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How is the slope of a voltage vs time graph the current? 
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#1
Dec2313, 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. 


#2
Dec2313, 03:38 PM

PF Gold
P: 2,019




#3
Dec2613, 11:55 AM

P: 1,810

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: 11,819

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



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