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

by x86
Tags: current, graph, slope, time, voltage
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x86
#1
Dec23-13, 01:25 PM
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.
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Dadface
#2
Dec23-13, 03:38 PM
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Quote Quote by x86 View Post
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.
skeptic2
#3
Dec26-13, 11:55 AM
P: 1,810
Quote Quote by x86 View Post
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

mfb
#4
Dec26-13, 12:20 PM
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How is the slope of a voltage vs time graph the current?

Quote Quote by x86 View Post
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.
psparky
#5
Dec27-13, 11:11 AM
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Quote Quote by mfb View Post
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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