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Homework Help: Period & peak current

  1. Aug 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the value of ω such that the peak value of i(t) is 2A.


    2. Relevant equations

    Ipeak = (Irms *(2)1/2)

    Zc = -j/ωC

    I = V/Z

    3. The attempt at a solution

    Irms = 2A/(2)1/2

    Irms = 1.41A

    1.41A = 25V (?) /( ω * 0.002F)

    ω = 25V / (1.41A * 0.002F)

    ω = 8865 r/s
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2013 #2
    This isn't the impedance of the capacitor. There's something missing (remember brackets for the denominator aswell).

    There isn't anything in the problem that calls for RMS values, so you could use peak-amplitude phasors instead. You just have to be consistent.

    I assume you're using the relationship |I| = |V|/|Z|, where I and V are the current and voltage phasors, respectively, and Z is the equivalent impedance of the load. Two things though:
    1. You're mixing up RMS and peak values. 25 V is the peak amplitude of the voltage source.
    2. The RHS denominator isn't the magnitude of the equivalent impedance Z. Where does the 10 Ω factor in?
  4. Aug 22, 2013 #3
    fixed that now

    So they're both in phase then, and angles are 0 for both?

    So it's really (or can be)

    2A = 25V / [1/(j* ω * 0.002F)]

    2A [1/(j * ω * 0.002F)] = 25V

    2A/(j * ω * 0.002F) = 25V

    2A / 25V = (j * ω * 0.002F)

    ω = 40
  5. Aug 22, 2013 #4
    No, the current and voltage would only be in phase if the equivalent impedance of the load was purely real. In your case, it has an imaginary component due to the contribution from the capacitor.

    What is the equivalent impedance of your load? I see a resistor in your schematic aswell and:

    Assuming you're using this relationship:
    |I| = |V|/|Z|

    shouldn't it be the magnitude of the equivalent impedance instead?

    Also, solving for ω:
    You'd get ω = -j*40 (disregarding units), which should tell you that you've taken a wrong turn somewhere. You can't just discard something from an expression because it doesn't fit with your picture of what it should be :wink:.
    Last edited: Aug 22, 2013
  6. Aug 22, 2013 #5
    How can you get a phasor if you aren't even allowed to know what the capacitor impedance is?

    Ok, let me try again:

    I = V/Z

    Z = V/I

    V/I = Z

    25V / 2A = (-1j/ω * 0.002F) + 10Ω

    12.5Ω = (-1j/ω * 0.002F) + 10Ω

    2.5Ω = (-1j/ω * 0.002F)

    ω = -1j/(2.5Ω * 0.002F)

    ω = -1j/0.005

    ω = -200j

    -j = 1/j (which is probably the main thing causing me problems here)

    ω = (1/200)

    ω = 0.005/js
  7. Aug 22, 2013 #6
    Let's take a step back for a second. Have a look at the relationship:
    \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}}
    Here ##\mathbf{V}## and ##\mathbf{I}## are phasors and ##\mathbf{Z}##, the impedance, is a complex number that relates them.

    For your circuit, if we choose the voltage phasor as the reference, we have:
    \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = 2\angle\arg(\mathbf{I}) \, \mathrm{A} = \frac{25\angle0^\circ}{10 - j\frac{1}{0.002 \omega}} \, \mathrm{A} \qquad (1)
    That's one equation in two unknowns, so we can't use it alone to determine ω uniquely. We could, however, use this relationship instead:
    |\mathbf{I}| = \left|\frac{\mathbf{V}}{\mathbf{Z}}\right| = \frac{\mathbf{|V|}}{\mathbf{|Z|}} \qquad (2)
    where ##|\cdot|## denotes the absolute value/modulus/magnitude of a complex number. You've seen that notation before?

    (2) should give you one equation in one unknown, ##\omega##.
    (1) & (2) should give you two equations in two unknowns, ##\omega## and ##\arg(\mathbf{I})##, if you cared about the phase angle of the current (which you don't in your problem).

    Try writing (2) again, but this time make absolutely sure you're using the magnitudes of the voltage and current phasors and the impedance.

    Just to make it clear, if the equivalent impedance of your circuit is given by Z = R + j*X, what is the absolute value/modulus/magnitude of the impedance?
    Last edited: Aug 22, 2013
  8. Aug 22, 2013 #7
    So use ( r2 + j2 )0.5 with the total Z then (where r is the real resistance & j the imaginary)?

    Sorry to ask, but what do you mean by "arg(I)" in the left hand side for your equation (1)? Is it the phase angle of the current ?

    So the current angle is unknown too and you can NOT just say Z = 25V / 2A? Are you also saying the impedance angle is 0?
  9. Aug 22, 2013 #8
    Well, sort of. There's no such thing as real resistance and imaginary resistance. The real part of the impedance is called the resistance. The imaginary part is called the reactance. j is the imaginary unit.

    For an impedance Z = R + j*X, where R is the resistance and X is the reactance, we have:
    |Z| = √(R2 + X2)

    Yes. arg(I) is a function that maps the phasor I to its phase angle. Arg is short for argument:

    If by Z you mean the magnitude of the impedance, then sure you can.

    No. I don't know what has given you that idea.

    All I did in (1) was insert the known values for your circuit and I used symbols for the parts that were unknown.

    25 V is the magnitude of the voltage phasor. 2 A is the magnitude of the current phasor. What is the magnitude of the equivalent impedance of your circuit?
  10. Aug 22, 2013 #9
    You know I meant the components there :P Guess I should be more clear too.

    Alright so, then 12.5 is still the magnitude of the total impedance. I kind of see where I messed up now ;/

    So you still have the magnitudes of the current, voltage, and impedance?

    12.52 = (102 + j2)

    156.25 - 100 = j2

    56.25 = j2

    j = 7.5 ("magnitude" of the imaginary component)

    7.5 = j * (-1/(0.002F * ω))

    ω = -j1/(7.5 * 0.002F)

    ω = 0.015 ?
  11. Aug 22, 2013 #10
    The equivalent impedance of your circuit ##\mathbf{Z}## is given by:
    $$\mathbf{Z} = R + jX = (10 - j\frac{1}{0.002\omega}) \, \Omega$$
    What is ##|\mathbf{Z}|## then?
  12. Aug 22, 2013 #11
    Don't I already have ω ?

    It would be 10 - 7.5j
  13. Aug 22, 2013 #12
    You know ω in the sense that it's the solution to:
    7.5 Ω = 1/(0.002*ω) Ω

    But that's not the equation you wrote and your value for ω isn't correct. Let me show you what I've been getting at:
    \mathbf{Z} = R + jX = (10 - j\frac{1}{0.002\omega}) \, \Omega, \, R = 10 \, \Omega, \, X = -\frac{1}{0.002\omega} \, \Omega\\
    |\mathbf{Z}| = \sqrt{R^2 + X^2} = \sqrt{10^2 + \left(\frac{1}{0.002\omega}\right)^2} \, \Omega
    Since you know ##|\mathbf{V}|## and ##|\mathbf{I}|##, you have everything you need to substitute into:
    |\mathbf{I}| = \frac{\mathbf{|V|}}{\mathbf{|Z|}}
    Which gives you one equation in one unknown, ω.
  14. Aug 22, 2013 #13
    Seems I didn't take the inverse right at the end right ; /, otherwise everything else is exactly what you did pretty much.

    ω = -j1/(7.5 * 0.002F)

    ω = 1/0.015

    ω = 66.667
  15. Aug 23, 2013 #14
    You got the right result, but there's still something off about the way you went about getting it. I'll get straight to the point:

    There should be no -j factor in this equation. j is the imaginary unit and is defined such that j2 = -1. The RHS of the equation is a complex number, but the LHS is supposed to be a real number, so you can tell there's something wrong with it.

    If we go back to:
    Here you're using the symbol 'j' as a variable, which is bad form because you might confuse it with the imaginary unit. If you really do mean to use 'j' as a symbol for the imaginary unit then:
    12.52 = (102 + j2) = (100 + (-1)) = 99

    Which is why I've been trying to get you to use 'X' as a symbol for the imaginary part of the impedance (the reactance).

    Following your procedure:
    |\mathbf{Z}|^2 = R^2 + X^2 \Rightarrow X^2 = |\mathbf{Z}|^2 - R^2 \Rightarrow \sqrt{X^2} = |X| = \sqrt{|\mathbf{Z}|^2 - R^2} = \sqrt{12.5^2 - 10^2} \, \Omega = 7.5 \, \Omega\\
    |X| = \left|-\frac{1}{0.002\omega}\right| \, \Omega = \frac{1}{0.002\omega} \, \Omega = 7.5 \, \Omega \Rightarrow \omega \approx 66.7 \, \frac{\mathrm{rad}}{\mathrm{s}}
    The difference isn't trivial. You might think I'm being nitpicky, but if you don't correct it now it's bound to get you into trouble later when doing calculations with complex numbers. You'll get all sorts of odd results, and that's the headache I'm hoping to spare you.
  16. Aug 23, 2013 #15
    Haha, alright thanks. How about I use a capital J for that then ?
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