Answer check to average power (simple)

Discussion in 'Homework Help' started by ihaveaquestion, Sep 20, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    I've checked my work at least 4 times now and I see absolutely nothing wrong, but apparently my answer is wrong.

    http://img23.imageshack.us/img23/2991/imgdi.jpg

    The answer in the back of the book is 11.3 W through R1.. I'm obviously not getting that.... what's up?

    Thanks in advance.
     
  2. t_n_k

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    Mar 6, 2009
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    Taking a different approach I calculated the impedance seen by the source.

    I get Z=4.6-j1.2 ohms

    This gives an RMS current of 2.38A from the source.

    So the current in the 1 ohm resistor is 2.38A rms.

    Hence power in 1 ohm, P = 1x(2.38)^2=5.66 Watts

    So we have even more confusion with 3 different answers!
     
  3. t_n_k

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    Mar 6, 2009
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    Interestingly 5.66 W is close to half of 11.3W .... this might be worth considering - maybe the book is wrong ....???
     
  4. t_n_k

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    Mar 6, 2009
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    I have run a simulation of the circuit to check values - 5.66W looks about right. Have you double-checked that you have correctly transcribed everything from the book problem statement to your working notes?
     
  5. ihaveaquestion

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    May 1, 2009
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    I have double and triple checked, haha...
     
  6. ihaveaquestion

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    I've caught the book being wrong at least 3 times now....

    I still can't find out what's wrong with my approach though...
     
  7. The Electrician

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    Oct 9, 2007
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    I get the same result that tnk does, calculating the impedance seen by the source.

    One error I see in your calculations is in the 4th equation down under the heading "For I1".

    You have:

    j6 I2 = -12.3 + 10.3 + I1(1 + j6)

    You divide left and right side by j6 and get:

    I2 = -1.72 - j2.05 + I1(-1 + j0.167)

    but you have sign errors; the result should be:

    I2 = +1.72 + j2.05 + I1(+1 - j0.167)

    This propagates to your final result causing a substantial error. There may be other errors, but I stopped looking when I found that one.
     
  8. ihaveaquestion

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    May 1, 2009
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    The line I think you're looking at is 6i I2 = -12.3 + 10.3i + I1(1+6i)

    I just plugged into my calculator (-12.3 + 10.3i) / 6i for the part without I1

    and it gives me -1.72 - 2.05i

    Then plugging in again (1+6i) / 6i for the I1 part

    it gives me -1 + 1/6i

    So I2 = -1.72 - 2.05i + I1(-1 + 1/6i) which is my reasoning for having that written
     
  9. The Electrician

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    Let's try doing it by hand.

    We know that 1/(6i) = -i/6

    then, for example, taking just the real part of (-12.3 + 10.3i) we would have:

    -12.3/6i = -12.3 * -i/6 = (12.3/6)i = (2.05)i = 2.05i, not -2.05i

    And for the imaginary part:

    10.3i/6i = 10.3i * -i/6 = -(10.3i/6)i = -(1.72)i*i = 1.72, not -1.72

    So your calculator is saying that (1+6i) / 6i = -1 + 1/6i

    Let's do it part by part.

    (1+6i) / 6i = (6i + 1)/6i = 6i/6i + 1/6i

    Where does the -1 come from? 6i/6i = 1, not -1.

    Next consider the second part. 1/6i = -i/6 = -0.167i, not +0.167i, as you have in the image attached to post #1.

    Unbelievable as it may seem, apparently your calculator is getting the signs wrong. What kind of calculator are you using?
     
  10. ihaveaquestion

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    May 1, 2009
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    What you're saying makes sense Electrician... very frustrating.... it's a Casio 115ES
     
  11. ELECTRONERD

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    May 26, 2009
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    I have the exact same calculator. :rolleyes: Funny it isn't working like it should...
     
  12. The Electrician

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    What do you get if you calculate 6i/6i?
     
  13. ELECTRONERD

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    May 26, 2009
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    For some reason I can't use "i" in my calculator. It isn't working...I need to know what settings the calculator should be on...
     
    Last edited: Sep 21, 2009
  14. ihaveaquestion

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    May 1, 2009
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    I just pressed the Mod/Setup button and pressed 2 for complex numbers.
     
  15. ihaveaquestion

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    How stupid... you have to put parenthesis around (6i) to get the right answer... i.e. (1+6i) / (6i)
     
  16. ihaveaquestion

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    May 1, 2009
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    Pretty sure I'm still not getting right answer with right calculations...
     
  17. ELECTRONERD

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    May 26, 2009
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    Ok, I have the CMPLX on the screen but I press A and then the <Shift> ENG button to get i, but it won't come out.
     
    Last edited: Sep 21, 2009
  18. ELECTRONERD

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    May 26, 2009
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    Nevermind, I figured it out. When you do (1 + 6i) / (6i) I get 1 / (-1/6i). Is that what you get? I do have the right "i" don't I? :D
     
  19. The Electrician

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    You should get (1 + 6i)/(6i) = 1/(6i) + 6i/6i = -.0167i + 1 = 1 - 0.167i

    I redid the calculations under the heading "For I1"

    When you get down to the place where you had the sign error, you should have:

    I2 = +1.72 + j2.05 + I1(1 - j0.167)

    Then substituting I2 = (j6 I1)/(2 + j4) you should have:

    j6 I1 = (2 + j4)(1.72 + 2.05 + I1(1 - j0.167)

    and finally:

    I1 = (-4.76 + j10.98)/(-2.67 + j2.33) = (3.05 - j1.44)

    The magnitude of I1 is 3.375 amps.

    The average power is 1/2*(3.375)*(3.375) = 5.695 watts

    Your numbers may be slightly different due to rounding error from using only 3 digits in the arithmetic, but this is essentially the same result tnk and I got using the impedance seen by the voltage source.
     
    Last edited: Sep 24, 2009
  20. The Electrician

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    Anyone who is going to do a lot of complex arithmetic should have a calculator that isn't going to have weird behavior that trips you up like this Casio. Imagine if you were trying to solve a problem on an exam and didn't know about this peculiarity!

    I would recommend something like a TI89 or HP50. You can get those on eBay for a quite reasonable price. Instead of the current production HP50, get the just barely obsolete HP49G+ for less money, but the same performance.
     
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