Verifying Answers for 2 questions

Discussion in 'Homework Help' started by circuit_boy, Sep 18, 2010.

  1. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    Good day fellas, I have some homework questions that I would like to check with the veterans in this area.

    If possible, can anyone use multisim to verify? or PSPICE?

    Here are the pictures of my questions and workings.

    Question 1: http://img690.imageshack.us/i/79806671.jpg/
    Answer to Question 1: http://img831.imageshack.us/i/imgmp.jpg/

    Question 2: http://img833.imageshack.us/i/48920688.jpg/
    Answers to Question 2 Part 1 :http://img545.imageshack.us/i/img0001e.jpg/
    Answers to Question 2 Part 2 :http://img94.imageshack.us/i/img0002te.jpg/

    Here are my answers for Question 2, if the solution is too long for you to bother with =D
    a) Vth = 13V, Isc = 13A, Rth = 1Ohm
    b) Vx = 12V
    c) Vx = 12V
    d) 8 watts, 2 watts, and 12 watts
    e) Independent sources 16V and 10V absorb power, dependent source 1.2Vx delivers power

    Feel free to give me any inputs! Your help is really appreciated!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Side comment, we have albums here at this site, and you can attach up to 6 images per post. It is the preferred way to do images (no ads, no chance of malware).
     
  3. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    I'm new here and I'll make sure to keep that in mind! Thank you, and mind evaluating my work? =D
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Q1. Mostly right except for the RMS value which is simply 5A.

    Q2. All looks OK but I suggest the 16V source is delivering rather than absorbing power and the controlled current source is absorbing power.

    The 16V source supplies 262.4W, the controlled current source absorbs 86.4W, the 10V source absorbs 154W and the resistors absorb 22W.
     
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  5. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    Thank you for correcting me on the RMS value!
    Is it alright if you would run a simulation using multisim? For the second question that is. I tried re-calculating using mesh analysis to find the direction of the current. And found out that 16v and controlled current source is supplying, while 10v is absorbing, how did you come up with your answer? Thank you!
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I don't think I need to run a simulation of the circuit - humbly submit I have sufficient confidence in my ability to calculate the answer.

    The reasoning for my conclusions about which sources are absorbing and which are delivering power is as follows:-

    You have correctly found the voltage Vx to be 12V. This means the current source is driving 14.4A (1.2*Vx) in the direction indicated (left to right).

    The potential drop across the current source is 6V - as constrained by the 16V and 10V DC sources. The left hand side of the controlled source is more positive than the right hand side so if we were to mark the most positive terminal of the source it would be the left hand one. So the current direction into the controlled source and the controlled source polarity are such that there is a power flow into the controlled source. If the source were delivering power the current would be flowing out of the more positive terminal - as would be the case with either a voltage or current source.

    The net flow of current out of the positive terminal of the 16V DC source is 16.4A - 2A into the resistor and 14.4A to the controlled current source.
    Since this 16.4A is flowing out of the 16V DC source positive terminal this source must be delivering power into the circuit.

    By similar reasoning the total current flow into the 10V DC source is 15.4A - 1A from the resistor and 14.4A from the current source. But in the case of the 10V source the current is flowing into the positive terminal meaning this source is absorbing power.

    Remember that the nett power balance for the overall circuit must be consistent - power absorbed must equal power delivered.
     
    Last edited: Sep 19, 2010
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  7. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    I understand most of the part, except for the current that flows into the controlled current source. Current from the 16V source is entering the controlled current source through the negative terminal, isn't it suppose to be delivering energy to the circuit by your concept? Since current coming out of the positive terminal would be the same as current going into the negative terminal.

    And through mesh analysis, I have found that the direction of the current produced by the controlled current dependent source is going clockwise in the upper loop.

    Please do correct any misconceptions I have made, I would like to learn more, thank you!
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As I said in my last post the two DC voltage sources constrain both the magnitude & polarity of the potential difference across the current source. All that the current source does is set the current in that branch. The polarity of the voltage drop across the source in this case is positive on the left and negative on the right - not negative on the left as you have stated or rather assumed.

    That's quite true but don't assume just because the current is flowing in that direction that the current source polarity is positive on the right - it's not 'obliged' to behave like a voltage source. At the risk of being annoyingly repetitious, I'll restate that in this case the two voltage sources have control over the current source polarity & potential difference. All the current source wants to do is set the correct current.

    Also did you consider my assertion that the power input must balance the power losses in the circuit? If you can resolve this issue in your mind (& on paper) I believe you will gain some additional insights.
     
    Last edited: Sep 19, 2010
  9. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    One last thing, the controlled voltage receives a 14.4A from the 16V source, since the power absorbed is 86.4W, V = IR will give us a solution that V = 6 for the controlled current source. Mind explaining how is the 6 found? I've tried to recalculate it several times, but to no avail
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's one of those things that can be immediately obvious or frustratingly not so obvious!

    I guess you haven't digested my point about the two DC sources constraining both the magnitude & polarity of the current source.

    Perhaps try to get the gist of the following line of reasoning ....

    Take the common 'rail' joining the negative sides of the 16V and 10V sources as the reference node.

    The node at the positive side of the 16V source will have a node voltage of 16V (relative to reference). This is the same node as the left hand side of the controlled current source. So the left hand side of the current source is at +16V. The node at the positive side of the 10V source will have a node voltage of 10V (relative to reference). This is the same node as the right hand side of the controlled current source. So the right hand side of the current source is at +10V.

    The potential difference between the left and right hand sides of the current source is therefore 16V-10V = 6V.
     
    Last edited: Sep 20, 2010
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  11. circuit_boy

    Thread Starter New Member

    Sep 18, 2010
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    Thank you so much, now I have comprehended this question. Your help has helped me tighten up some lose ends!
     
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