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Graphing the energy in an capacitor, coil and resistor

  1. Apr 28, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    I need to graph the current energy on a resistor, capacitor and inductor they are all separate it is not a RLC circuit.

    This is what I know
    • u(t) = sqrt(2)*220*sin(314*t+π/3)
    • R = 10 Ω
    • L = 3 mH
    • C= 3 mF
    I have already calculated the current for each component, now I need to calculate the energy on each component and then draw its graph.

    The currents are
    • Ic(t) = 293.52*sin(314*t+2π/3) current through the capacitor
    • Il(t) = 330.29*sin(314*t-π/3) current through the in inductor
    • Ir(t) = 31.11*sin(314*t+π/6) current through the resistor


    I want to know how can I represent the energy as a function of time for each of these components induvidual so that I can graph them then and get an diagram?
    And how to calculate the total energy on each component?

    I repeat this is not an RLC circuit, each of these components is induvidualy connected for itself.

    If anybody can help me out I will be grateful really, srry for my bad english.
     
  2. jcsd
  3. Apr 28, 2016 #2

    Simon Bridge

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    The graph is easy - plot E vs t.
    To calculate the energy on a component, you have to answer: what does "energy on a component" mean?

    A capacitor stores energy in it's electric field, and inductor stores it in the magnetic field ... resistors do not store energy, but do dissipate it.
    The amount and rate of energy stored or dissipated changes from instant to instant.
    Note: The rate that energy changes is called power.
     
  4. Apr 29, 2016 #3
    I think the best way to approach this would be to find the power dissipated/stored by the devices and then integrate the power to find the energy vs. time.

    Do you know how to calculate the power dissipated or stored?
     
  5. Apr 29, 2016 #4
    hint: the power dissipated by a resistor is (i^2)*R. You can plug in your current for the resistor and then integrate with respect to time from t=0 to t to find power vs. time.
     
  6. Apr 29, 2016 #5
    The energy stored on a capacitor is 0.5*(Q^2)/C. You can integrate your current to find Q and then plug this into the equation for energy stored by the capacitor.

    The energy stored by an inductor is 0.5*L*(i^2). For this one simply plug in your current.
     
  7. May 2, 2016 #6
    I did the calculations for the resistor and I got this p(t)=4839.58-4839.58cos(628t+π/3) W

    Now I got the energy (power) dissipated from the resistor by intergrading p(t) and I got this at the end W = 483.958 - 4839.58*(1/628)*sin(628t+π/3)

    When I put in the interval from the integral from t=0 to t I get that the total energy is 477.3 J

    And how do I now make a graph do I just plug in time values in the period in this function W = 483.958 - 4839.58*(1/628)*sin(628t+π/3) ?
     
  8. May 2, 2016 #7

    Simon Bridge

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    The current Ir(t) - and probably others, is not correct.
    Ir(t)=u(t)/R = [sqrt(2)*220*sin(314*t+π/3)]/10 = ...
    you got: 31.11*sin(314*t+π/6) ... how does dividing by 10 change the phase angle?
    (Tweek: do not round off irrationals until the final calculation ##22\sqrt{2}\approx 31.11##)

    Your integration is not right. You started out with
    p(t)=4839.58-4839.58cos(628t+π/3) W

    Which is an equation of form:
    ##p(t)=A-A\cos (\omega t + \delta)## ... and you want to use the fact that ##p(t)=\frac{d}{dt}E(t)##

    For some reason you got: W = 483.958 - 4839.58*(1/628)*sin(628t+π/3)
    ... first, you used "W" already in the power equation: do not use the same letter for two different things.
    This equation is ##\cdots = (A/10) - (A/\omega)\sin(\omega t + \delta)##
    The integral of cosine is sine, well done, but what usually happens when you integrate a constant?
    What you did there was ##\int A\; dt = A/10##.

    This was an indefinite integral.
    The indefinite integral should have an arbitrary constant ... so what happened to the "+C"?

    You probably didn't think you needed it because you were doing the indefnite integral on the way to doing a definite integral ... what you said, basically, was that the energy dissipated at time t is given by
    ##E(t) = \int_0^t p(t')\;dt'##


    Lets look at this strategy in a context you are probably more familiar with.
    Imagine you are given the acceleration function for a vehicle and you are asked to find the velocity-time graph for this vehicle.
    How would you go about it ... well, here is what you did:

    You said ##v(t)=\int_0^t a(t')\; dt'## ...
    so if the acceleration was a constant, then ##v(t) = at## which seems OK until you realise that the correct suvat equation is ##v(t)=u+at## ... so what went wrong?

    Once you have v(t) ... you need to graph it.
    Usually you only need to sketch the graph - this is where you find out the turning points and roots etc to guide the sketch.
    You could plug in a lot of values for t and get lots of v out and plot the points and hope you have enough to join up to make a reasonable graph .... but it is better to use your understanding of functions to draw a sketch. Like if you wanted to draw a parabola, you'd find the turning point and the roots and just sketch a reasonably parabola-shaped curve through those. In fact you will have the sum of a sine with a line ... so figure out what that looks like. It may help to put dotted lines in to show the voltage vs time function on the same axes.
     
    Last edited: May 2, 2016
  9. May 9, 2016 #8
    To correct myself the phase on ir is π/3 i typed it wrong in my first post as I see now.
    I solved the problem but my profesor told me to calculate the active power thorugh this formula P=1/T ∫p(t) dt, I was calculating it through P=U*I*cos φ, this is not a problem for me now to calculate the active power through the integral formula for the resistor, but how do I calculate reactive power on the coil and the capacitor without using this formula Q=U*I*sin φ?
     
  10. May 9, 2016 #9
    You know that power = voltage*current, which applies to all devices. You also presumably know the terminal relations for your devices (i.e. V = -Ldi/dt for an inductor and i = Cdv/dt or v = (1/C)*integral(i*dt) for a capacitor).

    So, first differentiate your current for the inductor and multiply by -L. Then multiply that result by the current through the inductor and that is your P(t). Then integrate P(t)dt from 0 to t to find energy as a function of time for the inductor.

    Next, apply a similar procedure for the capacitor but instead of differentiating the current you are integrating it. Then multiply by 1/C (note there is no negative sign for a capacitor). Then multiply that result by your current through the capacitor to find power. Integrate that to find energy.

    If you can do those two things applying i^2R for the resistor will be easy.

    There are only two things you need to know here I think. The terminal relations for your devices and P = V*I. You also need to know that energy is power integrated over time.

    Try to break the problem up into pieces.
     
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