Simple circuit with inductor and capacitor.

Discussion in 'Homework Help' started by cdummie, Oct 6, 2015.

  1. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    Capacitor and inductor are connected in series, C is unknown, L is unknown but it's internal resistance is known (R=20Ω). These two elements are charged with sinusoidal voltage generator with variable frequency, effective value is constant. For f1=580kHz and f2=612kHz current through the circuit is 0.707 times it's maximal value. Resonant frequency is f0=595.8kHz. Find L and C.

    First, i am not sure what it means "it's internal resistance is known (R=20Ω)". Does it means that ωL=R=20Ω, or i can think of it as a resistor added is series with inductor and capacitor?

    Then, i know that Imax=0.707Ieff and i know that:

    i(t)=Imax*cos(ωt+φ) and ω=2πf and when i have frequencies f1=580kHz and f2=612kHz i(t)=Imax*0.707 but i don't know how this can help me.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    The fact that they are telling you that the response is 0.707 at those two frequencies means those are the two cutoff frequencies, upper and lower.

    The fact that they tell you that the resistance is 20 ohms means that is the series resistance of the circuit.

    So you are given the resonant frequency, so you can find the LC product, which helps to simplify the calculations that follow.
    You are given the upper and lower cutoffs, so you can find the particular L and C.

    Also be aware that you may get a slightly different value for C and L depending on if you use the lower frequency or the upper frequency to solve for C because these upper and lower frequencies are obviously not the exact values for the given resonant frequency. They should be close though.
     
    Last edited: Oct 6, 2015
  3. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    I would certainly hope they are close enough. If you find the center frequency of 580 kHz and 612 kHz you get 595.785 kHz, which I would certainly think is close enough to 595.8 kHz as to not be a concern.
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    If I understand OP correctly, 20 Ohm is for inductor only. Which makes me think that 20 Ohm is XL, the reactance of the inductor, the units make sense.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    Yes, the 20 ohms is for the inductor, it is the series resistance of the inductor and thus it is handled in the same manner as having a pure inductor and a 20 ohm resistor in series. Thus the 20 ohm is in series with the whole circuit because the inductor is in series with the capacitor.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    Yes that is why i said they should be close though.

    When we see whole numbers or even numbers with only one decimal place we know it's not exact or an attempt at being exact with the usual 16 digits because the transcendentals will never come out to that nice or a number. We'd see something like 580.23856439459... and then some (that's fictitious but that's just to illustrate).
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    Think about this -- it is specifically stated that the inductance is unknown but that it's internal RESISTANCE is 20 Ω. The frequency at which this is true is not given but then the center and both upper and lower cutoff frequencies are given. So if this 20 Ω is the reactance (as opposed to the resistance, as stated), then what frequency is that 20 Ω at?

    The 20 Ω is the parasitic resistance of the inductor windings.

    @cdummie : Draw the circuit and determine the impedance as a function of frequency.

    Since the currents are the same at the upper and lower frequencies given, that means that the impedances must be the same at those two frequencies. So set them equal and see what the relationship between L and C has to be.

    Then leverage the fact that if you know the ratio of the currents at two frequencies, that you know the ratio of the impedances at those two frequencies. This gives you another equation relating L and C.

    Two equations, two unknowns.
     
  8. crutschow

    Expert

    Mar 14, 2008
    13,013
    3,233
    You know the resonant frequency, the upper and lower -3dB points (the circuit Q) and the series resistance.
    Based upon the formula for the circuit Q you can then determine the value of L and C.
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    The square root function is not transcendental.

    It would be a trivial matter to specify a set of frequencies in which the center and cutoff frequencies are represented by whole numbers. For example, consider a resonant frequency of 210 kHz with upper and lower cutoff frequencies of 126 kHz and 350 kHz. Or pick, say, 730.8 kHz as the center frequency. Now pick a lower frequency of 487.2 kHz and an upper frequency of 1096.2 kHz. Those are in exact relationships since the geometric mean of 487.2 kHz and 1096.2 kHz is 730.8 kHz exactly.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hello again,

    Yes but now try finding a cap and inductor value that is also a whole number or near that. If we start with a whole number for one thing, we end up with an irrational number for another thing. So we get a number with a large number of digits like 5.14837439394 rather than 5.1 or like that. This is the same thing that happens a lot with transcendentals. It was a simple point.
     
  11. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    You keep moving the target. First it was that you will get a different value if you use one cutoff frequency versus the other. Now it's that you'll get a capacitor/inductor value that is exactly that value.

    The values in the original problem will yield results consistent to 3 or 4 sig figs. Try finding a cap or inductor whose tolerance is that tight.
     
  12. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    First, i am sorry it took me so long to respond, now, i drew the circuit, but i am not quite sure what you meant by "determine the impedance as a function of frequency." Do you mean this:

    Z = R+jωL - j/ωc = R + j(2πfL - 1/2πfC) ?

    Now, currents are same for f1=580kHz and f2=612kHz so:

    Z1= R + j(2πf1L - 1/2πf1C) and
    Z2 = R + j(2πf2L - 1/2πf2C) since currents are same, as you said( they are in series so the voltage is same too) then impedances are equal so i have:

    R + j(2πf1L - 1/2πf1C)= R + j(2πf2L - 1/2πf2C)

    2πf1L - 1/2πf1C = 2πf2L - 1/2πf2C

    2πL(f2-f1) = 1/(2πC)[1/f2 - 1/f1]

    2πL(f2-f1) = (f1-f2)/[(2πC)f1*f2]

    2πL(f2-f1) = -(f2-f1)/[(2πC)f1*f2]

    2πL=-1/(2πf1f2C)

    and that is the relation between L and C that i have.

    And, the part that i don't understand at all is, how do i know the ratio of the currents at given frequencies?
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    I don't have time to go through your work in detail, but you made a claim that is patently false and so might need to reconsider what you did in light of that. When you say that "they are in series so the voltage is same too," that is simply not the case. Elements in series have the same current (that's what it means for them to be in series) but the voltages across them can be anything. Elements in parallel have the same voltage (that's what it means for them to be in parallel) but the currents through them can be anything.
     
  14. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    Oh, you are right, i don't know what i was thinking, but then, if currents are the same, why impedances have to be the same?
     
  15. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    The impedances of two elements don't have to be the same just because they are in series. In addition, the impedance of an inductor can never be the same as the impedance of a capacitor since the first is always positive and the second is always negative.

    But under certain conditions you can require them to be equal in magnitude.

    In order to maximize the current in a series circuit, you want to minimized the magnitude of the total impedance. You can't change the impedance of a resistor -- it is what it is. But since the impedances of the inductor and capacitor have opposite sign, the magnitude of the total impedance of those two elements is the difference between their magnitudes. If you want to minimize the total impedance, what does that say about the relationship you want between the magnitudes of the individual impedances of those two elements?
     
  16. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    If i want to minimize total impedance i should make the impedance of inductor same as impedance of capacitor, just with opposite signs, that way, their sum will be zero and i would have only resistance, is this correct?
     
  17. WBahn

    Moderator

    Mar 31, 2012
    17,746
    4,795
    Correct. So that is the point at which you will have maximum current and this IS the resonant frequency, which was given to you. So that allows you to determine a relationship between L and C that the values must satisfy.
     
Loading...