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A.c circuit calculation

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A source of e.m.f 240V and frequency 50 HZ is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V.

    Calculate the:
    (i) potential difference across the capacitor,
    (ii)capacitance of the capacitor,
    (iii) inductance of the inductor



    2. Relevant equations



    3. The attempt at a solution

    I got the answer to (i)
    as 245V

    that of (ii) I got to this extent C= 1/2450(pi)
    = 1/7700
    = 0.000129870
    aprox. 0.0001299
    = 1299*10^-4 uf

    my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
    What really happened?
     
  2. jcsd
  3. Nov 10, 2012 #2

    ehild

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    C = 0.0001299 F is correct. 1 F = 106 μF, so C=0.0001299*106=130 μF

    ehild
     
  4. Nov 10, 2012 #3
    I really want to understand it, is it
    106 or 10-6
     
  5. Nov 10, 2012 #4

    ehild

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    "micro" is small. microfarad is a small unit, one millionth, 10^(-6) (1/1000000) of a farad. 1 F = 1000000 microfarad.

    ehild
     
  6. Nov 12, 2012 #5
    What is the relationship between charge and capacitance. I know of this formular:
    Q = CV
    where Q = quantity of charge in columb; C = capacitance in uf and V = potential difference between either of the plates in volts.

    Suppose C = 2.4uf, can we say that C = 2.4*10-6 since 1 microfarad uf = 10-6farad.

    By the way one farad is equal to what?
    Farad is the unit of what?
     
  7. Nov 12, 2012 #6
    Why would you say that the voltage in part (i) is 245 volts? I think you should look back at Kirchoff's Rules, unless you can justify that 245 volt answer better.
     
    Last edited: Nov 12, 2012
  8. Nov 12, 2012 #7

    ehild

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    Two parallel metal plate can store charge, and when charged, there is a certain potential difference between the plates.

    Assume there is +Q charge on one plate and -Q charge on the other plate and the potential difference is V. The ratio C=Q/V is called capacitance.

    The unit of capacitance is farad (F). 1 F is the capacitance if 1 coulomb charge is stored at 1 volt potential difference.

    C(farad)=Q(coulombs)/V(volts)

    1 F = 106μF, or 1μF=10-6 F. 2.4 μF=2.4 10-6 F.

    ehild
     
  9. Nov 12, 2012 #8
    The first highlited write up is for a.c circuit:
    (1) I got the answer to (i)
    as 245V

    that of (ii) I got to this extent C= 1/2450(pi)
    = 1/7700
    = 0.000129870
    aprox. 0.0001299
    = 1299*10^-4 uf

    my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
    What really happened?







    (2) The second highlited write up is for capacitance of a capacitor:
    (2)
    What is the relationship between charge and capacitance. I know of this formular:
    Q = CV
    where Q = quantity of charge in columb; C = capacitance in uf and V = potential difference between either of the plates in volts.

    Suppose C = 2.4uf, can we say that C = 2.4*10-6 since 1 microfarad uf = 10-6farad.

    By the way one farad is equal to what?
    Farad is the unit of what?
     
  10. Nov 12, 2012 #9
    Right, but you should justify why you got the 245 volt across the capacitor...
     
  11. Nov 12, 2012 #10

    ehild

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    Read the problem and try to find the voltage across the capacitor from the data given. Say if you get different result.

    The OP had no problems with question i, he does not need to show his derivation.

    ehild
     
  12. Nov 12, 2012 #11
    Well, I can't get it, but I'd like to solve this problem for my own studying. Ought I start a new thread to do that?
     
  13. Nov 12, 2012 #12

    ehild

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    Well, try to solve the problem as you suggested. You have resistor, capacitor and inductor in series. How add up the voltages across them?

    ehild
     
  14. Nov 12, 2012 #13
    I tried thinking of it like a voltage divider-- I found the impedance of the series combination, and tried to take emf*(Zc)/(Ztot), but since we don't know C or L, I couldn't solve it that.

    I know that the phase relationship means we can't just straight up add the voltages to 240, but I can't figure out how to get around this.

    Although, the same I has to flow through every piece of the circuit the same, so I can find R (14 ohms)... but not L or Q on the capacitor.

    Yeah, I can't figure out how to calculate L or C so that I can use the series impedance.
     
    Last edited: Nov 12, 2012
  15. Nov 12, 2012 #14
    All I can think is that I can ignore the other circuit elements, and then I know that Vc=Ic*Zc... Ic=10 Amps, and Zc= 1/(j*w*C). But I don't know C!
     
  16. Nov 12, 2012 #15

    ehild

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    You can not ignore the other elements.

    There are three elements connected in series, and 10 A flowing through one element. What is the current flowing through the others?

    The magnitude of the impedance is voltage over current. Z=V/I. What is the resistance of the resistor and what is the impedance of the inductor then?

    While the ac voltage across a resistor is in phase with the current, the voltage leads the current by phase pi/2 across an inductor and lags behind the current by pi/2 across a capacitor. You can consider the ac voltages as phasors, and use vector addition to sum them when you apply Kirchhoff's voltage law.

    ehild
     
  17. Nov 13, 2012 #16
    Now I get 90 as the capacitor's voltage. Resistance is 14, inductive reactance is 5, because they each have 10 amps.

    I add these impedance as vectors and see that their combined impedance is 14.866. Then 240-(10*14.866)=90 volts left for the capacitor.

    Moving in the right direction.
     
  18. Nov 13, 2012 #17
    Also, I'm realizing that it doesn't make sense for voltage across capacitor to be 245. The max voltage across the source is 240... And that's the max. How could any single element possible have 245 across it, then?
     
  19. Nov 13, 2012 #18

    ehild

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    Not quite.

    There is pi phase difference between the voltages across the capacitor and inductor. So the vector sum of voltages is

    [itex]V=\sqrt{V_R^2+(V_L-V_C)^2}[/itex]

    You see that the resultant voltage can be smaller than the voltage across either the capacitor or the inductor. Assume, for example, that VR=50 V and both VL and VC are 10000 V, what is the resultant voltage?

    The combined impedance of a series RLC circuit is [itex]Z=\sqrt{R^2+(Z_L-Z_C)^2}[/itex]

    ehild
     
  20. Nov 13, 2012 #19
    But I don't know the capacitance, so I can't use Z=-j/wC, and I don't know the voltage obviously, so I can't use Z=V/I.

    How about this: I know the source gives V=240sin(wt); the resistor will have in phase current. Thus, its V is ZIsin(wt)=140sin(wt). The inductor's current leads by 90 degrees, so its voltage is ZIcos(wt). The capacitance lags by 90 degrees, so its current is -10cos(wt)/(wC). Is it possible to set those last 3 things equal to the source V?
     
  21. Nov 13, 2012 #20
    Unless I'm simplifying the leads and lags idea too much. The C and L lead/lag the voltages through THEM by those 90 degrees, but that doesn't mean they lead and lag the voltage source, right? Those could take on different angles than the source as a whole?
     
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