the terminals of the element in an ideal basic circuit element is
i = 0, t < 0;

i = 20e(-5000t) A, t $$\leq$$ 0

Calculate the total charge ( in microcoulombs) entering the element at its upper terminal.

2. Relevant equations

3. The attempt at a solution I'm just starting a Circuit Analysis course next week and I'm looking ahead in the text, trying to hit the ground running. There are no sample problems that even give me a starting place??
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 18, 2010

### rock.freak667

Current can be expressed at the rate of flow of charge, so that i =dQ/dt.

So you can integrate over time and get the total charge. Though why it time t < 0?

3. May 18, 2010

### kaiser0792

I suppose that is just a way of saying that there was no current flowing before the reference time, t = 0.

4. May 18, 2010

### kaiser0792

I tried integrating and came up with -0.04e-5000t + C coulombs.

Answer is supposed to be 4000 microcoulombs. My integration is a little rusty. Help?

5. May 19, 2010

### rock.freak667

The -0.04 should be 0.004, but remember your time is t≥0. So you are really integrating from 0 to ∞ so you need to compute

$$\left[ -0.004e^{-5000t} \right]_0 ^{\infty}$$

6. May 19, 2010

### kaiser0792

Thanks for the help, I knew the Integral to be solved and the limits of integration, what I was missing was the negative exponent of e. You helped me, thank you.

7. May 19, 2010

### kaiser0792

Thanks rock.freak, I was overlooking the negative exponent of "e" when I was integrating.
You helped me, thanks. Sometimes you just need to bounce it off someone else.