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Circuit Help

  1. Oct 14, 2007 #1

    pka

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    Hello there! I'm new to this forum and not quite sure if this question has been asked before. I've posted a link to the picture of the circuit I am trying to solve for. The circuit is rather simple but I cannot get it. =/ I need to find the voltage across the capacitor. I need to figure out the correct technique in order to do this because I have tried a few different ways and they do not give me the correct answer. I'm more concerned with how to do problems like these (multiple resistors, capacitors and inductors) because I seem to be doing something wrong and...I'm becoming more confused. This is what I've done:

    I've tried Kirchoff's circuit laws and this is what I've come up with:
    Vs = VR1 + VC1 and VC1 = VL + VR2 + VR3

    I solved for the impedance in the inductor (ZL) and for the impedance in the capacitor (ZC). I get ZC = -j and ZL = j10.

    I figured that I could use the voltage divider rule to solve for the voltage across the first resistor (VR1) which would then be:

    VR1 = Vs*[ (ZR1) / (ZR1 + ZL + ZC + ZR2 + ZR3) ]
    This gives me 16.3 - j2.44

    Solving for the voltage across the capacitor I get VC1 = Vs - VR1.
    This then gives me 83.7 + j2.44

    This isn't the correct answer and I'm pretty sure not the correct way to approach this problem, lol. I don't think I've done the voltage divider rule correctly. In fact, I'm a little confused as to using the voltage divider rule. Can I still use it even though this circuit is made up of series and parallel components?

    I have not tried doing a mesh current analysis just yet, but that to me seems to confuse me and take me in a circle just as much as the voltage analysis does.

    Any help anyone can offer would be greatly appreciated! And many thanks in advance.


    Circuit:

    [​IMG]
     
    Last edited: Oct 14, 2007
  2. jcsd
  3. Oct 14, 2007 #2

    dlgoff

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    ZL+ZR2+ZR3 are in parellel with ZC.

    Welcome to PF. Normally if this is homework help, it should be posted Homework & Coursework Questions section.
     
  4. Oct 14, 2007 #3
    This looks like a perfect candidate for nodal analysis. One essential node, so one equation w/ one variable.

    Voltage division doesn't act the way you think it does. All elements have to be in series. You would need to do voltage divider for R1 and Zeq of [Zc || (ZL + R2 + R3)].

    Typically, though, it's not the best idea to transform an element whose voltage or current you're trying to figure out. Your best course of action is to use nodal analysis on Vc, as this gets you the answer you're looking for in about 2 minutes.
     
    Last edited: Oct 14, 2007
  5. Oct 15, 2007 #4

    pka

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    Hey! Thank you for moving the thread where appropriate (whoever did it). I just wanted to respond and say thanks for the help but I still seem to not be getting the correct answer which is supposed to be (in polar notation, A|_ (theta) where A is the amplitude and theta is the phase shift) 10.01 |_(-83.2 degrees) Volts.

    I have tried several different things...I don't know if my complex math skills are what is causing the problem but I have tried a mesh/voltage analysis. I designated the first loop to be called Ia (the part of the circuit where the voltage source is) and Ib (the part of the circuit where the inductor is).

    My equations are as follows:

    Vsource = VR1 + VC1
    Vsource = Ia*ZR1 + (Ia - Ib)*ZC1

    VC1 = VL + VR2 + VR3
    (Ia - Ib)*ZC = Ib*ZL + Ib*ZL + Ib*ZR2 + Ib*ZR3

    Solving for Ia I get from the second equation I get:
    Ia = Ib* ( ZL + ZR2 + ZR3 + ZC) / ZC

    When I substitute that into the first equation and solve for Ib, my final equation is this:
    Ib = Vsource / [ ( (ZL + ZR2 + ZR3 + ZC)(ZR1 + ZC) / ZC ) - ZC ]

    So far, is this correct or are my equations off somewhere?

    The value I get for Ib is -40 + j141. The value I get for Ia is -34.5 + j3.4.

    With Ia and Ib I can calculate VC1 (I think, lol) from:
    VC1 = (Iz - Ib)*ZC.

    Plugging in the values for all knowns I get VC1 to be -2.74 - j34.3

    But...this is also not what I am supposed to be getting. Can critique my work and tell me what it is I am doing wrong???

    Many thanks again for any help anyone can offer!
     
  6. Oct 15, 2007 #5
    Why do you want to do this with mesh or voltage divider?

    There's one essential node here and it just so happens to be the node with the exact voltage that you're looking for. Just write out a KCL equation and be done with it.

    [tex]\frac{V_c_1-100}{10}+\frac{V_c_1}{-j}+\frac{V_c_1}{50+j10}=0[/tex]

    You should probably go back and bone up on your basic DC circuit analysis skills. This circuit should SCREAM nodal analysis to you.

    edit: I wasn't trying to be rude above. But you should get used to the fact that AC circuit analysis works exactly like DC, except that you're using complex numbers. Just use what you learned in DC circuits and this stuff becomes quite easy.
     
    Last edited: Oct 15, 2007
  7. Oct 15, 2007 #6

    pka

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    You are good. Thank you so much for helping me with this problem. What was confusing me was the right hand side of the circuit. I wasn't sure if I could turn the series part into an equivalent impedance and then solve for the current across the capacitor. Also, I get confused with solving problems like these. But thank you for the help! I know what I am doing now and I'm going to practice more of these problems.

    I was wondering though, if I wanted to find the current across one of the other resistors (R2 or R3) or even the inductor, could I or would I have to then use mesh?
     
    Last edited: Oct 15, 2007
  8. Oct 16, 2007 #7
    You could use mesh to verify your answer if you like, but it's not needed to get the current. You can simply divide V_c1 by the impedance to get the current through the inductor, r_2 or r_3 (it's all the same current). The sooner you just forget that you're working with complex numbers the better. You find current the exact same way as you did in DC, you divide the voltage by the impedance (resistance) it's going through. So V_c1/(50+J10) is the current on the right. And the current on the left is (V_c1-100)/10.
     
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