Solve Integration Problem: Charge Flowed After 5 Seconds

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In summary, the conversation is about finding the charge flowed through a capacitor after 5 seconds, using the formula I = Ioe^-(kt). The individual is struggling with the calculations and someone points out their mistake of using e^0 instead of e^-1. The problem is solved and the individual is grateful for the help.
  • #1
happyjoe
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Hey everyone. This has been really bugging me for 4 hours now, and I cannot figure out where I am going wrong. If anyone can help me that would be greatly appreciated.

The current flowing through a capacitor is:

I = Ioe^-(kt)

Io = 2 amps
k = 0.2s
T = time

Find the charge flowed after the first 5 seconds


Q = [tex]\int[/tex] I dt


5
[tex]\int[/tex] 2e^-(0.2t) dt
0

= 1/-0.2e^-(0.2t)

= -10e^-(0.2t)

5
[tex]\int[/tex] -10e^-(0.2t)
0


[ -10e^-(0.2t)] = {-10e^-(0.2x5)} - {-10e^-(0.2x0)}


= -10e^-1 - 0

Then I end up with a negative number for Q . If anyone could point me in the right direction i'd be thankful .
 
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  • #2
happyjoe said:
[ -10e^-(0.2t)] = {-10e^-(0.2x5)} - {-10e^-(0.2x0)}


= -10e^-1 - 0

Your mistake is: e^0 does not equal 0, e^0 equals 1.
 
  • #3
Wow, I can't believe I made a mistake that simple, lol

Thanks for your help! You really helped me out there!
 

1. What is integration and why is it important in science?

Integration is a mathematical process used to find the total or cumulative value of a function. In science, integration is important because it allows us to calculate the total amount of something over a period of time or space, providing a better understanding of a system or process.

2. How do you solve an integration problem?

To solve an integration problem, you first need to identify the function and the limits of integration (the start and end points for the calculation). Then, you can use integration techniques such as substitution, integration by parts, or trigonometric substitution, to find the antiderivative of the function. Finally, you can plug in the limits of integration and solve for the total value.

3. What does the "charge flowed after 5 seconds" mean in an integration problem?

In an integration problem, "charge flowed after 5 seconds" refers to the total amount of electric charge that has passed through a point in a circuit after a time period of 5 seconds. This can be calculated by integrating the current (which is the rate of flow of charge) over the time interval.

4. What units are used in an integration problem involving charge?

In an integration problem involving charge, the units used will depend on the units of the function being integrated. For example, if the function is electric current (measured in amperes), the units of charge will be coulombs (C) as current is the rate of flow of charge. If the function is electric field (measured in volts per meter), the units of charge will be coulombs per meter (C/m).

5. Can integration be applied to other areas of science besides physics?

Yes, integration can be applied to other areas of science besides physics. In chemistry, integration can be used to calculate the total moles of a substance produced in a reaction over a period of time. In biology, integration can be used to determine the total amount of a substance in a biological system over a period of time. Essentially, integration can be used in any field where a quantity changes over time or space.

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