Get the Voltage from the I(A)-Time diagram

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Discussion Overview

The discussion revolves around determining the voltage from an I(A)-time diagram related to the discharge of a chemical battery and a capacitor. Participants explore how to calculate energy based on current and time, while addressing the implications of assumptions made regarding resistance and voltage.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose using the area under the I(A)-time diagram to calculate charge (Q=I(A)*t).
  • Others argue that without knowing the voltage, it is impossible to determine the energy of the battery and capacitor.
  • A participant mentions having a chemical battery and a capacitor, questioning if they can find the energy and whether they can assume V=I*R with R=1 ohm.
  • There is a challenge regarding the assumption of R=1 ohm, with questions about why not use different resistance values.
  • Some participants suggest that more information is needed to find the voltage, such as additional references or measurements.
  • A participant notes that the diagram is from an electric car and mentions the total discharge capacity of the battery.
  • Another participant suggests using the charge (Q) and capacitance (C) to determine the change in voltage, but emphasizes the need to know C.
  • One participant clarifies that I*R represents a voltage drop through the battery, not the battery voltage itself, and discusses the concept of delivered energy during discharge.

Areas of Agreement / Disagreement

Participants express differing views on how to approach the problem, particularly regarding the assumptions about resistance and the need for additional information to determine voltage. No consensus is reached on the best method to calculate energy from the given data.

Contextual Notes

Limitations include the lack of a voltage curve and the ambiguity surrounding the resistance value to be used in calculations. The discussion also highlights the dependency on specific definitions and the need for further context regarding the battery and capacitor in use.

papm
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The area under the diagramm is the capacity.Q=I(A)*t(s)
How can I can get the Voltage so tha I get the energy of the battery and capacitor?
 
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Welcome to PF.
I assume you are talking about the discharge of a chemical battery.
papm said:
How can I can get the Voltage so tha I get the energy of the battery and capacitor?
Because only current and time are specified, it is not possible to know the voltage of the battery.
Does the word "capacitor" you use actually refer to the capacity of the battery in kW*hours, or do you have a capacitor in the circuit.
 
Thank you for the reply
I have a chemical battery and a capacitor (over 45(A)).
Can I find the energy of them?
Can I assume V=I*R with R=1ohm?
Because I(A) get straight to 45(A) and we don't have a curve.
 
papm said:
Can I assume V=I*R with R=1ohm?
Where does the idea of 1 ohm come from, why not 1 milliohm, or 1k ?
 
Baluncore said:
Where does the idea of 1 ohm come from, why not 1 milliohm, or 1k ?
Don't know I just assumed to find out the energy.
Maybe becouse we get the I=45(A) immediately?
 
We need more information.
You need to find the voltage from other references or make a measurement.
 
This exercise was on last exams.
The only information was the diagramm that is from a electric car with a battery and and a capacitor for acceleration. There is no more information.
 
That diagram will contain the detail.
 
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papm said:
Because I(A) get straight to 45(A) and we don't have a curve.
That's 50A isn't it? That graph can tell you the net charge into/out of the capacitor. Then use Q = C.V to determine the associated change in voltage, but you do need to know C.

There'll be a graph describing the battery, and you use that to determine the battery voltage.
 
  • #10
This is cycle of a functioning electric vehicle [as translated from Greek]
Let's say this is total discharge battery= 117.6 Ah .A usual battery of 12V 120 Ah could present this discharge process. However a 24 V battery could present the same.
It could be one like this:12V/120Ah/115F51 JIS Car batteries, Sealed lead acid batteries -in my opinion.
 
  • #12
By the way, I*R it is not the battery voltage-except in short-circuit case-but it is the voltage drop through the battery and so I^2*R it is not the delivery power but the battery losses.
The delivery energy it is ʃV*I*dt .Since in the discharge time interval [up to- let's say 1.8 V from 2.2 V rated] you may consider the delivered voltage as constant and equal to the rated and then the delivered energy will be Vrated*Q.
 

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