Help with parallel complex number resistor-inductor

Discussion in 'Homework Help' started by Starky, Apr 25, 2012.

  1. Starky

    Thread Starter New Member

    Jun 12, 2009
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    So I'm working on a problem and have gotten a little stuck, and figured I'd ask you fine gentlemen to help me out.

    I'm doing some thevenins on an AC RLC, and have gotten stuck applying parallel resistance to complex numbers.

    The values are R1 = 120, and Z1 = 95 + j12.5

    So am I right in thinking that I need to convert these to polar form before I can can use 1/(Z/1)+(1/Z2).

    Which gives 120 <0 and 95.819 <7.496

    Do I just do 1/(1/120)+(1/95.819) giving 53.277 <7.496?

    Is that correct?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    First, write your expressions so that they match what you mean, instead of forcing others to guess at what you meant to write. Remember, multiplication and division take precedence over addition and subtraction.

    So when you write 1/(Z/1)+(1/Z2), what you are saying is [1/(Z/1)] + (1/Z2) which is Z + (1/Z2). Now, I have no idea what this Z is that you are using, since you've only mentioned R1 and Z1.

    Similarly, 1/(1/120)+(1/95.819) is 120 + (1/95.819)

    Please repost, being more careful with your notation. Also, please carry your units in your work and show how you got your answers, because your problem appears to be sloppiness in working with complex numbers, so only by seeing the steps you took can we identify where the sloppiness crept in and the error got made.
     
  3. Starky

    Thread Starter New Member

    Jun 12, 2009
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    Sorry about that, was getting pretty late when I wrote that, anyway...

    Anyway I have a resistor and an inductor in series.

    The resistor is 120 Ω, the Inductor is 95 +j12.5

    I'm trying to solve these as parallel resistor using the standard 1/R = 1/R1 +1/R2... method with impedances.

    Making the 120 ohm resister 120 +j0 and converting that to polar form gives:
    |Z1| = √(120^2 + 0^2)
    and
    θ = arctan(0/120).
    = 120 Ohm at 0°

    Likewise 95 + j12.5 gives:
    |Z2| = √(95^2 + 12.5^2) = 95.819
    And phase angle:
    arctan(12.5/95) = 7.496°

    Giving Z values of:
    Z1 = 120 at 0°
    Z2 = 95.819 at 7.496°

    So using Zt/1 = 1/Z1 + 1/Z2 to give:

    1/(1/Z1 + 1/Z2)
    =1/ (1/120@0° + 1/95.819@7.496)

    1/120 = 8.333*10^-3
    1/95.819 = 10.436*10^-3
    1/7.496° = -7.496°

    (8.333*10^-3) + (10.436*10^-3) = 18.769*10^-3

    Finally:
    1/18.769*10^-3 @-7.469°
    = 53.279 @ 7.469°

    Hope this makes a bit more sense,
    Thanks.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    This doesn't make sense.
     
  5. Starky

    Thread Starter New Member

    Jun 12, 2009
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    Series was an error and I could not edit it by the time I caught it.
    I meant parallel - given I'm doing Thevenin's - removing the power supplies and working out the 2 values as parallel. Basically following the example on http://www.allaboutcircuits.com/vol_1/chpt_10/8.html - but with the complex numbers.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    As I suspected, the problem isn't the circuit theory, the problem is being able to work with complex numbers.

    When you add (subtract) two complex numbers, you have to add (subtract) the real parts to get the real part of the result and add (subtract) the imaginary parts to get the imaginary part of the result.

    And, again, I strongly recommend that you carry your units through out your work. Most mistakes you make (though not this one) will affect the units and scream out that the enswer is wrong.
     
  7. Starky

    Thread Starter New Member

    Jun 12, 2009
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    Aren't all the units in this case just Ohms? The unit doesn't change as far as I can see.

    Maybe I'm misunderstanding what the website tutorials explain on complex numbers I was basing my work on the examples found here: http://www.allaboutcircuits.com/vol_2/chpt_2/6.html.

    Look at how they deal with the values and magnitude angles there, I still can't see where I've gone wrong.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Yes, all the starting numbers should have units of ohms. So they should all be written with units of ohms. But not all of the numbers in your example have units of ohms because some of the numbers are the recipricols of resistance and therefore have units of mhos, siemens, or inverse-ohms (take your pick which name you like best).

    As you go through and perform the math, you are taking a lot of recipricols. A very easy mistake for someone to make is to not take a recipricol where they should have or take one where they shouldn't have. If you don't track the units throughout your work, then you will probably not catch the mistake. But if you do, then you will probably catch the mistake immediately. Even if you only catch it later on, it will be a lot easier to walk back to where the units stopped working out and, when you do, that is almost certainly where your mistake lies.

    So follow along and see if this makes sense:

    The resistor is Z1 = 120Ω, the inductor is Z2 = 95Ω +j12.5Ω

    Because these are connected in parallel, the equivalent impedance is:

    <br />
\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2} <br />
\frac{1}{Z} = \frac{1}{120\Omega} + \frac{1}{(95+j12.5)\Omega} <br />

    Now, taking recipricols is most easily accomplished in polar form. Thus:

    <br />
Z_1 = (120+j0)\Omega = 120\angle0^\circ\Omega<br />
Z_2 = (95+j12.5)\Omega = 95.82\angle7.50^\circ\Omega<br />

    The recipricols of each of these is then:

    <br />
\frac{1}{Z_1} = 0.00833\angle 0^\circ \text{S}= 8.33\angle0^\circ \text{mS}<br />
\frac{1}{Z_2} = 0.01044\angle -7.50^\circ \text{S}= 10.44\angle -7.50^\circ \text{mS}<br />

    Notice how the units are no longer ohms, they are siemens (for which the symbol is an uppercase S).

    So now we have

    <br />
\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2} <br />
\frac{1}{Z} = 8.33\angle0^\circ \text{mS} + 10.44\angle -7.50^\circ \text{mS} <br />

    Notice how the units allow us to do a quick sanity check at this point. The left hand side has units of 1/Z, or conductance. The right hand side has two terms both of which have the same units (not only siemens, but specifically milliseimens), so that they can be added together at all. Furthermore, the sum has units of millisiemens which are units of conductance and, hence, are consistent with the left hand side.

    Up to this point, you are fine. It is your next step where you mess up. You appear to have added two complex numbers by adding their magnitudes and adding their angles. You can't do this. Convert them to rectangular form

    <br />
\frac{1}{Z_1} = 8.33\angle 0^\circ \text{mS} = (8.33 + j0)\text{mS}<br />
\frac{1}{Z_2} = 10.44\angle -7.50^\circ \text{mS} = (10.35 - j1.362)\text{mS}<br />

    and then add the real and imaginary components separately.

    <br />
\frac{1}{Z} = (8.33 + j0)\text{mS} + (10.35 - j1.362)\text{mS}<br />
\frac{1}{Z} = (18.68 - j1.362)\text{mS}<br />

    To take the final recipricol, convert back to polar form:

    <br />
\frac{1}{Z} = (18.68 - j1.362)\text{mS} = 18.73\angle -4.17^\circ \text{mS}<br />
Z = \frac{1}{18.73\angle -4.17^\circ \text{mS}} = 53.4\angle@4.17^\circ \Omega<br />

    If this seems pretty cumbersome, that's only because it is. You might be thinking that there has to be a better way. Well, usually you can leverage the algebra is you work the problem symbolically and only plug in actual values at the end. It also helps to have a firm understanding of complex arithmetic (only gained through practice).

    Here's how I would have approached it:

    <br />
Z_1 = R_1<br />
Z_2 = R_2 + jX_2<br />

    with

    <br />
R_1 = 120\Omega<br />
R_2 = 95\Omega<br />
X_2 = 12.5\Omega<br />

    We then crank the algebra:

    <br />
\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2} <br />
\frac{1}{Z} = \frac{1}{R_1} + \frac{1}{R_2 + jX_2}<br />

    Put everything over a common denominator in order to add the fractions:

     <br />
\frac{1}{Z} = \frac{R_2 + jX_2}{(R_1)(R_2 + jX_2)} + \frac{R_1}{(R1)(R_2 + jX_2)}<br />
\frac{1}{Z} = \frac{(R_1+R_2) + jX_2}{(R_1)(R_2 + jX_2)}<br />

    You can now just flip the result

     <br />
Z = \frac{(R_1)(R_2 + jX_2)}{(R_1+R_2) + jX_2}<br />
Z = R_1\frac{R_2 + jX_2}{(R_1+R_2) + jX_2}<br />

    Note that the above is simply the equivalent of the formula for parallel resistors, namely

     <br />
Z = \frac{Z_1Z_2}{Z_1+Z_2}<br />

    and we could have started from this point. This is actually the point at which a very common mistake gets made, namely writing this as the sum over the product. If you make this mistake, it messes up the units; but you can only detect that if you carry the units throughout your work.

    Now multiply both numerator and denominator by the complex conjugate of the denominator

     <br />
Z = R_1\left(\frac{R_2 + jX_2}{(R_1+R_2) + jX_2}\right)\left(\frac{(R_1+R_2) - jX_2}{(R_1+R_2) - jX_2}\right)<br />
Z = R_1\frac{(R_2 + jX_2)[(R_1+R_2) - jX_2]}{(R_1+R_2)^2 + X_2^2}<br />
Z = R_1\frac{[R_2(R_1+R_2) + X_2^2] +jX_2R_1}{(R_1+R_2)^2 + X_2^2}<br />

    At this point, do a sanity check on the units and then plug in the values, noting that R1 + R2 is 215Ω.

     <br />
Z = 120\Omega\frac{[95\Omega(215\Omega) + (12.5\Omega)^2] +j(12.5\Omega)(120\Omega)}{(215\Omega)^2 + (12.5\Omega)^2}<br />
Z = (53.25+j3.88)\Omega = 53.4\angle 4.17^\circ \Omega<br />

    Notice that this approach didn't require a single polar-rectangular conversion in either direction (except to express the result in both forms).
     
  9. Starky

    Thread Starter New Member

    Jun 12, 2009
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    0
    Wow, thanks for such a great explanation, I'm fairly sure I understand that - I'm going to work through a few more exercises and compare to simulated/calculated results to make sure I do.

    I see what you mean about carrying units, I'll defiantly have to keep that in mind. I'm not formally educated in mathematics (except for GCSE's/highschool - 10 years ago), and most practical uses I've used just cheat and take shortcuts with rough numbers being acceptable (The if something blows stick a larger resistor in method).

    So again, thanks for taking the time to help me out.
     
    Last edited: Apr 27, 2012
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