Thevenin Worksheet Problem #40

Discussion in 'Homework Help' started by Red Bow Tie, Jul 8, 2012.

  1. Red Bow Tie

    Thread Starter New Member

    May 3, 2012
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    Am I correct in re-drawing the unbalanced bridge as 2 voltage dividers in parallel connected by the 2.7K resistor? Removing RL leaves me a circuit with 3 loops. Calculating branch currents and applying them to find Vth does indeed give me 7.6922 Volts.
    But, how do I calculate Rth for this circuit? If I short out the 9V supply, it (I think) leaves me with the 1.5K and 3.3K in parallel. Is this correct? The 2.7K and 1K being in series produces a 3.7K that will be in parallel with the equivalent of the 1.5K and 3.3K? Now matter what I do I cannot come up with the textbook answer of Rth=766.84 ohms. I come up with roughly 806 ohms.
    I'm obviously reading the resistors series/parallel combinations incorrectly. Can someone help me to correctly deduce Rth?
     
  2. mlog

    Member

    Feb 11, 2012
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    36
    Would you happen to have a schematic diagram or a figure that you could upload for us? Thanks.
     
  3. Red Bow Tie

    Thread Starter New Member

    May 3, 2012
    12
    1
    This is taken directly from the worksheet of Thevenin's Theorem on this site

    Observe the following circuit:
    [​IMG]


    Note that it is not reducible to a single resistance and power source. In other words, it is not a series-parallel combination circuit. And, while it is a bridge circuit, you are not able to simply analyze the resistor ratios because it is obviously not in a state of balance!
    If you were asked to calculate voltage or current for any component in this circuit, it would be a difficult task . . . unless you know either Thévenin's or Norton's theorems, that is! Apply either one of these theorems to the determination of voltage across the 2.2 kΩ resistor.
    Hint: consider the 2.2 kΩ resistor as the load in a Thévenin or Norton equivalent circuit.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This imagine shoudl help

    [​IMG]
     
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  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes the answer seems to be for an entirely different problem.
    The values I have are
    VTh=3.375V
    RTh=1.719k

    Ok I missed it was for the 2.2k case.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes we are. I was in autopilot. I took it to be the standard bridge problem with the voltage across the 2.7k being of interest. You have the matter well in hand.
     
  8. Red Bow Tie

    Thread Starter New Member

    May 3, 2012
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    Thanks so much for the help. I had the exact same re-written circuit as your top image but somehow could not see how to re-write it into your bottom image. Thanks especially for including how you got to the bottom image re-writing the circuit yet again.
     
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