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Complex current calculation in Parallel R L Circuit

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the complex current in this circuit
    http://imageshack.us/photo/my-images/51/figure1j.jpg/

    2. Relevant equations
    I calculated the inductor's impedance and found that impedance = 12.57 Ohms
    so I used it in the equation for the Current.
    I need to write the answer in terms of a + bj (complex numbers)


    3. The attempt at a solution

    I tried to find the current with the Ohm's law I = V/Z where Zt = 1/1/Zr+1/Zl
    I am not sure is that is the right way to calculate this.
    Thanks
     
  2. jcsd
  3. Nov 25, 2012 #2

    gneill

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    Staff: Mentor

    Which current are you trying to find? There are two parallel branches: one consisting of a capacitor of unknown value, and one consisting of the SERIES connection of an inductor and a resistor. So it looks like there are three different currents to choose from.
     
  4. Nov 25, 2012 #3
    Take a look of the question it is said the complex current of the circuit

    http://imageshack.us/photo/my-images/696/figure1u.png/
     
  5. Nov 25, 2012 #4
    I couldn't see the circuit, but as for bringing it to terms of a + bj... Impedance of the resistor is just R, the inductor is Z = jωL so the admittance is Y = -j1/L (same as 1/jωL but I keep complex numbers in the numerator).

    any reason I couldn't see the circuit?
     
  6. Nov 25, 2012 #5

    gneill

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    From the problem statement in your second image (yes, for some reason your images are not being displayed in your posts; I hit "quote" to obtain the full text of the post and copied the URL from out of the IMG tags) you need to find the impedance of the load algebraically to start with. That means ignoring the numerical component values to begin with and deriving an expression for the total impedance symbolically.

    So using L, R, and C for the component values, can you write an expression for the impedance?
     
  7. Nov 25, 2012 #6
    So Ztotal must be equal to 1/(1/Zr+1/Zl) + Zc?
     
  8. Nov 25, 2012 #7
    The resistor and inductor are not in parallel.

    The capacitor is in parallel with the series resistor + inductor.
     
  9. Nov 25, 2012 #8

    gneill

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    No, the resistor is in series with the inductor (so their impedances add) while the capacitor is in parallel with their sum.

    It occurs to me that it might be simpler to write the currents of the two branches separately (since they are in parallel they share the same potential difference of E = 220V), then sum the results. This will tame the algebra which might otherwise get quite hairy.
     
  10. Nov 25, 2012 #9
    Oh, I see the stupid mistake .. So the total impedance of the Resistor and Inductor is 8+12.57=20.57Ohms
    But how to get rid of the Capacitor.. because they ask me about the capacitor in the second part of the question.
    It seems that the final equation should be I= V/1/(1+Zc+1/Zr+l)
     
  11. Nov 25, 2012 #10

    gneill

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    Remember that impedances are complex values. The impedance of the inductor is 12.57j Ohms, and doesn't add to the real component of the resistor impedance.
    Forget the numbers for now. Do the math symbolically. If the capacitor has value "C", and the power supply has voltage "E", what's the (complex) current through the capacitor branch? Do the same for the Jacuzzi branch, using R and L for the component values.

    Separate the currents into their real and imaginary components and add the 'like' components together to arrive at the total current.
     
  12. Nov 25, 2012 #11
    You can't add them together like that. R is a *real* resistance while L is an *imaginary* reactance. The series impedance is 8 + j12.57 with the imaginary reactance being separated from the real resistance by the 'j'.

    This has to do with the current and voltage relationships that these devices impose across them. A resistor is governed by V=IR, where the current and voltage are always in phase (there is no time delay between the two waveforms -- if you place 10 volts across a 1 ohm resistor, 1 amp flows through it immediately, not 10 seconds from now). An inductor is governed (in the steady state**) by the same equation: V = I (jwL) but the resistance is complex. This is because a time delay is introduced in the V/I relationship and this is summarized by the phase displacement introduced by the complex 'j'. If you apply V volts across an inductor, the current through the inductor does *not* change immediately but causes the current to change gradually according to di/dt = v/L. This time delay is the phase seen in complex impedances.

    I don't know what they are asking about the question but the I expression you have looks to be missing some brackets.

    The impedance of the series R/L is ZL = R + jwL and the impedance of the cap is Zc. So you need the parallel combination of those two to find the impedance of the circuit seen by the source.

    BTW -- it is much easier not to make algebraic errors and to sanity check your work if you wait to plug in actual values until the very end. It will save you much hassle :)
     
  13. Nov 25, 2012 #12
    The thing I come up with is the following I=V/1/(1/Zc+1/(R+R+jwL)) isnt the whole jwL = Zl or it is Zl=R+jwl

    and I still dont get how to get rid of Zc becuase otherwise we will have 2 unknowns..
     
  14. Nov 25, 2012 #13
    That will be the total current delivered by the source (see the equation I marked with ******* below which is equivalent for the impedance calculation)

    but you have an extra R there. The inductor impedance by itself is jwL. The resistor has impedance R. The sum is R+jwL


    Maybe I chose an unfortunate symbol. The impedance of the inductor alone is jwL. The impedance of the resistor in series with the inductor (which I called ZL or Z load) is R+jwL

    The voltage E appears across the series R/L so the current through R/L can be found independently of the capacitor and is E/(R+jwL)

    The total current supplied by the source is the sum of the current through the capacitor and the current through the series R/L. I(total) = E/(R+jwL) + EjwC.

    You could also find the total current by finding the total impedance seen by the source. This is Z(total) = (R+jwL) || (1/jwC) ********* and I(total) = E / Z(total)


    How you go about answering the question will depend on what information is given and what is asked. I can only see the circuit diagram though :)
     
  15. Nov 25, 2012 #14
    So, for the first part a) the total current doesnt have anything to do with Zc..
    so its I=E/R+jwL
    and doing the nasty algebra I came to this E(R-jwL)/R^2-jwL^2
    220(8+j12.57)/166,0049
    can someone tell me how to remove the j from the above equation since i know that Zr+l = 8+j12.57
    and j^2 =- 1
     
    Last edited: Nov 25, 2012
  16. Nov 25, 2012 #15

    gneill

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    Actually, it does have something to do with Zc. You want to find an expression for the total current which includes the capacitor so that you can apply the hint given for part b.
    However, you can proceed by finding the currents for the two branches separately, then add the results to find the total.
    You should make liberal use of parentheses to make the interpretation of your equations clear. Presumably you meant,

    I = E/(R + jωL)
    by which presumably you mean:

    I = E(R - jωL)/(R2 - (jωL)2)
    The j in the denominator goes away when you carry out the squaring of the (jωL) term. The j in the numerator stays as is.

    This value for I is the current in the Jacuzzi branch. Now, what's the current in the capacitor branch?
     
  17. Nov 25, 2012 #16
    so we have (ER - jEwL)/(R^2 + wL^2) and this gives us (220*8+j220*12.57)/166,0049

    I think in the second part of the question which is Now that you have the complex current, what value for the shunt capacitor will minimise
    the total current? Hint: Set the imaginary part to 0

    we have to find the Zc?
     
  18. Nov 25, 2012 #17

    gneill

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    Note the added parentheses in order to make the mathematical operations clear.

    You don't need any numbers yet! Just the symbols. Remember that the problem says you want to arrive at a formula you can apply to Bob's friend George's similar (but not the same) Jacuzzi. For that you need a formula containing symbols only.
    You need to find the expression for the current WHICH INCLUDES THE CAPACITOR before you can do part b.

    You have an expression for current of the Jacuzzi branch. Now write an expression for the current in the capacitor branch.
     
  19. Nov 25, 2012 #18
    Zt=(RjwC)/(R+jwC-w^2LCR) ?
     
  20. Nov 25, 2012 #19

    gneill

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    That doesn't look right. But why are you going back to the impedance? Just write the expression for the current through the capacitor; Since the two branches are in parallel, you can simply add the two currents to find the total.
     
  21. Nov 25, 2012 #20
    isnt it when they are in parallel 1/Zt = 1/Zc + 1/Zr+l ?
    so the current through cap will be I = V/Zc?
     
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