HNC Electrical Resistor & Capacitor Question

Discussion in 'Homework Help' started by JimmyB, Jun 1, 2010.

  1. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Hi Can anyone help me with this Q.

    A Resistor & a Capacitor are connected in series across a 150V a.c supply. When the frequency is 40Hz the current is 5A, & when the Frequency is 50Hz the current is 6A. Find the resistance and capacitance of the resistor and capacitor repectively??

    I do not know where to start with this one, I am doing a distance learning course so it is hard to get help..

    Thanks for any help...
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    With the magnitude of the current you can find the total impedance of the circuit at each frequency.

    Then use (R^2)+(Xc^2)=Z^2

    to get two equations with two unknowns.
     
  3. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Thanks, I did get the total impedances I just didnt know what to do with with them, does this question need a simultaneous equation calculation? Thanks for the help, I thought I knew this area quite well but this one has got me!!
     
    Last edited: Jun 1, 2010
  4. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Hi Mik3, I am usure where to go from the impedances are you able to help any further?

    Thanks
    Jimmy
     
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    You have two equations and two unknowns, thus you need to make simultaneous equations calculation.
     
  6. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    I'm still not able to get a simultaneous equation from there,

    R and X have no coefficients is it still possible??
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Start with .....

    \frac{150}{\sqrt{R^{2}+X_{1}^{2}}}=5

    \frac{150}{\sqrt{R^{2}+X_{2}^{2}}}=6

    where

    X_{1}=\frac{1}{\omega_{1}C}

    X_{2}=\frac{1}{\omega_{2}C}

    \omega_{1}=2\pi f_{1}

    \omega_{2}=2\pi f_{2}

    f_{1}=40 Hz

    f_{2}=50 Hz
     
  8. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    I'm sorry TNK I still dont see it... is this going to be substitution??
     
  9. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Yes, you can use substitution or any other method for solving simultaneous equations. If there are no coefficients then the coefficient is one.
     
  10. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    I cant substitute it too many squar roots and squars for me mik3, just cant get my head round this one
     
  11. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    {R}^{2}+\frac{1}{2*pi*40*C}={30}^{2} (1) for 40 Hz

    {R}^{2}+\frac{1}{2*pi*50*C}={25}^{2} (2) for 50 Hz

    solve (1) for {R}^{2}

    {R}^{2}={30}^{2}-\frac{1}{2*pi*40*C} (3)

    sub (3) into (2)

    {30}^{2}-\frac{1}{2*pi*40*C}+\frac{1}{2*pi*50*C}={25}^{2}

    Then find C. I will let you find it.
     
  12. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Not sure how to set out the calculations on here but is the answer

    C= 28.3 micro F??

    But also from your equations 1 & 2 shouldnt the Xc part of the equation be squared??
     
  13. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    mik3 does that sound about right??
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    The correct answer in not 28.3mF it is R=11.6Ω; C=144uF.
     
  15. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    I know, Xc has to be less than Z,
    are you able to show me the workings?
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Here is a detailed solution for C ....

    X_{1}^{2}+R^{2}=(\frac{150}{5})^{2}=30^{2}=900 \ \ (1)\\<br />
X_{2}^{2}+R^{2}=(\frac{150}{6})^{2}=25^{2}=625 \ \ (2) \\<br />

    Subtract eqn (2) from eqn (1)

    X_{1}^{2}-X_{2}^{2}=900-625=275

    (\frac{1}{\omega_{1}C})^{2}-(\frac{1}{\omega_{2}C})^{2}=275

    (\frac{1}{\omega_{1}})^{2}-(\frac{1}{\omega_{2}})^{2}=275C^{2}

    \frac{1}{(2\pi f_{1})^{2}}-\frac{1}{(2\pi f_{2})^2}=275C^{2}

    \frac{1}{251.33^{2}}-\frac{1}{314.16^{2}}=275C^{2}

    swap sides

    275C^{2}=1.5831E^{-5}-1.01323E^{-5}=5.7E^{-6}

    C^{2}=\frac{5.7E^{-6}}{275}=2.0727E^{-8}

    C=\sqrt{2.0727E^{-8}}=1.44E^{-4}=144 \mu F
     
  17. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Thanks tnk, easy when you know how, dont think I would have ever worked that one out though.. do you know of any other good text books to buy, I have 'Hughs electrical & electronic, tenth edition book' that is basicaly what i'm learning from... thanks J
     
  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This is really applied maths - perhaps you just need to practice solving equations of various types.
     
  19. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Yes I know what you mean, but I really learn from examples which doesnt always help!
     
  20. JimmyB

    Thread Starter Member

    Jun 1, 2010
    38
    0
    Hi tnk, are you able to show the working of the substitution method as mik3 started to show.. I can only get so far but am having trouble isolating 'C'

    Thanks for your time J..
     
Loading...