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RCL Circuits

  1. May 29, 2004 #1
    Hi all!

    I'm a student in APU University, Cambridge, England. One of my modules is Analogue Electronics, Exams are a few weeks away and i wanted to get an idea of how well ive grasped some of the concepts. We have been given a sheet to work through but have no one to mark it as all lectures are off site.

    just wanted to make sure i was on the right track. heres part of it:-

    Circuit Diagram - http://www.mwdata.plus.com/diagram.JPG

    a) Show that the capacitor C1 has a capacitance of about 13nF.

    so C1= 1/(2*Pi*f*Xc)


    =13nF? i think thats ok

    b) Calculate the magnitude of the combined impedance of the capacitor C1 in series with R1.

    so Z total= Zr1+Zc1?

    Z total = 14kOhm?

    c) Calculate the current I1 in the circuit.

    I1 = Ir1+Ic1?



    total = 8.75mA?

    d) Calculate the voltage Vc1 across the capacitor C1.

    so Vc1= Ic1 * Xc

    Thank You In Advance.

    Jack Wood.
  2. jcsd
  3. May 29, 2004 #2
    a) is correct. d) is correct except you got the wrong value of Ic.

    b) Not quite. If you draw a vector diagram of the voltages in the circuit you will have something like this:
    Code (Text):
    |   /
    |  /
    | /
    Where the vertical vector is the voltage of the capacitor, the horizontal vector is the voltage of the resistor and the diagonal vector, which represents the vector addition of the other two, is the voltage of the source. Thanks to Ohm's law you can divide the magnitude of those vectors by I, which is the current throughout the circuit (it's the same for all components). Now the orthogonal vectors represent the impedance of the capacitor and the resistance of the resistor and the last vector represents the impedance of the whole circuit, which is what you are looking to find. Through Pythagoras you see that Z2 = Xc2 + R2 and from here you find Z (10kOhm).

    c) VRMS = IRMSZ. You have both VRMS and Z, so you can see how IRMS = 0.3mA.
  4. May 29, 2004 #3
    Sorry for the crude drawing, I'm not on my main computer right now.

    Attached Files:

  5. May 29, 2004 #4
    Excellent. yes that rings some bells now.

    now on to the next sheet, any problems with that and ill be back :).



    EDIT: thanks for the diagram :biggrin:
  6. May 29, 2004 #5
    back again :)

    diagram http://www.mwdata.plus.com/diagram2.JPG

    a) Calculate the Inductance (H) of the inductor L1

    can't seem to figure out how to do it.

    L1 = VL1 / XL1?

    It's late at night and my brain hurts, ill be seeing these equations in my sleep. lol.


  7. May 30, 2004 #6
    No, VL1 / XL1 is the current through the inductor. This is what you're looking for:

    [tex]X_L = \omega L = 2\pi fL[/tex]

    And you want to extract L. :smile:
  8. May 30, 2004 #7
    ok, im going to sound like a total n00b here but, what do you mean by extracting L? i understand how the do the calculation but what is wL? and how do i get L to do 2*Pi*f*L?

    sorry to be thick :)

  9. May 30, 2004 #8
    [tex]\omega L[/tex] is [tex]X_L[/tex] (definition). [tex]\omega[/tex] is also equal to [tex]2\pi f[/tex], so [tex]X_L = 2\pi fL[/tex]. By "extract L" he means "solve for L", i.e. [tex]L = \frac{X_L}{2\pi f}[/tex].
    Last edited: May 30, 2004
  10. May 30, 2004 #9
    Sorry, I don't study in English so I don't always get the terms right. :smile: Thanks Muzza.
  11. May 30, 2004 #10
    perfect, thanks man!

  12. May 30, 2004 #11
    If you find any more interesting questions feel free to post them, I myself have a final exam about this in about a week's time so I could use the exercise. :smile:
  13. May 30, 2004 #12
    Oddly enough, I also have an exam concering RCL-circuits (among other things) coming up in a week.
  14. May 30, 2004 #13
    It's that time of year I guess.
  15. Jun 1, 2004 #14
    Yeah i got one more to get through.

    here goes:-


    not knowing too much about diodes i can't do much of it so here it is. cheers once again for the help :).

    a) Calculate the value of the DC supply Voltage (V5) in the circuit/

    b) Calculate the current through the zener diode in the circuit .

    c Calculate the power dissipated in both the zener diode and the resistor the of the circuit.

    using P=I*R?

  16. Jun 1, 2004 #15
    Sorry, no idea. I replaced electronics with a technology project so I never learned about diodes either. :smile: (We were taught RCL in physics class.)
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