Calculating Charge and Current

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Homework Help Overview

The discussion revolves around the concepts of electric current and charge, specifically focusing on the relationships defined by formulas involving current (I), charge (Q), and time (T). Participants are exploring calculations related to batteries, including current flow, charge transfer, and battery capacity in amp-hours.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the formula I = ΔQ/ΔT to calculate current and charge. Questions arise regarding the conversion of amp-hours to coulombs and the implications of current ratings on battery usage time. There is also exploration of the relationship between watts, joules, and seconds.

Discussion Status

Some participants have confirmed the correctness of initial calculations, while others are questioning their understanding of the relationships between current, charge, and time. There is ongoing clarification regarding the conversion of units and the implications of battery capacity on usage time.

Contextual Notes

Participants are navigating through potential misunderstandings about unit conversions and the definitions of physical quantities, such as amp-hours and their relation to coulombs. The discussion reflects a learning process with varying levels of confidence in the calculations presented.

FeDeX_LaTeX
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Hello;

I just wanted to clarify whether I am using this formula correctly. By definition, current is the rate of flow of charge. In other words, its how much charge flows per second. One amp (1 A) is equal to one coulomb per second (1 C/s). Charge and current are related by the equation: I = ΔQ/ΔT.

1) A battery supplies 10 C over a period of 50 seconds. What is the current?

My answer: Given the formula above, I = 10 coulombs / 50 seconds = 0.2 A. Is this correct?

2) Another battery is connected for 2 minutes and provided a current of 0.4 A. How much charge flowed?

My answer;

0.4 = Q/120
Q = 0.4*120 = 48 C

3) A car battery has a capacity of 24 Ah (amp hours). If it provides a current of 48A how long can it be used for? How much charge (in coulombs) does it contain?

My answer: Not completely sure about this question, but I think that the car battery can be used for 2 hours. As for the latter question, I don't understand... how do I convert from amp hours to amps?

Thanks.
 
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Your answers to 1 and 2 are correct. An amp-hour is current multiplied by time. What physical quantity do you get when you do that?
 
Hello;

Yeah I thought about that but I ended up getting 1 amp per 150 seconds, which isn't much help... but 1 amp is 1 C/s. How can I use that in my formula?

Also, I said that the battery can be used for 2 hours, but is it actually 0.5 hours? If the battery has a capacity of 24 Ah, so if 48*time = 24, 24/48 = 0.5?
 
Once more, what physical quantity do you get when you multiply amps with hours?
 
Oh... 1 coulomb. How could I not see that...

So is the answer 2 coulombs?
 
No. To get coulombs, you need to multiply amps by seconds, not hours.
 
2 hours = 3600*2 = 7200 seconds.

So the answer is 7200 coulombs?
 
Why are you multiplying by 2? What is the rating of the battery in amp-hrs?
 
Whoops. If it provides 48A, then it will be using twice as much current and thus will last for half the time, so 0.5 hours. 0.5 hours in seconds is 1800 seconds, so is it 1800 coulombs?
 
  • #10
FeDeX_LaTeX said:
Whoops. If it provides 48A, then it will be using twice as much current and thus will last for half the time, so 0.5 hours. 0.5 hours in seconds is 1800 seconds, so is it 1800 coulombs?
One more time. To get Coulombs you multiply amps with seconds. You have 1800 s, how many amps does the battery provide during this time?
 
  • #11
48*1800 = 86400 coulombs
 
  • #12
Yup.
 
  • #13
Little bit unrelated, but Watts work in the same way, correct? Watts = Joules / seconds, so to find J I'd have to multiply watts by seconds...?
 
  • #14
Precisely.
 

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