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Homework Help: Reduce a polar quantity by a percentage

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data

    reduce a current in its polar form 95 -46.37° by 20%

    2. Relevant equations

    3. The attempt at a solution

    When dividing a polar number by a scalar one you just divide the magnitude by the scalar, the phase will remain unchanged, so to reduce the polar value by 20% is just some division and multiplication, that is, 95 -46.37° minus 95 -46.37°/ 100 * 20 so according to the rule during the multiplication and division the phase will remain unchanged but is this correct?

    So the answer I have is 76 -46.37°

    But if the magnitude is reduced by 20% does the phase also get reduced too?

    thanks for any help
  2. jcsd
  3. Jul 25, 2009 #2


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    Your answer is correct. Imagine a linear system that causes a phase shift of -46.37° in an incoming signal.
    If you decrease the amplitude of your incoming signal by 20%, the amplitude of the output will be reduced by the same percentage, but the phase shift will be unchanged.
  4. Jul 27, 2009 #3
    Thank you for the reply, I appreciate it, I'm even more confused now as the full question relates to impedance/ admitance triangles (Z²=R²+X² / Y²=G²+B²) and asks for the new power factor of the system if the current drops 20%, however as the power factor is derived from the cos of the phase angle then as this remains the same then the power factor remains the same??
    The only way I can get a sensible answer is to say that the Conductance G of the circuit has not altered for the drop in Amps through it (is this a fair comment?) Then I can derive a new phase angle value and therefore power factor value but the phase angle as said above doesn't change with the drop in Amps???

    Thanks again
  5. Jul 27, 2009 #4


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    Unless the question is ill formulated, the power factor wont change.
  6. Jul 27, 2009 #5
    Ah cool so its a bit of a trick question then?
    What it asks is:
    What value of capacitor is required in parallel to the load to reduce the supply current of 95 L-46.37º by 20% and the new power factor with the capacitor in circuit.

    So from my answers I have devised a value for the capacitor but the power factor remains the same, therefore it is a trick question and the supply current drops but the power factor infact remains the same?

    Thank you again for your help i understand this much better now
  7. Jul 27, 2009 #6


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    No, it is not a trick question. You have originally stated the question wrongly.
    Now, with the correct statement we can work on it.
    Clearly your original load has a resistance and an inductance in parallel. From the total value of the current and the angle you can calculate the currents in the resistor and in the inductor.
    Now, if you put a capacitor in parallel, the current in the resistor will not change, but the current in the reactive part of the circuit will.
    Since you know the new value of your hypotenuse and one of the legs of your triangle has not changed, you can calculate the new leg and the new power factor.
  8. Jul 28, 2009 #7
    Ok Cel, thank you, I should of specified the question better but I thought i had the thing sussed out once I worked out how to reduce the polar current by 20%

    The question is related to the admittance triangle Y²=G²+B²

    The load on the circuit is represented in polar form by 3 admittances, I have resolved these into their conductances G and susceptances B and from the triangle relationship worked out the total admittance Y.
    The values are:
    Y = 0.2159 S
    G = 0.149 S
    B = -0.1563 S

    Then again from the triangle relationship Φ = arctan (B/G) have worked out the phase angle of the admittance to be L-46.37º

    Now using the value of Y and the known Voltage (also in polar form 440 L0º) used I=YV to get the value of I in its polar form 95 L-46.37º.

    And as the cos of the phase angle is the power factor (from true power =Voltage supply x total current x cosΦ) this works out at +0.69

    I am happy with all the above I think?!

    Now it asks me (hence the original question of the post) to reduce the supply current by 20% by putting a capacitor in parallel with the load, what size capacitor is needed and what is the new power factor.

    So this is where I now get confused:

    I can easily derive a new value for Y using Y=I/V = 0.1727 L-46.37º

    If the phase angle of the current is not altered when it drops 20% only the magnitude then the above calculation is true, but now all the derived formula use the -46.37º as the angle in all the relationships, as I believe the next steps would be (and these are the steps that the text sort of leads you down):

    Calculate the new value of conductance G=Ycos -46.37º

    Plug the values of Y and G into Y²=G²+B² to get a value for B

    then use the difference between the original circuit value for B and the new value for B in the equation

    Capacitance = Bc/2Πf - this is the equation it is leading you to use as we have a value of 50Hz for the frequency f.

    Then this gives the capacitor value C

    I guess the part I am really not understanding is the original post about the 20% reduction in current not affecting the phase angle of that current which then leaves the new admittance triangle with the same angle??

    the good thing to come out of this so far is that I understand all the concepts much better!

  9. Jul 28, 2009 #8


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    No, I said that the phase would not change because you did not state the whole problem. With the complete statement, phase will vary. The new value of Y will be 20% smaller in magnitude. The phase will change and is is yet unknown. The conductance G is unchanged, so you must alter B in order to have the new Y.

    B must be changed by the addition of a capacitor in parallel to the load. What is the conductance of this capacitor to change the value of B?
    Knowing the conductance and the frequency you can obtain the capacitor.
  10. Jul 28, 2009 #9
    Thanks again Cel.

    Ok, so if the conductance of the circuit does not change then it is pretty straightforward....but I had considered this before as a worked example in the text had conductance not changing but this was for a circuit which did not have a new component added, whereas this one has a capacitor added (for which I am not given any data whatsoever) so the conductance will change wont it?
    A worked example in the text seems to suggest that you use the difference in the original and recalculated value of B to calculate the Capacitor value using C=Bc/2Πf.

    So if I put my figures in:

    Original current I=95 L-46.37º Amps

    Reduce it by 20%

    I=76 Amps (magnitude only changes not the phase angle)

    Now we can get a new value of Y from Y=I/V Y=76 L46.37º/ 440 L0º

    Y=0.1727 S

    G remains unaltered so from Y²=G²+B²

    B is now = 0.0872 S

    And from B=YsinΦ the new phase angle works out at 26.90º

    Now using the difference in values of B and putting it in C=Bc/2Πf

    C = 0.000775 or 775pF

    Then from Φ = arcos of power factor

    power Factor now = 0.89

    I would be happy to submit that as an answer as I understand the concepts, the value of the capacitor looks ok and the power factor has indeed improved.
    I actually have these answers from one of my rough attempts but couldn't grasp how the conductance of the circuit had not changed with the inclusion of a new component, the text doesn't ellude to this anywhere so would it be such a small change that for practicle purposes it could be ignored?

    Thank you again
  11. Aug 4, 2009 #10
    Hello Alfie

    how did you get on with that question ive had a similar problem relating to it as you did but im now on the next part which asks for the size of capacitor to bring the circuit to unity, but my calcs infer it is a smaller sized capacitor than the one that reduces it by 20%


    regards ian
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