Capacitive reactance question

Discussion in 'Homework Help' started by metelskiy, Jan 21, 2011.

  1. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    I have the following problem:
    A 5 μF capacitor is connected across a voltage source = 170sin(377t) V.
    What is capacitive reactance?
    Here is my approach (f is not given):
    I know, to find capacitive reactance i use formula X_C=\frac{1}{2\cdot\pi\cdot f\cdot C}
    So that would be \frac{1}{2\cdot\Pi\cdot 0Hz\cdot 5uF} ? I don't even need to touch V_s?
     
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  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    In generic terms a voltage source could also be written as

    Vs = Vp sin(ωt)
     
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  3. mbohuntr

    Active Member

    Apr 6, 2009
    413
    32
    Reactance is for AC voltage, therefore will have a frequency. 0hz will not work. Both inductors and capacitors will behave similar to a resistor at various frequencies, The capacitor offers less reactivity at higher frequency, and the inductor prefers lower frequencies. Things like radios, remotes and speakers use this theory to only respond to specific frequencies. When the capacitative reactance is equal to the inductive reactance, the circuit is said to be at resonance. The 377t seems like it might be time?? perhaps there is a frequency there...

    http://www.allaboutcircuits.com/vol_2/chpt_3/2.html
     
    Last edited: Jan 21, 2011
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  4. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,790
    Did it escape your notice that
    Code ( (Unknown Language)):
    1.  
    2. 2∏*60 ≈ 377
    3.  
    So your capacitive reactance is
    Code ( (Unknown Language)):
    1.  
    2. 1 / (377)(5e-6) = ? Ohms
    3.  
    BTW (1/0) would approach infinity so that cannot be the answer
     
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  5. mbohuntr

    Active Member

    Apr 6, 2009
    413
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    That's good insight papa, I didn't look into the 377t, We were always given the frequency. I've never seen it written with a "t". I was thinking the t was for time, and 1/t = f.
     
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  6. nathalie

    New Member

    Jan 9, 2011
    14
    1
    You need to use the angular frequency, ω which from the given equation is equal to 377.

    Dont forget ω = 2 ∏f


    :)
     
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  7. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    Papabravo, thanks for clearing me on this that f=60Hz
    X_C=\frac{1}{377\cdot 5uF}=1.8kohms

    On the same problem (A 5 μF capacitor is connected across a voltage source = 170Vsin(377t) what is the equation of current and there are multiple answers:

    a. i = 0.32 sin(377t+90°)A
    b. i = 0.32 sin(377t-90°)A
    c. i = 0.5 sin(377t+90°)A
    d. i = 0.5 sin(377t-90°)A
    e. i = 0.5 sin(377t+180°)

    But I don't get any of those answers. Since I=\frac{V}{R}
    So I=\frac{170Vsin(377t)}{1.8kohms}=94.4V
    What am I doing wrong?
     
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  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    I'm unsure how you ended up with 0Hz for frequency. :confused:

    170 is the amplitude, while 377t is frequency. The function will give you the amplitude and phase at time t

    2 \cdot \pi \cdot \omega = f
    Z_c = \frac{1}{\omega \cdot C

    The 2π is added to your equation to get degrees per second instead of radians per second for frequency.
     
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  9. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    Thanks I already figured out that f=60Hz, for some reason i didnt figure it out at first.
     
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  10. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,790
    Obviously calculation is not your strong suit. Remember a uF is one one-millionth of a Farad.

    (377 * 5e-6)^-1 = 1 / (377 * 5 * 10^-6) = 1 / (.001885) ≈ 531 Ohms

    The actual impedance of the capacitor is a complex number which is 531 Ohms at an angle of -90°. You could also write 0 - j531

    so you would have
    Code ( (Unknown Language)):
    1.  
    2. 170*sin(377t + 0°) / (531, -90° )
    3. ≈ 0.32sin(377t + 90°)
    4. That is
    5. {170 / 531, 0° - (-90°)} = {0.32, +90°}
    6.  
    which corresponds to answer (a.)

    Remember ELI the ICE man which tells you that current I leads voltage V in a capacitor.
     
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