
#1
Dec2313, 01:25 PM

P: 86

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
Dec2313, 03:38 PM

PF Gold
P: 2,004





#3
Dec2613, 11:55 AM

P: 1,784

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
Dec2613, 12:20 PM

Mentor
P: 10,864

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



#5
Dec2713, 11:11 AM

P: 661

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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