Voltage vs Time Graph: Understanding & Deriving Relationships

AI Thread Summary
The discussion focuses on deriving relationships from a voltage vs. time graph, particularly in the context of a battery and resistor circuit. Participants clarify that if the battery's voltage decreases linearly from 1.5V to 0V over three hours, the current can be calculated using Ohm's Law (I=V/R). Integration of the voltage function provides a method to determine the total charge delivered over time, resulting in a calculation of milliamp hours. However, the utility of this calculation is questioned, suggesting the need for further clarification on the original question. Overall, the conversation emphasizes understanding the relationship between voltage, current, and time in electrical circuits.
megr_ftw
Messages
68
Reaction score
0
What are you able to derive or integrate from a voltage vs. time graph? Like if I have a battery and a resistor and want to know the charge after so many hours or seconds.
We went over this in class but could someone elaborate the relationship?
 
Physics news on Phys.org
A battery across a resistor, voltage is constant, graph of voltage versus time is a horizontal straight line. Are you sure that your real question isn't about a capacitor?
 
I had a problem the teacher gave us and it was a graph of time(x-axis) vs. voltage(y-axis) and it had the line as a negative slope where the voltage of the battery was 1.5V and it crossed the y-axis at 3 hours. they gave us a resistor of x ohms.
So there is no capacitor in the problem and how should I treat it?

*note this is not a homework question, just a general one
 
You're question is a bit confusing... I'll try to help...
I'll assume:
-You're saying the battery's voltage is reducing linearly from 1.5V to 0V over a three hour duration.
-You're circuit is a battery with a resistor across it.
-You said something about charge, so I'll assume current.

Current, I=V/R

If you integrate that function, you get 0.5*(1.5V)/R*time. For example, let's assume that R=10 Ohms:
Then you could calculate: 0.5*1.5/10*3hr = 225mAh (milli amp hours).

But this is actually a useless calculation. Can you elaborate on your question?
 
elliotr said:
You're saying the battery's voltage is reducing linearly from 1.5V to 0V over a three hour duration.

I think the person needs:
decreasing 1.5 volts in 10800 seconds, for a slope of -0.0001389 volts per second
V = (-0.0001389 V/s) t + 1.5 V
 
Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as ##L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)## and then calculates ##\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)##. I have problems with the last step. I might have written ##\frac {dA_x} {dt}...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Back
Top