1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: RLC AC circuit analysis.

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data


    A1 = 5A, A2 = 4A, A3 = 3A. V = 120V

    a) Given the circuit above, find A4

    b) Find the angle between E and IT. Draw a diagrammatic (?) and find the power in the circuit.

    c) Find the values of R, L and C.

    2. Relevant equations

    3. The attempt at a solution


    IT is 5A.
    R has 120V across it, and 4A through it. Therefore ZR = 30Ω∠0°
    L has 120V across it, and 3A through it. Therefore ZL = 40Ω∠90°
    C has 120V across it, and xA through it.

    I'm not sure how to proceed here. Can I use admittance, somehow?

    I'm thinking I should be able to find the angle of the current through the R and L components (the current should be 0° and 90°?) Whatever's left goes through the capacitor (at -90°?). How can I figure out how to compute this?

    Attached Files:

    Last edited: May 3, 2014
  2. jcsd
  3. May 3, 2014 #2


    User Avatar

    Staff: Mentor

    The phasor for the current through the resistor lies on the real axis of the complex plane. That of the inductor lies along the negative imaginary axis, and that for the capacitor along the positive imaginary axis. That is to say, if we were to assign current phasors you'd have

    ##I_2 = A_2##
    ##I_3 = -jA_3##
    ##I_4 = jA_4##

    Now, A1 is the magnitude of the total current phasor. Can you write an expression for the magnitude of the total current in terms of the current phasors? Can you solve for A4?

    By the way, your calculations for R and L are fine.
  4. May 3, 2014 #3
    Isn't it the other way around? Inductor is positive, capacitor is negative?

    To be honest, no. I'm not sure I understand what you're asking.

    I understand phasors. I'm not sure I understand the magnitude of the totalt current (A1) in terms of the current phasors.

    I'm sorry I can't show more effort. The only thing I've done is writing the expression you lined up, but to me, IT = I1, but at the same time, I'm confused about the phase angle of the total current.

    So what I've done is 5A = 4A - j3A + jxA

    jxA = 1A + j3A which is 3.16A at 71.57°

    I know this doesn't make any sense at all. The real part should be 0.
  5. May 3, 2014 #4


    User Avatar

    Staff: Mentor

    Consider the impedances of inductors and capacitors. Ignoring the magnitudes, inductors are +j while capacitors are -j. But by Ohm's law currents are given by I = E/Z; the impedance is in the denominator. But 1/(+j) = -j, and 1/(-j) = +j. So the sign of the imaginary component of the current is the opposite of that of the impedance.

    You don't care about the phase angle of the total current at this point. The value you're given is the magnitude. The total current phasor is given by the sum of the individual current phasors (the branches are in parallel so the currents add). Their phasors are specified in complex form: one is real and the other two imaginary. How would you take the magnitude of their sum? (How would you write the magnitude of a complex number?)
  6. May 3, 2014 #5
    Alright, I understand.

    The magnitude of a complex number would be the square root of the imaginary part squared, plus the real part squared.

    I'm sorry, but I still can't figure this out. One of the imaginary parts is what I'm after. The way I see your explanation at this point is A4 = ## √(4^2 + 3^2) ## = 5. Which is wrong.
  7. May 3, 2014 #6


    User Avatar

    Staff: Mentor

    A4 is not the total current. A4 is the capacitor current. What is the expression for the magnitude of the total current A1?
  8. May 3, 2014 #7
    A1 = A2 + A3 + A4, I would assume.

    More precisely, since we're working with complex numbers:

    A1 = ## √(A2^2 + A3^2 + A4) ##

    My problem with this logic, which is my own, is that A2 and A3 adds up to A1. A4 must then be 0, which doesn't make much sense.

    I'm confused, that's for sure.
  9. May 3, 2014 #8


    User Avatar

    Staff: Mentor

    That's not right. The reason that I suggested looking at the currents instead of their magnitudes was to point out that only one of them is associated with a real value and the other two are imaginary, and to fix the appropriate +j or -j to the imaginary quantities. It's okay to work with the magnitudes (A1 through A4) provided you remember which ones correspond to real and imaginary quantities.

    A2 is the magnitude of the real current flowing in the resistor. A3 is the magnitude of the imaginary valued current flowing through the inductor. A4 is the magnitude of the imaginary valued current flowing through the capacitor. One is real the other two imaginary. In sum there is a total real value and a total imaginary value. The magnitude of the whole is what you are looking for.

    A2 and A3 do not add up to A1. A2 and A3 and A4, taking into account their phase angles, add up to A1 if you are talking about phasor magnitudes. Treat them like complex numbers and add them accordingly.
  10. May 3, 2014 #9
    So 4 is the total, and only, real part. Since I1, the total current, is 5, the only imaginary part that satisfy that "equation" is 3. Therefore it has to be +3 or -3.

    The known imaginary part is -3. Therefore the imaginary part of A4 has to be 0 or +6.

    Am I making sense here?
  11. May 3, 2014 #10


    User Avatar

    Staff: Mentor

  12. May 3, 2014 #11
    Oh dear, that's a relief. Sometimes my brain refuse to make sense of things.

    I'm glad there's the infinite space of the Internet between us sometimes. I don't have to face my embarrassment.

    Thanks a lot for helping me, I really appreciate it.
  13. May 4, 2014 #12
    New day, new possibilities.

    For my b) problem: Find the angle between E and IT. Draw a diagrammatic (?) and find the power in the circuit.

    The angle of E is 0°. I'm just assuming this. Is E "always" said to be 0° because phase angle is relative?

    The angle of IT is 4A [itex]\frac{+}{-}[/itex] j 3A
    IT = 5A [itex]\frac{+}{-}[/itex]∠36.87°

    The phase angle between E and IT is therefore [itex]\frac{+}{-}[/itex]∠36.87°

    For the diagrammatic (is this the correct term?), I've drawn the attached picture below. Is this correct? It shows that the current is at the phase angle I found above - assuming that is correct. I don't see any usefulness for the drawing, however.


    Is there any way I can use it to find the power in the circuit?

    Attached Files:

  14. May 4, 2014 #13


    User Avatar

    Staff: Mentor

    Theoretically the reference angle is arbitrary, but by convention it is almost always provided by the source powering the circuit, and usually its voltage waveform.

    There will be only one phase angle for the current. It can't be both + and - some angle.

    You should have the magnitudes of each of the branch currents now along with their phase angles. An arithmetic sum will give you the net current in complex form (its magnitude should be A1). Find the phase angle of that complex current value.

    You've drawn a diagram showing the source voltage phasor and (once you've decided on a single phase angle) the current phasor. I suppose that the real power might be found as the dot product of the current vector with the voltage vector. Maybe your book or class notes have some other suggestion?

    Another possibility is that they want you to draw the current triangle showing the magnitude of the current on the hypotenuse and the real and imaginary currents as the legs of the right triangle. Scale the sides by E (essentially multiplying current by voltage) to turn it into a power triangle.

    Usually I just take the complex product P = EI*, where I* is the complex conjugate of the current. This yields P as the complex power, giving you both the real and imaginary parts of the power in one operation.
  15. May 4, 2014 #14
    I assumed it can be 2, therefore I made two calculations (putting them into one). One if the current through the capacitor is 0, and one if the current through the capacitor is 6A.

    Thank you. The magnitude of the power is 600W right? 120V * 5A. The complex conjugate will change the phase angle from + to - or - to +.
  16. May 4, 2014 #15


    User Avatar

    Staff: Mentor

    Ah. You might want to make that more obvious in the work that you hand in. And perhaps use different colors for the two solution vectors.

    That would be the Apparent power, not to be confused with the Real power or the Reactive power. Only Real power is actually consumed by the load (turned into work or heat).
  17. May 4, 2014 #16
    As everything else on this forum, that's actually not a bad idea.

    Yes, of course, the power triangle. The calculations I've done with the real power has always been with given resistance. Earlier I found the resistance to be 30Ω

    Will the real power in the circuit then be P = I2 * R = 52A * 30Ω = 750W?
  18. May 4, 2014 #17


    User Avatar

    Staff: Mentor

    The total current is a complex value with magnitude 5. The real part of the total impedance won't be R but rather the real component of the total impedance (something closer to 20 Ω --- you can calculate a more accurate value). The real power will depend upon the phase angle of the current with respect to the voltage, and is in fact the basis of what is called the power factor which is the cosine of the phase angle.

    As an alternative, knowing that the resistor is the only component that will dissipate real power you might choose an alternative formula for calculation power that takes advantage of the known voltage across the resistor or the given current magnitude that flows through the resistor (A2).
  19. May 4, 2014 #18
    Lets go with the easier suggestion first.

    The voltage is 120V, the current is 4A at 0° thus the power will be 480W?

    Can the power have an angle? I would assume yes, the combined angle of the current and voltage, depending on the magnitude of the two?

    In this case, both the voltage and current is at 0° therefore, the power will be at 0°?
  20. May 4, 2014 #19


    User Avatar

    Staff: Mentor

    Yup. That's the real power that the circuit dissipates.

    Sure in the general case power is a complex quantity.

    Yes, it's purely real power as is to be expected for a resistor.

    You might want to confirm that can find the same real power via other methods. Maybe the P = EI* approach.
  21. May 4, 2014 #20
    I did the dotproduct and got the same answer.

    P = EI* = 120V + j0V * 4A - j0A = 480W + j0W = 480W ∠0°?
  22. May 4, 2014 #21


    User Avatar

    Staff: Mentor

    Okay, can you try one more? Take the total circuit current I (the sum of all branch currents) in complex form and calculate EI*.
  23. May 4, 2014 #22
    I can always try.

    ITotal = I2 + I3 + I4

    = (4A + j0A) + (0A - j3A) + (0A + j6A)

    = 4A + j3A

    = 5A ∠36.87°

    The complex conjugate of 4A + j3A is 4A - j3A, thus the total current will be 5A ∠-36.87°

    P = EI* = (120V + j0V) * (4A - j3A) = 480W - j360W = 600W ∠-36.87°

    Is that the apparent power, as you mentioned earlier? That's my first thought at least. My second thought (ironically) is that I've done it all wrong.
  24. May 4, 2014 #23


    User Avatar

    Staff: Mentor

    You've done splendidly. That P is the complex power phasor, its magnitude (600 VA) is the apparent power. It's components are the real power (480W) and the reactive power (360 VAR). Note the unit variations that are used for each. They are used to distinguish between the apparent, real, and reactive values when presented, although they are all equivalent to Joules/second in the final analysis.
  25. May 4, 2014 #24
    Well then, I've gained a greater understanding of the power triangle and how it works. The word power magically triggers the watt unit in my brain, unfortunately, but I will most certainly keep that in mind.

    The last problem is pretty straight forward. If I get any trouble I'll post here. If not, I'll take a short break and start my next problem.

    Thanks for taking your time. Really.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted