Help in transformation theorems 2

Discussion in 'Homework Help' started by holybrings, Apr 4, 2007.

  1. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    its almost the same with the 1st question, but juz want to ask that should i compress the last 3 resistor 1st?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Ok, this one is a little bit more of a challenge but still you will be using the same basic steps that you employed in the previous one.

    Once again you will be able to write the expression for voltage across the 3 ohm resistor as a function of i.

    hgmjr
     
  3. hgmjr

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    Jan 28, 2005
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    If you know the voltage across the 3 ohm resistor you automatically know the voltage across the 6 ohm resistor in parallel with it.

    You then have all you need to determine the current flowing in the 6 and 3 ohm resistors. The sum of those two currents give you the current flowing in the 8 ohm resistor. Know the current in the 8 ohm resistor you can calculate the voltage across the 8 ohm resistor (Ohm's Law).

    I am going to stop here and let you digest the content and give me some of you initial calculations.

    Tell me what you get for the voltage across the 3 ohm resistor, the current in the 6 ohm resistor and the voltage across the 8 ohm resistor.

    Once we have those values established we can proceed to the next step.

    hgmjr
     
  4. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    the 3 ohm V=3i
    then the 6 ohm is A=0.3i correct?
     
  5. hgmjr

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    3i is correct for the 3 ohm resistor but 0.3i is not the correct current for the 6 ohm resistor.

    Remember, the voltage is 3i and the resistance is 6. Ohm's Law should help you figure this one out.

    hgmjr
     
  6. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    should be 0.5A correct?
     
  7. hgmjr

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    Jan 28, 2005
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    There you go.

    Now you have the current in the 6 ohm and the current in the 3 ohm.

    You should be able to get the current in the 8 ohm from that info and then you should be able to calculate the voltage across the 8 ohm resistor.

    hgmjr
     
  8. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    the 3 ohm current juz only i right?
    add up 1.5A? correct?
     
  9. hgmjr

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    Jan 28, 2005
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    Yes, 1.5i is the total current flowing in the 3 and 6 ohm resistors. It therefore follows that 1.5i amps are flowing in the 8 ohm resistor since it shares the current path with the combination of the 3 ohm and the 6 ohm resistor.

    Now you know the current in the 8 ohm resistor you can compute the voltage drop across the 8 ohm resistor.

    hgmjr
     
  10. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    means the voltage drop across the 8 ohm is 12v?
     
  11. hgmjr

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    Jan 28, 2005
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    Yes the voltage drop is 12i volts. You must remember that it is critical to keep the "i" in the resulting calculated value.

    You are now ready to calculate the total voltage (in terms of i) across the entire leg of the circuit that includes the 8 ohm resistor and the parallel 6 and 3 ohm resistors.

    You calculated the voltage across the 3 ohm resistor to be 3i volts and that is the same voltage across the 6 ohm resistor. With the voltage across the 8 ohm resistor and the voltage across the 3 and 6 ohm resistor you have all that is needed to compute the voltage across this leg of the circuit.

    What do you get for that voltage in terms of i?

    hgmjr

    hgmjr
     
  12. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    juz add up all the 3 voltage?
     
  13. hgmjr

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    Jan 28, 2005
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    Nope.

    Remember that the voltage is the same across the 3 ohm and the 6 ohm resistor. You only need to account for that voltage once. So the total voltage is made up of the voltage across the 8 ohm resistor and the voltage across the parallel combination of the 3 and the 6 ohm resistor.

    This takes a bit of getting use to but you will get the hang of it after you have done enough problems.

    If you haven't already recognized what we are actually doing here, it is just the successive application of Ohm's Law, Kirchoff's Voltage Law, and Kirchoff's Current Law.


    hgmjr
     
  14. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    means that the voltage is 15i?
     
  15. hgmjr

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    Jan 28, 2005
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    You got it.

    With the value 15i volts, you now know the voltage across the 10 ohm resistor. Ohms Law will allow you to calculate the current in the 10 ohm resistor in terms of i.

    You will then have the current in each of the three main legs of the circuit and you can use the same technique you used on the earlier problem. That technique being to set up the equation with all the currents (not the voltages) you have in the three legs on one side of the equation and the value of your current source on the other side of the equation and then you can solve for i just as you did previously.

    You are so close to completing this one.

    hgmjr
     
  16. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    15i=I(10)
    I=1.5i? is it?
     
  17. hgmjr

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    Jan 28, 2005
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    Yelp. Once again you have the right answer for the current in the 10 ohm resistor.

    As I said earlier, you have all the info you need to set up your equation and calculate i just as you did on the previous problem.

    hgmjr
     
  18. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    the 1.5i is the 1st 10 ohm or the 10 ohm at the back?
     
  19. holybrings

    Thread Starter Active Member

    Apr 4, 2007
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    is the back one, but how about the 1st 10 ohm?
     
  20. hgmjr

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    Jan 28, 2005
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    It is true that the 8 ohm resistor in series with the parallel combination of the 3 ohm and the 6 ohm resistor is 10 ohm, however I was referring to the individual 10 resistor on the left side of your circuit.

    hgmjr
     
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