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

by x86
Tags: current, graph, slope, time, voltage
 P: 94 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.
PF Gold
P: 2,019
 Quote by 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.
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.
P: 1,810
 Quote by x86 A lot of websites say that if you take the slope of a voltage vs time graph, you get the current.
Which websites?

 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 V = J / C then 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.
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

Mentor
P: 11,819
How is the slope of a voltage vs time graph the current?

 Quote by x86 A lot of websites say that if you take the slope of a voltage vs time graph, you get the current.
That is true for capacitors if you include capacitance as constant factor.
PF Gold
P: 752
 Quote by mfb That is true for capacitors if you include capacitance as constant factor.
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.

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