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Homework Help: Calculate the total current

  1. Jul 21, 2005 #1
    A coil of self-inductance 0.7H is joined in parallel with a non-inductive resistance of 50 ohm. Calculate the total current, the current throught the wattless and power components when connected to a supply of 200V at a frequency of 50Hz.

    Here is my steps:
    Impedance Z= root [ 50^2 + (2pi*50*0.7)^2 ] = 225.5 ohm
    Total current = 200/ Z = 0.9756A

    Am I right? as for the other two questions, i really don't know how to solve them. :confused: Please help me with it. Thank you!

  2. jcsd
  3. Jul 21, 2005 #2


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    You have treated the resistor and coil as a series combination. The problem says they are in parallel.


    When you do the first part correctly, you will be able to find the total current. You can find the current in each branch, and therefore in each device using the individual impedences and the applied voltage.
    Last edited: Jul 21, 2005
  4. Jul 21, 2005 #3
    the equation for total impedance Z you used above would apply to a series circuit with coil inductance 0.7H and resistor resistance 50 ohms.
    however, in this case, these 2 components are in parallel, and you would first need to compute Ztotal using (complex impedances):

    [tex] \frac{1}{Z_{total}} \ = \ \frac{1}{Z_{coil}} \, + \, \frac{1}{Z_{resistor}} [/tex]

    and then use:

    [tex] Total \ Current \ Magnitude \ = \ \frac{200 \ Volts}{|Z_{total}|} [/tex]

    an easier method for this parallel circuit is to sum the currents thru each branch of the parallel circuit:

    [tex] \mbox{Current Thru Coil} \ = \ \frac{200}{2 \pi (50)(0.7)\mathbf{j}} [/tex]

    [tex] \left ( \ \ \mbox{Current Magnitude Thru Coil} \ = \ \left | \, \frac{200}{2 \pi (50)(0.7)\mathbf{j}} \, \right | \ = \ \frac{200}{2 \pi (50)(0.7)} \ \ \right ) [/tex]

    [tex] \mbox{Current Thru Resistor} \ = \ \frac{200}{(50)} [/tex]

    [tex] \mbox{Total Current Magnitude} \ = \ \, \left \Large | \, \mbox{Current Thru Coil} \ + \ \mbox{Current Thru Resistor} \, \right | [/tex]

    this also answers the last 2 questions of your problem.
    (*** Clarifications/corrections added from suggestions by OlderDan ***)
    Last edited: Jul 21, 2005
  5. Jul 21, 2005 #4


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    The currents are not in phase. You cannot simply add them. Your first equation is valid only if complex impedences are used. I fixed the link in my earlier post. Check it out.
  6. Jul 21, 2005 #5
    Hello olderdan and geosonel, thank you for your help. I realise I have treated the problem wrongly. However, when I follow your method, i get the total current to be 4.9095 ohm; and the answer given is 0.8865A. Yet, the other two answers are 0.8642A and 0.1964A respectively, which do not add up to 0.8865A...What do you think? :)
  7. Jul 21, 2005 #6
    OlderDan -- your comments are, of course, correct.
    the entire Msg #3 was originally intended to incorporate COMPLEX impedances and use complex representations. clarifications/corrections have been added based on your suggestions.

    Clari --
    changes were made to Msg #3 to better indicate the COMPLEX representations of the various quatities. do you know how to use these complex representations? specifically, the total current would be equal to:
    total current = (complex current thru coil) + (complex current thru resistor)
    = (-0.9095j) + (4)

    total current MAGNITUDE = | (-0.9095j) + (4) |

    [tex] \ = \ \sqrt{(0.9095)^{2} + (4)^{2}} \ = \ 4.102 \ amps [/tex]

    the individual current magnitudes must be combined like shown above because the current thru the coil and current thru the resistor are out of phase (which is why they are represented by complex quantities).

    apparently the book is wrong about the answer. either that or you may have made a careless error in copying the problem. you might double check the problem.
  8. Jul 21, 2005 #7


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    I think you have not treated the impedence as complex when calculating the equivalent impedence, and the unit of "ohm" in your result should be "amp". I see no way of getting the answers you have been given. A 50 ohm resistor by itself connected to a 200 volt supply would give a 4 amp current. A parallel path through the 0.7H inductor would give a current of 0.9095 amp at 50 hz that is 90 degrees out of phase with the resistor current. You cannot simply add these two (which is effectively what you did by not calculating the equivalent impedence correctly) because they are out of phase, but even when added correctly they still add up to about 4.10 amps, which is a lot more than the answers you are given.

    EDIT: Late to the party again :smile: See geosonel's reply for more detail on the calculations

    Please check the problem and make sure you are stating it correctly. The ratio of the currents given for the separate currents is about right. Is there something else in series with the parallel combination? Do you have the voltage right?
    Last edited: Jul 21, 2005
  9. Jul 23, 2005 #8
    I think your answers are right, while those given by my teacher are wrong, so i believe i know how to sort it out now. ^^ Thank you very much!! :smile:

    To geosonel, i haven't learned anything about the RLC circuit in parallel in fact, so i know nothing about complex representations. What is j in:
    To olderdan, i have stated out the problems clearly, and there's certainly nothing in series with the parallel combination.
  10. Jul 23, 2005 #9
    Clari -
    we didn't know your mathematical background when suggesting the "complex representation" method for AC circuits. unfortunately, it's difficult to provide a complete tutorial on AC circuit methods in this forum.

    if you're interested in learning what the "j" signifies in the previous msgs and how the "complex representation" method works, try the tutorial given in the URL link below. (this tutorial takes 8 pages, including the introduction). this link also provides good info on many other electronic circuit topics.
    Last edited: Jul 23, 2005
  11. Jul 23, 2005 #10


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    I have done the problem with very basic method by taking voltage V = v(max)sin(wt)
    taking the total instantaneous current then finding the rms value of it using integration method and got the same answer 4.102 A as by geosonel
    Last edited: Jul 23, 2005
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