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Help with AC Circuit containing a motor

  1. Nov 24, 2014 #1
    Hello, this is not a homework question, but is a past year exam question which I'm stumped at. Any help will be appreciated.

    1. The problem statement, all variables and given/known data

    We're designing a current reduction system for an AC motor which is modeled as an inductor ##L##in series with a resistor ##R##. This motor is connected to 230##V_{rms}## at ##50hz##. An RMS meter is used to measure the current the motor takes. We need to find:
    (1) Motor Impedance
    (2) RMS current is measured as 15A. Voltage is ##v_a cos(\omega t)## and current is ##i_a cos(\omega t + \theta)##. Write down ##\omega##, ##v_a## and ##i_a##.
    (3) Voltage is now ##v_a e^{j\omega t}##, write down current and hence obtain an equation that relates ##R## and ##L##.
    (4) A variable capacitor is added in parallel that reduces current to a minimum of 12A when C = 0.1F. Write down the total impedance and find L.

    2. Relevant equations
    ##V=IZ##
    ##Z_t = R + X_l##

    3. The attempt at a solution
    (1) ##Z = R + j\omega L##
    (2)
    ##i_{peak} = \sqrt 2 \times 15 = 21.21 A##
    ##v_{peak} = \sqrt 2 \times 230 = 325.269 V##
    ##\omega = 2\times \pi \times 50hz##
    (3) This is the part that I don't really get
    Using ##V=IZ## we have
    ##i(t) = \frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} cos(\omega t + \theta)## which I think can also be written as ##\frac{v_a}{\sqrt{R^2 + (j\omega L)^2}} e^{j\omega t + \theta}##
    where ##\theta## would be ##tan^{-1} \frac{R}{\omega L}##

    Is this the relationship they are looking for?

    (4) Capacitor is added in parallel to existing impedance ##R+j \omega L##. Hence ##\frac{1}{Z} = \frac{1}{R + j\omega L} + j\omega C##

    This is as far as I've gotten. Could someone guide me along please? Thank you.
     
  2. jcsd
  3. Nov 24, 2014 #2

    gneill

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    Staff: Mentor

    Hi docsxp, Welcome to Physics Forums.

    Can you verify the value of that capacitance? 0.1F seems awfully large to place across a 230V mains. I'd expect it to pass a current of several thousand amperes! I suspect that the units are not Farads but perhaps milli-Farads (mF).
    I think that by expressing the voltage as ##v_a e^{j\omega t}## they mean to treat the voltage as a phasor (although I don't know why they chose to use the peak value rather than the RMS value).

    In that case you can write ##I = \frac{v_a}{Z} = \frac{v_a}{R + j \omega L}##. Then taking magnitudes,

    $$I_a = \frac{v_a}{\sqrt{R^2 + \omega^2 L^2}}$$

    You can define a quantity ##X_o = \frac{v_a}{I_a}## (it'll come in handy later) and then write an expression relating R and L to it. Note that when you take the magnitude of the impedance you don't include the "j" in the imaginary term.
    You can proceed as before and write an expression for the magnitude of the current phasor with the capacitor in play. Then rearrange as before to find the magnitude of ##E/I## (you have both for the second case: 230 V and 12 A RMS. Use peak if you wish). Define ##X_1 = \frac{v_a}{I_1}## where ##I_1## is your "new" current of 12 A. You should find that you can make use of the ##X_o## expression defined above to simplify things a bit and eliminate the R variable.
     
  4. Nov 24, 2014 #3
    When you use ##|Z| e^{j\Phi}## notation do you not use ##V_{peak}##? Do you still use ##V_{rms}##?
     
  5. Nov 24, 2014 #4
    So I attempted the question again and I'm still confused. When you say that I have to use ##v_a## do you mean the magnitude or do I have to treat it as a phasor? If we're focusing on magnitude then,

    ##i_a = \frac{v_a}{\sqrt{R^2 + (\omega L)^2}}##
    ##21.21 = \frac{325.269}{\sqrt{R^2 + (\omega L)^2}}##

    And for the capacitor added, the overall impedance of the circuit, in terms of magnitude becomes ##|Z| = \sqrt{R^2 + (\frac{1}{\omega C} - \omega L)^2}##

    Am I correct? Because the phasors for capacitors and inductors point in the opposite directions.

    When we define ##X_0## do we use phasor notation?
     
  6. Nov 25, 2014 #5

    gneill

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    Staff: Mentor

    When working with AC phasors the usual convention is to use the RMS values for AC voltage and current.

    When RMS values are used, calculations of power (I*V) are straight forward.
     
  7. Nov 25, 2014 #6

    gneill

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    Staff: Mentor

    When dealing with phasor values for calculation you use the magnitudes of the phasors along with their phase angles. Here you have measured values of current using an RMS current meter, so only the magnitudes are available to you.

    Your formula for the overall impedance magnitude is not correct. Your formula would apply if the capacitor was in series with the R and L, but the capacitor in this case is connected in parallel with the RL branch. How do impedances in parallel combine?

    The given values of voltage and current are phasor magnitudes. So ##X_o = 230 V / 15 A## would yield the magnitude of the impedance, that is, ##X_o = \sqrt{R^2 + \omega^2 L^2}##.
     
  8. Nov 25, 2014 #7
    Like resistors in parallel. ##Z = \frac{1}{\frac{1}{j\omega L} + j\omega C}## But this will give rise to a very complex impedance.

    Unless we leave it in ##\frac{1}{Z}## form.

    I will try again with the corrected formula.
     
  9. Nov 25, 2014 #8
    gneill, thanks for sticking with me for this long. Anyway I did the question again.

    Using the combined impedance ##Z = \frac{1}{\frac{1}{j\omega L} + j\omega C}## we can actually calculate after simplification. However this system will be most efficient when there is no complex reactance, that is to say, in the phasor diagram I want to achieve a phasor lying entirely in real. The resistor here already is in the real, but the inductor is introducing complex reactance.

    Hence I did ##\frac{1}{j\omega C} = -j\omega L## Because I want Im(Z) = 0

    ##\frac{1}{\omega C} = \omega L##
    ##L = 0.000101H##

    Is this the correct answer?
     
  10. Nov 25, 2014 #9

    gneill

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    Staff: Mentor

    No that doesn't work. You'll find that the real part of the impedance is influenced by the complex parts of the separate impedances and vice versa. Your impedance is given by:

    $$ Z = \frac{1}{\frac{1}{R + j\omega L} + j\omega C} $$

    Simplify it and take the magnitude. Note that the magnitude of this impedance will be equal to V/I for the 12 Amp case. Yes, it's a bit of a math grind...

    You should see an opportunity to substitute in the Xo that was defined earlier, and R can be replaced using the same expression.
     
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