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Homework Help: Circuits Question

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img133.imageshack.us/img133/1101/28162945xv9.jpg [Broken]

    2. Relevant equations

    p = iv
    i = dq/dt
    w = intergral of p dt = intergral of vi dt

    3. The attempt at a solution

    I don't really understand how to do this problem. First of all, I know that the area under the graph between some time interval is equal to the charge.

    So for the part A of the question, I got the following, but I dont even know if its correct.

    a) @ 1s = 2C, @3s = -1C, @4.5s = 0C

    I don't know how to do the rest.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 15, 2009 #2


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    For 1 you've got the general idea, but your answer for 3 sec is incorrect.

    For 2 recognize that P(t) = V*I(t) = 2*I(t)

    For 3 the energy absorbed then is as before evaluated as the integral at the times they ask.
  4. Jan 15, 2009 #3


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    One thought however.

    P = |V*I|

    The negative current will still result in positive power.
  5. Jan 15, 2009 #4
    @LowlyPion, thanks for the reply, but I am still a bit confused.

    Part A)

    @1s would the charge equal to 2C or 0C? Why is 2C? Shouldn't it be 0C because in the graph @1s, the current drops from 2A to 0A.

    @3s, why is my answer -1C incorrect? What is the correct answer and do you do get that answer/

    Part B

    You said I just use the equation P = iv. According to your reply, v=2V. Did you get 2V from the circuit image on the left shown next to the graph?

    Part C

    I am not sure how to solve for this.
  6. Jan 15, 2009 #5


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    The area under the origin is negative area as regards to figuring charge entered. But you have already entered 2C by 1sec, so by 3sec then you have removed 1C leaving 1C not -1C. A small point, but one your grader will take seriously.
    Yes. Your picture shows a 2V source.
    As you did in Part A. The integral of the function from Part B. Except as I added, this time you want the sum of the absolute areas.
  7. Jan 15, 2009 #6

    Part A) What you said doesn't make sense. The question is asking for the amount of charge entered exactly at t=3s. This mean, you don't get the area under the graph from the previous time frame. You just have to get the area under the graph exactly at t=3s. Am I correct?

    Part B) I am not sure if I did this correctly, but can you take a look.

    @1s = 2(2) = 4W
    @2.5s = 2(-1) = -2W
    @4.5s = 2(0) = 0W
    @5.5s = 2(1) = 2W

    I am not sure why you said "negative current will result in positive power"

    Part C)

    W = ingegral of P. But there is no function here with respect to t. So calculate the energy.

    I think, but now sure, do I just multiply the y axis (current) of the graph by 2V. This would give me a new graph. Then I sum the areas under the graph. If I do this, I get 4J-4J+2J = 2J.
  8. Jan 15, 2009 #7


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    I think the problem is ambiguous. Since little is known about what's in the box, and the voltage source might be a rechargeable battery for instance, then I guess that you can say there is net negative energy if the current is flowing backwards. So I guess just disregard my concerns about negative power.

    In general though I think of power usually going to heat, in which case I would normally count it as always positive - just looking at the box without regard for positive or negative current. But they do ask for net energy, so I guess your answers would be appropriate.
  9. Jan 15, 2009 #8
    Thanks a lot. I will ask my prof about this question to see what he says. I'll let you know what he said.
  10. Jan 15, 2009 #9


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    At the very least he will know you were thinking about the problem.
  11. Jan 15, 2009 #10
    I didn't ask the professor, but I ran into the soultions and you were correct, I was wrong.
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