Exam revision help.

Discussion in 'Homework Help' started by Uyet123, May 4, 2015.

  1. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    Can you help how to do this question please, I have no idea what impedance is, and my lecture notes does not show what impedance is either. Can you tell me the steps on how to solve this problem.

    New_Picture.jpg
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Did you google impedance?

    impedance of resistor = R
    impedance of inductor = jωL
    impedance of capacitor = 1/(jωC)

    Now go ahead and show your work.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    How about your text?

    How about the Great Google?
     
  4. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    We don't use textbooks on this module, just the lecture notes:). I tried Google but no luck.
     
  5. WBahn

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    Mar 31, 2012
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    You tried Googling "impedance" and got no relevant hits?

    Gee. When I Googled that word the very top hit was:

    http://en.wikipedia.org/wiki/Electrical_impedance
     
  6. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    well, I do not know if I can use that formulas since my tutor did not teach it nor it is in any of our lectures.

    Anyways, here are my answers:

    i. 36 ohms
    ii. 0.05 Amps
    iii. 36 ohms
    iv. 0.17 Amps (should this be the current for the 50 ohm resistor?)
    v. 0.04 Amps(again current for the middle resistors?)

    I used principle of superposition to solve the problem, correct me if I am wrong.
     
  7. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    for question ii is it the current when I short the voltage from the right hand side?

    for i and iii is it the total resistance for each when I short either one of the voltages? Since there are only resistors, and you said impedance of resistors is as is, I just get total resistance?
     
  8. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    how can your impedance be 36 ohms when you have a 50 ohms resistor in series...?
     
  9. WBahn

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    Mar 31, 2012
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    It doesn't really do to just post your answers -- you need to show your work so that we can see what you are doing right and what you are doing wrong.

    In the meantime you need to start asking if your answers make sense. In particular, what is the maximum value that the resistance could be for (i) and what is the minimum value that the resistance can be for (iii)? Do both of those agree with your answers?

    Then ask if the current in the leftmost branch is 50 mA and the current in the rightmost branch is 170 mA, how can the current in the middle branch possibly be 40 mA?

    Always, always, ALWAYS ask if the answer makes sense.
     
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  10. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    Ok for part (i) it asks for impedance seen by the left supply, so I think I have to short the supply from the right and get the total resistance : (1/140 + 1/50)^-1 = 36 Ohms

    (ii) for this part, I think its just the total current if I short the rightmost voltage supply, so I = V/R = 2/36 = 0.05 Amps

    (iii) same as part i but I short the leftmost supply and work with just the rightmost supply, so total resistance: 50 + 60 + 80 = 190 Ohms

    (iv) current in the right most branch, I think its the total current for the 50 Ohm resistor?
    I(A) = middle resistor and right resistor is in parallel so voltage is the same, then to solve for current in 50 ohm resistor: I = V/R = 2/50 = 0.04 Amps
    I(B) = if I short left most branch, resistors are in series, so current is the same: I = V/R = 5/190 = 0.02 Amps
    I(total) = 0.02 + 0.04 = 0.06 Amps

    (v) works the same as part iv: shorting right most branch: I(A)= 2/140 = 0.01 Amps
    I(B) = shorting left most branch: 0.02 Amps
    I(total) = 0.01 + 0.02 = 0.03 Amps


    Here are my answers with explanation. Tell me if I get something wrong
     
    Last edited: May 5, 2015
  11. WBahn

    Moderator

    Mar 31, 2012
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    That is correct.

    So you are saying that the rightmost source has absolutely no effect on the current drawn from the other supply? Does that make sense.
    You also need to indicate polarity, since 50mA flowing upward is very different than 50mA flowing downward. Don't make people, including yourself later, guess.
    You also need to start tracking your units -- 2/36 is a number, not a current. 2V/36Ω is a current.

    If the leftmost supply is shorted, then how do you claim that the three resistances are in series. Does every electron that goes through the 50Ω have to go through the other two resistors? If not, then they are not in series.

    What is I(A) and (B). DEFINE YOUR TERMS!
    Again, what is your reasoning for claiming that all the resistors are in series AFTER you put a short (i.e., a 0Ω resistor) across two of them?
    Again, what is the direction of the 60 mA.

    Also, why did you feel you could find the current in the leftmost supply by just taking it's voltage divided by the resistance it sees with the other supply turned off but yet you didn't believe you could do the same thing for the rightmost supply?

    Now ask if your answers are consistent. If (ii) says you have 50 mA flowing upward in the left branch and (iv) means you have 30 mA flowing upward in the right branch, wouldn't that require that you have 80mA flowing downward in the center branch? Or if the right branch was flowing downward, then you would have 20mA flowing downward in the center branch. Or, for the other two possibilities for your current polarities in the left and right branch, you could have 20 mA flowing upward or 80 mA flowing upward. Do you begin to see why current direction matters?

    Also ask if your answers make sense for the problem given. If you have 30mA in the center branch, then you have a voltage across the center branch of (30mA)(140Ω)=4.2V (polarity unknown since you didn't specify it). But you have a 2V voltage across that same branch. Does that make sense?
     
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  12. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    Yeah, because it only asks for the current drawn from the left supply voltage? Correct me if I am wrong.

    I don't know much about polarity, but how do you determine if its flowing down or up?

    Ok, so the current will not flow through the middle resistors since it will prefer the shorted circuit from the left thus making the total resistance only 50 Ohms.

    I(A) means the current on the 50 ohm resistor when I short the right supply and I(B) means the current on the 50 ohm resistor when I short left supply.

    Going by my new answer for part iii. I(A) = V/R = 2V/50 ohms = 0.04 Amps
    I(B) = 5v/50 omhs = 0.1 Amps
    I just add them together = 0.14 Amps.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    To explore that line of thinking, let's use a simplified example:
    2batt.png
    How much current is flowing in the left battery?

    The circuit you have been given is a bit ambiguous. The voltage sources are shown as being sinusoidal sources and so the voltages that each produce alternate between positive and negative over time, so the stated voltages, the 2V and the 5V, are the amplitudes of the sinusoidal waveforms. However, we don't know if the waveforms are sin(wt) or (cos(wt). But that really doesn't matter because we are only looking for the magnitude of the resulting currents so as long as we have the relative phase of the two sources correct we will get the correct answer. Normally one end of an AC source is labeled as positive to serve as a reference polarity, but that is not the case here. However, the orientation of both sources is the same, so it is reasonable to assume that the polarity of the top of each source is the same. So without loss of generality, we can simply declare the top of each source to be the positive terminal. We can declare the current to be positive in which ever direction we want, as long as we are then consistent with that choice.

    1.png

    Correct.

    Again, what are the polarities of the two currents. When you calculate I(A), you get a current going downward in the 50Ω resistor and when you calculate I(B) you get a current that is going upward in the 50Ω resistor. So one of them is negative relative to the other.

    Always ask if the answer makes sense. What if the two voltage sources were the same? Would there be any reason for current to flow in the 50Ω resistor at all? What would your approach yield? Are those two answers consistent with each other.
     
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  14. Uyet123

    Thread Starter New Member

    Jan 25, 2015
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    so if I assume downward is negative then: I(total) = -I(A) + I(B)

    Thanks for the help.
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    Yes, you are getting it. Keep up the struggle (it's how we tend to learn best).
     
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