# How is the slope of a voltage vs time graph the current?

1. Dec 23, 2013

### x86

A lot of websites say that if you take the slope of a voltage vs time graph, you get the current. However, the math tells a different story.

where V = voltage, J = joules, C = coulombs, A = amperes, s = seconds

A = C / s
C = A * s
V = J / C
J = C / V

if we take the slope of Voltage vs Time, our unit is:

V/s = J/(C * s) = J / (A * s^2) = (C * V) / (A * s^2) = (C * V) / (C /s * s^2) = V/s

No matter what I do, I can never get the unit ampere.

How is it mathematically possible that the slope of a voltage vs time graph has the unit of the current? I don't get it.

2. Dec 23, 2013

The slope does not have the units of current. The units are V/s (which can be expressed in different ways) as you have shown. Perhaps you misread the information on the websites or you looked at websites giving wrong information.

Last edited: Dec 23, 2013
3. Dec 26, 2013

### skeptic2

Which websites?

If V = J / C then J ≠ C / V

You forgot to consider capacitance.

First the variables need to be used properly.
Type______Symbol______Unit
Voltage...........E...............V
Current...........I................A
Charge...........Q................C
Capacitance....C................F
time...............t................s

Q = C * E
∴ E = Q / C
E / t = Q / (C * t)
E / t = I / C

4. Dec 26, 2013

### Staff: Mentor

That is true for capacitors if you include capacitance as constant factor.

5. Dec 27, 2013

### psparky

I agree with that if you are referring to:

C*dv/d(t)=i(t)

Then yes, the slope of the voltage multiplied by the capacitance will give the current at any given time thru the capacitor.