# Why does i(t)=dq(t)/dt give the value of AC current only?

1. Nov 11, 2011

### gikiian

I know the mathematical and geometrical reason, but does there exist a physical interpretation behind this?

Thanks

2. Nov 11, 2011

### f95toli

I am not sure what you mean.
That formula works for DC as well. An "exotic" example would be a single electron pump where the (dc) current is given by the number of electrons pumped per second...

3. Nov 11, 2011

### 2milehi

I don't believe it does. Say that a constant 10 Coulombs of charge per second is flowing through a conductor. This would be equivalent of 10 Amps of current. The rate of change in Coulombs per second is zero. So the equation i(t)=dq(t)/dt would yield zero also.

4. Nov 11, 2011

### Andrew Mason

!? The rate at which charges move is dq/dt. That is the current. You appear to be confusing dq/dt with d2q/dt2. The rate of change of charge flow is not the same as the rate of charge flow.

AM

5. Nov 11, 2011

### JeffKoch

No, it is 10 amps of current.

6. Nov 11, 2011

### 2milehi

Say at t = 2 seconds there is 10 Coulombs flowing in the conductor, at t = 3 seconds there is 10 Coulombs flowing in the conductor. A simplified way of figuring dq/dt would be (10 - 10) / (3 - 2) = 0. A slope of a horizontal line is always zero.

d2q/dt2 is the rate of change of the rate of change.

Last edited: Nov 11, 2011
7. Nov 11, 2011

### Studiot

There is or there are?

But no, there are 10 coulombs per second flowing.

If you re-examine your units you will understand what everyone is telling you.

8. Nov 11, 2011

### cmos

It helps if you think in terms of current DENSITY at first:
J = ρ.v = ρ.(dx/dt)

Now consider a cross-section of your conductor and perform a surface integral, the result:
I = (dq/dx) (dx/dt) = dq/dt

So the proper interpretation of the dq/dt term is the rate of charges that cross a certain reference surface. So even for a DC current of, say, 10 A, then every second 10 C cross that surface.

9. Nov 11, 2011

### 2milehi

OK I get it now. I was thinking in terms of current flow, not charge flow. At t=2, 20 coulombs have flown through a conductor. At t=3 30 coulombs have flown through a conductor. So (30 - 20) / (3 - 2) = 10 coulombs per second. The coulomb count is accumulative.

Sorry for the confusion (and bad grammar) - watching a two-year old can cause a lack of concentration.

Last edited: Nov 11, 2011