1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electrical energy converted to heat in coil

  1. Mar 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Given a coil with N turns, a radius of r, a resistance of R', and an induced current i'(t) running through the coil, determine the totale energy converted to heat in the coil for t>=0.


    EDIT:
    BnZkW10.png
    The circuit on figure 4 consists of a resistance R and a capacitor with capacitance C. The voltage over the capacitor is V0 for the time t=<0 with the shown polarity. A small coil with N turns and a radius r is placed in the same plan as the large circuit, at a distance d from the wire ab. The coils resistance is R'. At the time t = 0 the switch is turned on. Both circuits are stationary and we assume that ONLY the straight wire ab, which is closest to the coil, generates a magnetic field. We can ignore the self-inductance from both circuits.

    a.) Determine the current i(t) through the large circuit at the time t>=0.

    \begin{equation}
    i(t) = \frac{V_0}{R}e^{-\frac{t}{RC}}
    \end{equation}

    The wire ab is considered infinitly long and the magnetic field at the coil is assumed to be uniform (d>> r).

    b.) Determine the direction and size of the induced current i'(t) in the coil for the time t>= 0.

    \begin{equation}
    i'(t) = \frac{r^2N\mu_0V_0}{2dR^2CR'}e^{-\frac{t}{RC}}
    \end{equation}

    c.) Calculate the total energy converted to heat in the coil for t>=0.

    this is where the problem arises. The answer should be:

    \begin{equation}
    U = \frac{\mu_0^2N^2V_0^2r^4}{8R'R^3d^2C}
    \end{equation}

    2. Relevant equations
    I found the following, but it is for energy stored in a toroidal solenoid so I don't think it is correct:

    \begin{equation}
    U = \frac{1}{2}Li'(t)^2 = \frac{1}{2}\frac{\mu_0N^2A}{2\pi r}i'(t)^2
    \end{equation}

    3. The attempt at a solution
    Using the above equation:

    \begin{equation}
    U = \frac{\mu_0N^2r}{4}i'(t)^2
    \end{equation}

    But according to the answer this isn't correct.
     
    Last edited: Mar 18, 2016
  2. jcsd
  3. Mar 18, 2016 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Any formula for energy will have to involve time because the longer the duration that current flows, the more energy will be dissipated.

    Is there a possibility it may be a trick question? You have a current i'(t) flowing in a resistance R'.
     
  4. Mar 18, 2016 #3
    Well actually that formulation is my fault. This is the last problem of a 3 part problem, in the first two parts I found the current i'(t) that would be induced in a coil with a resistance of R', the last part simply asks what the total energy to heat conversion would be in the coil as a function of time.
     
  5. Mar 18, 2016 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    If current is i'(t), what exactly is capital I?
     
  6. Mar 18, 2016 #5
    Sorry, I should have been written as i'(t), will fix it.
     
  7. Mar 18, 2016 #6

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps remember that an ideal inductor is lossless. So no energy is converted to heat in the ideal part of a real inductor. What about losses in the other parts of a real inductor?
     
  8. Mar 18, 2016 #7
    Hmm... I think my explanation is causing more confusion than anything. Give me a few minutes to write the full original text in.
     
  9. Mar 18, 2016 #8
    I updated the question so all the details are there.
     
  10. Mar 18, 2016 #9

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Your equation 4 is for the energy stored in an inductor. Not the energy dissipated in the inductor.

    Consider a resistor R with current I(t) flowing through it. What is the simple equation for the instantaneous power dissipated in the resistor.
     
  11. Mar 18, 2016 #10
    Hey, thanks for replying. Was away when you sent that. The instantaneous power dissipated in the resistor is the product of the square of the current and the resistance over which the current runs?

    \begin{equation}
    P = I^2R \Rightarrow P' = i'(t)^2R'
    \end{equation}

    However this doesn't give the same result as the answer, this gives:

    \begin{equation}
    P' = \frac{r^4N^2\mu_0^2V_0^2}{4d^2R^4C^2R'}e^{\frac{t^2}{R^2C^2}}
    \end{equation}
     
  12. Mar 18, 2016 #11

    gneill

    User Avatar

    Staff: Mentor

    Power is not the same as total energy. Power is the instantaneous energy dissipation with respect to time. How might you find the total energy from the power?
     
  13. Mar 18, 2016 #12
    Intergrating over the power from 0 to t, at which point I get the same equation :) well sort of :( The problem is the integration of e^(x^(2)), I looked it up and it seems to involve 1/2 sqrt(pi) erf which isn't something I recall having had about in calculus. The answer sheet gives the equation without the sqrt(pi) erf part but otherwise seems to be the same as you can see from the OP. Am I making a mistake here?
     
  14. Mar 18, 2016 #13

    gneill

    User Avatar

    Staff: Mentor

    I think you'll find that the exponent of e in your equations should be negative. Also, note that ##(e^{-x})^2 \ne e^{(-x)^2}##. So you'll want to check the exponent of e in your power equation for two reasons.
     
  15. Mar 18, 2016 #14
    I ran it through wolfram alpha to check, and surely you were right about the exponent. I am however a little confused now because I get the following, keep in mind the i had to change the name of some things, like R' is now R_0, V_0 is now v etc.
    alw7DhA.png
    as you can see I now have a negative value and e is still present, which it shouldn't be.
     
  16. Mar 18, 2016 #15

    gneill

    User Avatar

    Staff: Mentor

    You've taken an indefinite integral, so you've ended up with an unknown constant. What are the integration limits for the time interval of interest?
     
  17. Mar 18, 2016 #16
    I see the problem :D its from 0 to infinity, not 0 to t. Also the t isn't multiplied on the fraction, its only present as an exponent, so t=0 doesn't make the entire thing 0 it just makes the e part equal to 1 and for t = infinity we get e expression becomes 0, which means we have 0 - (above expression without the e part), so double negatives cancel out. Sorry, I guess I'm just tired... Should have caught that mistake.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electrical energy converted to heat in coil
  1. Heat generated in coil (Replies: 22)

Loading...