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Magnitude of voltage across impedance

  1. Oct 4, 2010 #1
    1. The problem statement, all variables and given/known data

    The following three impedances are connected in series across a 100-V, 20-kHz supply:
    (1) a 12-Ω resistor, (2) a coil of 100 μH inductance and 5-Ω resistance and (3) a 390-nF
    capacitor in series with a 15 Ω resistor. Sketch the circuit diagram, impedance diagram and the phasor diagram; take the supply voltage V as the reference phasor.

    (a) Calculate the magnitude of the voltage V2 across the second impedance.

    (b) Calculate the magnitude of the voltage V3 across the third impedance.

    2. The attempt at a solution

    Ok, I'm pretty lost here, but this is what I have done so far...

    I'm trying to get the current amount flowing in the circuit, so thought I should get a total impedance Z for the circuit.

    First resistor = 12 Ohms
    For the coil = 5 + 2 * Pi * (20*10^3) * (100*10^-6) = 17.6 Ohms
    For the Capacitor = 1 / 2 * Pi * (20*10^3) * (390*10^-9) = 20.4 Ohms
    Second resistor = 15 Ohms

    Total Z for the circuit = 12+17.6+20.4+15 = 65 Ohms.

    That is as far as I have gone so far. Is that the correct way to get total Z?

    Next I will find the current by using I = V/Z ?

    I guess that would be 1.54 A ?
  2. jcsd
  3. Oct 5, 2010 #2
    Keep in mind that capacitors and inductors are complex values.

    X_l=j17.6 ohm
    X_c=-j20.4 ohm

    Total impedance is then Z=R+j(X_l-X_c)

    Sketch up the impedance diagram(complex plane) and you see why.
  4. Oct 5, 2010 #3
    Ok. I see that in this diagram:

    [PLAIN]http://macao.communications.museum/images/exhibits/small/2_4_4_1_eng.png [Broken]

    Does that mean my total Z is in fact 24.2 Ohms?

    I came to that conclusion by using 27 + (-2.8)

    Sorry if I'm not getting it, I'm finding it hard to get my head around these AC basics :blushing:
    Last edited by a moderator: May 5, 2017
  5. Oct 5, 2010 #4
    Nope. The total impedance is Z=27-j2.8

    You cant sum real and imaginary numbers. Just like vectors, the impedance consist of an angle and magnitude.

    Current is then: I=U/R=100+j0 / 27-j2.8=....A
  6. Oct 6, 2010 #5
    I've come up with a figure of 3.69 A

    How does that sound?

    Thanks for posting back by the way :)
  7. Oct 6, 2010 #6
    Wrong. It's supposed to be a complex number with both real and imaginary parts. (i.e I=2+j3 A or 3A <(angle) 36*)
    Using a scientific calculator does this job easy, calculating by hand requires using complex conjugate. Refer to your calculus textbook.

    Said in other words, the current has a phase angle in reference to the voltage source (leading or lagging current, Power factor and so forth)

    BTW: I have to correct my answer Z=32 - j2.8 ohm
    Didnt read the text thoroughly Total resistance is: 12 + 5 + 15 =32 ohm
    But still, your current must have a phase angle.
  8. Oct 7, 2010 #7
    Gah... Back to the drawing board.

    Was working on other parts of my assignment today, I'll give this another shot tomorrow.

  9. Oct 10, 2010 #8
    Ok... Had another attempt.

    Total Z = 32 - j7.8 (rather than 2.8 as before)

    I = 3 A /_ -13.7

    Magnitude of Voltage across coil = 40.8V

    Magnitude of Voltage across both 15 Ohm resistor & Capacitor = 75.9V
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