Calculating Charge and Current

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The discussion focuses on the calculation of charge and current using the formula I = ΔQ/ΔT, where current (I) is defined as the flow of charge (ΔQ) over time (ΔT). The user correctly calculates current as 0.2 A when a battery supplies 10 C over 50 seconds and finds that 48 C flows when a current of 0.4 A is maintained for 2 minutes. For a car battery with a capacity of 24 Ah providing 48 A, the user initially miscalculates the duration of use but later clarifies that it lasts 0.5 hours, equating to 1800 seconds and resulting in 86400 coulombs of charge. The conversation also touches on the relationship between watts, joules, and seconds, confirming that watts can be calculated by multiplying watts by time to find joules. Understanding these relationships is crucial for accurate electrical calculations.
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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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