R L C circuit

Discussion in 'Math' started by lil coob, Nov 12, 2009.

  1. lil coob

    Thread Starter New Member

    Nov 12, 2009
    7
    0
    Hi all,
    i am stuck on how to find the resistor and coil values on this circuit.
    i have been given the following:

    Voltage = 100v phase angle 0° at 20hz
    Current = 1.5A phase angle -30°
    Capacitor = 50μF

    any help would be much appreciated
    many thanks
     
    Last edited: Nov 12, 2009
  2. count_volta

    Active Member

    Feb 4, 2009
    435
    24
    Very easy. Phasor analysis.

    But what is the voltage 100V across and current 1.5A through? I assume the voltage 100V is the input voltage source since it has the frequency term. But what is 1.5A through?

    After you figure that out, convert everything to impedances and the voltage source to a phasor.

    ZL = jwL Ω and ZC = 1/jωC are the inductor and capacitor impedances respectively.

    where ω= 20hz *2π = 125.6 rad/s

    Do you know what I'm talking about? Not sure if you learned this yet.

    If you do phasor analysis on your circuit you can easily find the resistance and inductance values. It would also help if you gave us the schematic of your circuit.
     
    Last edited: Nov 13, 2009
  3. lil coob

    Thread Starter New Member

    Nov 12, 2009
    7
    0
    ye i can sort of see what you mean
    all it says is the current flowing in the circuit is 1.5A phase angle -30

    i have made a quick sketch from the data for you
     
  4. lil coob

    Thread Starter New Member

    Nov 12, 2009
    7
    0
    i expect the coil to be of of a greater size than the capacitor due to the -30° angle of the current, am i right in thinking that?
     
  5. neonstrobe

    Well-Known Member

    May 15, 2009
    70
    12
    Hi

    By size I think you mean impedance: yes the coil will have a greater impedance than the capacitor to give you a current lagging the voltage.

    It should be easy to calculate. The reactive parts of your circuit are as previous contributor said j(ωL-1/(ωC)). The phase angle is obtained from the imaginary part of the ac divided by the real which gives you the ratio -1/sqrt(3) in this case (remember triangles?). You know C so you should be able to find L and R.

    Hope this is a bit more of a clue.
     
  6. count_volta

    Active Member

    Feb 4, 2009
    435
    24
    Ok so its a series circuit.

    All you need to do is KVL using phasor analysis. Or even a voltage divider. I will go ahead and do it for you this time, but I want you to learn from this. Pay attention to what I do. Step by step.

    [​IMG]
     
  7. neonstrobe

    Well-Known Member

    May 15, 2009
    70
    12
    OK a bit more help.
    I mentioned that the phase angle is tan-1(Im /Re) which we know is -30 deg.
    SInce the total imaginary is jwL-j/(wC) we can set

    [wL-1/(wC)]/R=-1/sqrt(3)

    We know that the magnitude of the current is 100/1.5 = 66.67 ohms
    but this is made up from the reactive and resistive bits.

    The mag of Z=sqrt(R**2 + (wL-1/wC)**2) =66.67 ohms

    Substitute for wL-1/(wC) from the phase ....

    cheers ...mine's a pint
     
  8. mojojerry

    New Member

    Oct 27, 2009
    2
    0
    I am only a 2 yr electronic student, 30 years ago and never studied since. We only studied RCL circuits 1 quarter. I was never to smart anyway. This discussion sounds so intresting that I just have to make a fool of myself. I though that you had two have 2 of any 3 in order to find the third in a regular current, voltage, resistance circuit. If this is true in and RCL circuit as well? Would you not have to test for an actual voltage across the resistor and or inductor to see what the volt drop is and then R=V/I? If the volt drop on R1 is 50 v then R= 33.333 ohm. God only knows what phasor analysis is? never heard of this. It do sound intresting though. mojojerry
     
    Last edited: Nov 24, 2009
  9. zgozvrm

    Active Member

    Oct 24, 2009
    115
    3
    For a series RLC circuit, we know the following:

    E_T = E_R \: + \: E_L \: + \: E_C

    I_T = I_R = I_L = I_C

    Z_T = Z_R \: + \: Z_L \: + \: Z_C

    E_L leads E_R by 90° which, in turn leads E_C by 90°

    X_L leads R by 90° which, in turn leads X_C by 90°

    I_T and E_R are in phase with each other.



    So, first we have:

    <br />
$<br />
I_T = I_R = I_L = I_C = 1.5 \angle -30^\circ A = 1.299 - j0.75 A<br />
$<br />

    and we know

    $X_C = \frac{1}{2 \pi fC} \: = \: \frac{1}{2 \pi 20 \times 50 \mu F} \: = \: 159.155 \Omega$

    Which gives us

    $Z_C = 159.155 \angle -90^\circ \Omega \: = \: 0 - j159.155 \Omega$

    Now, we can find

    $Z_T =\frac{E_T}{I_T}$ and $ \: E_C = I_C \: \times \: Z_C$



    $Z_T = \frac{100 \angle 0^\circ V}{1.5 \angle -30^\circ A} \: = \: 66.667 \angle 30^\circ \Omega \: = \: 57.735 \: + \: j33.333 \Omega$

    and

    $E_C = 1.5 \angle -30^\circ A \: \times \: 159.155 \angle -90^\circ \Omega \: = \: 238.732 \angle -120^\circ V \: = \: -119.366 \: - \: j206.748 V$


    Now, for the tricky part:

    Since we know the angular relationship between E_R and E_C, we know that E_R is 90° ahead of E_C which turns out to be (-120°) + 90° = (-30°)

    Also, we know that E_L is 90° ahead of E_R, which is (-30°) + 90° = 60°

    Now, all we have to do is find the magnitude of those values: Since E_L and E_C are 180° apart, the sum of their angles must be either 60° or -120°, therefore it must be 60° so that, when added to [E_R] (which has an angle of -30°), we get a vector resulting with an angle of 0°.

    Also, the x-components of the sum of E_L, E_C, and E_R must be 100 and the sum of their y-components must be zero.

    This gives us:

    $E_{L+C} \: \times \: \cos{60} \: + \: E_R \: \times \: \cos(-30) =100$

    $\frac{E_{L+C}}{2} \: + \: \frac{E_R \: \times \: \sqrt{3}}{2} \: = \: 100$

    $E_{L+C} \: + \: (E_R \: \times \: \sqrt{3}) \: = \: 200$

    and

    $E_{L+C} \: \times \: \sin{60} \: + \: E_R \: \times \: \sin(-30) \: = \: 0$

    $\frac{E_{L+C} \: \times \: \sqrt{3}}{2} \: - \: \frac{E_R}{2} \: = \: 0$

    $\frac{E_{L+C} \: \times \: \sqrt{3}}{2} \: = \: \frac{E_R}{2}$

    $E_{L+C} \: \times \: \sqrt{3} \: = \: E_R $

    $E_{L+C} \: = \: \frac{E_R}{\sqrt{3}}$

    substituting, we get

    $\frac{E_R}{\sqrt{3}} \: + \: (E_R \: \times \: \sqrt{3}) \: = \: 200$

    $E_R \: + \: 3E_R \: = \: 200 \sqrt{3}$

    4E_R \: = \: 200 \sqrt{3}

    E_R \: = \: 50 \sqrt{3} = 86.603

    or, more correctly, $E_R \: = \: 86.603 \angle -30^\circ \: V \: = \: 75 \: - \: j43.301 \: V$



    From here, you can divide E_R by I_R to get Z_R which gives you $57.735 \angle 0^\circ \: \Omega$

    and, you can calculate E_L by:

    E_L \: = \: E_T \: - \: E_R \: - \: E_C \: = \: (100 \: - \: 75 \: - \: (-119.366)) + j(0 \: - \: (-43.301) \: - \: (-206.748)) \: V

    or

    $E_L \: = \: 144.366 \: + \: j250.049 \: V \: = \: 288.732 \angle 60^\circ \: V$

    Divide $E_L \: \div \: I_L$ to get Z_L

    $Z_L \: = \: 288.732 \angle 60^\circ \: V \: \div \: 1.5 \angle -30^\circ \: A = 192.488 \angle 90^\circ \: \Omega $


    This gives us R = 57.735Ω and XL = 192.488Ω
    Since XL=2∏fL, and f=20Hz, we can solve for L:

    L = \frac{X_L}{2 \pi f} \: = \: \frac{192.488 \Omega}{2 \pi \: \times \: 20Hz} \: = \: \frac{192.488}{125.664} \: = \: 1.532H
     
    Last edited: Dec 14, 2009
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    789
    I got different answers

    R=57.735 Ω and L=1.532H

    Solution attached as pdf
     
  11. zgozvrm

    Active Member

    Oct 24, 2009
    115
    3

    Yes, you're right. I inadvertently used the magnitude of E_R (86.603) as the value of the resistance, forgetting to divide by the current first.

    I made the appropriate corrections to my earlier post.
     
    Arcnetting likes this.
  12. Arcnetting

    New Member

    Dec 2, 2017
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  13. Arcnetting

    New Member

    Dec 2, 2017
    2
    0
    I am grateful to you for having taken the time to answer the question in detail. As a student, it has helped in other ways besides. I have a question for you, (hoping your still around). You associated the current phase with the resistor phasor in this circuit, which I thought the resistor would be in phase with the voltage at 0 degrees. Also, at this point in my studies I understood that an inductor is basically shorted at low frequencies. Not knowing if that only applies to resonant circuits where that information came up and that if it is true for a circuit would it affect the phase angle only resulting in the impedance difference equating to pure resistance?
     
  14. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,029
    1,082
    Hi,

    If i understand your question correctly that would only be a case of pseudo pure resistance not exactly pure resistance.
    I point out the difference here because it depends on the application whether or not the inductive part is significant or not, and the only way to get a theoretical pure resistance is through resonance. Anything other than that is still reactive, but i use the phrase pseudo pure resistance because to the rest of the circuit it may look very very nearly resistive only, but in another different circuit that same set of R, L, and C may result in a complete circuit response that depends more significantly on the value of L too. So except for the perfect case of resonance and assuming ideal components, it will be application specific on how we should interpret the pseudo pure resistance. OF course for pure DC it will be pure resistance too, not pseudo pure resistance but i am sure you know that.

    I get the same values as others R=57.7 and L=1.53 (units Ohms and Henries).
    I took the long road to the solution, where i first analyze the circuit for the current amplitude and phase shift and that gives me two equations. Solve the amplitude equation for the function R(L) and then substitute that into the equation for phase shift, then solve that for L, then substitute that L back into the amplitiude equation and solve for R, which gives me both R and L. Lastly subs both R and L back into the equation for current and calculate the amplitude and phase shift and make sure it comes out to 1.5 amps at -30 degrees.
     
  15. mufet

    New Member

    Jan 11, 2018
    8
    0
    use transfer functions
     
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