1. Not finding help here? Sign up for a free 30min 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!

Is there a faster and easier way to find these values in a circuit

  1. Jul 23, 2013 #1
    Is there any other easier way to get the answer for no. 2?

    Power supply voltage E=8+4j [V]; Current I=1-j [A].
    1) Find the power factor of the circuit as seen from the power supply.
    2)Find the values of x and y.

    There's a little mistake at the very last part of my working, but anyways, I want to know if there is an easier way.
     

    Attached Files:

    Last edited: Jul 23, 2013
  2. jcsd
  3. Jul 23, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    I can't think of a distinctly easier method. You have two unknowns so you have to come up with two equations. Equating the terms of the known impedance to those of the expression for the impedance seems logical. After that you bring to bear any algebra tricks you've learned. (hint: consider forging a new expression by taking the ratio of the terms)
     
  4. Jul 23, 2013 #3
    How to do that? :eek:
     
  5. Jul 23, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    First write out the two terms (real and imaginary) equated to their know values.
     
  6. Jul 23, 2013 #5
    That's what I did. But I still don't get the answer :/
     
  7. Jul 23, 2013 #6

    gneill

    User Avatar

    Staff: Mentor

    Looking at your work, I don't clearly see the two separate equations as such. Can you write them out?
     
  8. Jul 23, 2013 #7
    I got it!! There was a lot of miscalculations before. I got x=5, y=5/2. Correct?
     
  9. Jul 23, 2013 #8

    gneill

    User Avatar

    Staff: Mentor

    Your result looks fine.
     
  10. Jul 23, 2013 #9
    Thanks!!!!!I hope I pass this subject.
     
  11. Jul 23, 2013 #10
    Yes, there is an easier way.
    Once you’ve divided the complex source voltage by the complex current, and subtracted the (1+j8) you have the series equivalent of the other components. This type of calculations go back and forth between series and parallel configurations, so it is a good idea to develop formulas or even functions you can use for those conversions over and over again.

    Instead of using the method taught in class to multiply and divide complex numbers by converting back and forth from rectangular to polar, here is a short cut.

    The series value of Xs and Ys is 1-j2.
    Xp = (Xs^2 + Ys^2)/Xs = (1^2 +(- J2^2))/1 = 5
    Yp = (Xs^2 + Ys^2)/Ys = (1^2 +(- J2^2))/-j2 = -j2.5

    This shows the answer you arrived at for Yp should be –j2.5 instead of +j2.5.
     
  12. Jul 24, 2013 #11
    WOW!I didn't see that earlier. That is definitely a shortcut. By the way, in the diagram it already shows the negative sign with y, so all we need is the y value, without the negative sign. But I don't understand this part:

    How did you get to find the parallel values like that?
     
    Last edited: Jul 24, 2013
  13. Jul 24, 2013 #12
    I derived that about 20 years ago but I no longer remember the derivation. I haven't seen it published anywhere. Based on these formulas, can you figure out how to go from parallel to series?

    Here is another method. http://www.eetimes.com/document.asp?doc_id=1278134

    By the way you can use Excel as a pretty good complex arithmetic calculator, even if a little clunky. You first have to load it into Excel. To do that, click on the question mark inside the blue circle at the upper right in Excel and type "complex" into the search window and follow the instructions. This allows you to set up the formulas where all you have to do is plug in values. You can even create your own functions.

    The Smith Chart or rather a version of the Smith Chart called the Immittance Chart can also do the conversion graphically. An explanation of the Smith Chart and how to use it is a little long to include here. http://www.microwavesoftware.com/manuals/imchart.html
     
    Last edited: Jul 24, 2013
  14. Jul 24, 2013 #13
    Thanks for the info, but in exams, we're not allowed to use calculators and definitely not excel. So I guess, I'll just equate the terms then. But it really got easier when we subtract the known impedance from the total impedance, rather than summing all the impedance. Thanks!
     
  15. Jul 24, 2013 #14
    I do remember how I found the formulas for series to parallel conversion. They’re nothing more than a rewriting of this formula which is used for impedance matching. I got this first formula from a Motorola (now Freescale) RF Transistor Data book. It’s amazing that as simple as this is, it’s not more widely used.

    Xs = √ (Rp * Rs – Rs^2)

    Rp = (Xs^2 + Rs^2) / Rs

    It was probably a lucky guess that by replacing Rs with Xs would give Xp.

    Impedance matching is really nothing more than series to parallel conversion. Since the components in parallel always have higher values than the equivalent impedance in series, matching is done by selecting the series reactance that will convert the lower resistance to the higher resistance in a parallel configuration. The first formula is how you calculate that series reactance.

    For instance, suppose we want to match an impedance of 50 + j0 to a load of 10 + j15. First we would calculate the reactance needed to transform a series 10 + j15 to a parallel 50 +jX. We do that with the top formula and get plus or minus 20 Since we already have j15 ohms of reactance we only need j5 more. We get that by adding an inductor of j5 ohms in series.

    Now we convert to parallel and get 50 + j33.33. Now all we have to do is add a shunt capacitor with a reactance of –j33.33 to cancel the +j33.33 and we have our match.

    P.S. My suggestion to use a calculator or Excel to do the calculations wasn't meant for use during tests but for homework and later on the job.
     
    Last edited: Jul 24, 2013
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: Is there a faster and easier way to find these values in a circuit
Loading...