Voltage vs Time Graph: Understanding & Deriving Relationships

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SUMMARY

The discussion focuses on deriving relationships from a voltage vs. time graph in the context of a battery and resistor circuit. It clarifies that with a constant voltage across a resistor, the graph is a horizontal line, while a decreasing voltage scenario results in a linear negative slope. The calculation of charge over time is demonstrated using the formula I=V/R, leading to the conclusion that for a 1.5V battery over three hours with a 10 Ohm resistor, the charge is 225mAh. The discussion emphasizes the importance of understanding the voltage drop and its implications on current and charge calculations.

PREREQUISITES
  • Basic understanding of Ohm's Law (I=V/R)
  • Familiarity with voltage vs. time graph interpretation
  • Knowledge of charge calculation in milliamp hours (mAh)
  • Concept of linear functions in physics
NEXT STEPS
  • Study the integration of current over time in electrical circuits
  • Learn about the behavior of capacitors in RC circuits
  • Explore the effects of different resistor values on charge calculations
  • Investigate real-world applications of voltage vs. time graphs in electronics
USEFUL FOR

Electronics students, electrical engineers, and anyone interested in understanding the dynamics of battery discharge and circuit analysis.

megr_ftw
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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?
 
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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
 

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