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  • Question as requested by Woussko

    Find the voltage at Vo. There are several ways to do this, the most elegant of which is to use a theorm that begins with a T.
    Whats the answer, what is the Theorm
    Attached Files

  • #2
    Re: Question as requested by Woussko

    Any chance you mean Theorem rather than Theorm?

    Right now my brain won't go into gear. I would have to do it by cheating and I'm going to pass on that.
    My bet is someone (with a brain that actually does work) will come up with the answer.

    Please no hinting.
    Last edited by Woussko; 08-11-2008, 09:30 PM.

    Comment


    • #3
      Re: Question as requested by Woussko

      Um ya Theorem, spent too many cells finding a good question.
      I will post the name of the theorem tomorrow if no one gets it

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      • #4
        Re: Question as requested by Woussko

        I was sleeping throughout all my EE core classes i really should know this.... Ill see if i can come up with the answer by morning

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        • #5
          Re: Question as requested by Woussko

          actually doesnt kirchoff's law state that the sum of all voltages within a circuit are equal to 0? therefore regardless of the resistance the voltage would be -30, reversed polarity?

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          • #6
            Re: Question as requested by Woussko

            thevinem's therom
            sorry about the spelling


            cactus man

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            • #7
              Re: Question as requested by Woussko

              Thevenin is the theorem.
              Kirchoff's Current Law will also solve this but that law states that the sum of all currents at any node in a circuit must equal zero.
              Kirchoff's Voltage Law is the sum of all voltages about a closed loop in a circuit is equal to zero (ie the sum of the voltage drops across each resistor in the loop less the voltage supplied to the node equals zero)
              30 (-30) is not the answer.

              Comment


              • #8
                Re: Question as requested by Woussko

                As for terms, theory and math formulas I would totally blow this. I bet however I can come pretty close to the value of Vo

                WB knows the answers to all of this, I'm sure. Unlike brainless me the howler, he actually can get his brain into the correct mode of thinking.

                For fun sometime I'll cheat and make up the circuit as close as I can and will state the resistance of each resistor I use. I'll also use a lower power supply Voltage. No way am I going to try to obtain 1% or better wire wound power resistors to make up this circuit. Sorry but I really need the $$$ for other things.

                Wayne, Please no telling. I really want to figure out Vo on my own and see if what I come up with and what you already know Vo to be are close if not exactly the same value.
                Last edited by Woussko; 08-12-2008, 11:03 PM. Reason: be, not me --- spelling error

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                • #9
                  Re: Question as requested by Woussko

                  WB

                  Hang in there and sometime I'll post what I think is the correct Voltage Output for your circuit. I did simplify it some. When my old thinker works again if it ever does then I'll figure it out. Until then, unless someone else gets it right please no hints for "The Old Howling Hound" here. Thanks

                  Comment


                  • #10
                    Re: Question as requested by Woussko

                    I come up with Vo as 9.1 Volts and I didn't make up the circuit. I used my own crazy math formula. Am I even close? Sometime I will make up the circuit but in a simplified way. You don't really need all the resistors shown. Then once made I'll take some measurements. The only difference is the Vi will be 3.0 rather than 30 Volts.

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                    • #11
                      Re: Question as requested by Woussko

                      you are within 5%

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                      • #12
                        Re: Question as requested by Woussko

                        Originally posted by wbrooks View Post
                        you are within 5%
                        Thanks

                        I did it by simplifying the circuit down to 4 resistors and I'm sure you understand that part well. Then I did lots of rounding off of numbers. Sometime I'll work more on getting it just right. I was mostly trying to see if my shortcuts more or less work or not. Put simple my old brain has enough dead cells in it.
                        Last edited by Woussko; 08-14-2008, 08:49 AM.

                        Comment


                        • #13
                          Re: Question as requested by Woussko

                          The full solution, Kirchoff current method

                          Replace the right half of the circuit with the equivalent resistor.
                          The load leg has 3 series resistors 20,30, and 40 ohm on parallel with the middle 40 ohm resistor ((20+40+30)x40)/((20+40+30)+40)=27.7 Ohms



                          Find total current I=V/R = 30/(10+27.7+5) = 0.7 amps

                          Find the voltage drop across the 27.7 ohm equivalent resistor

                          V= IR = .7*27.7= 19.4V

                          Now go back to the original circuit knowing that the node voltage across the middle point is 19.4V



                          So I1 would be I=V/R= 19.4/40= 0.485 amps
                          Kirchoffs laws says that IT=I1+I2 so I2=0.7-0.485= 0.215 amps
                          Vo= I2*40= 8.6V
                          Attached Files

                          Comment


                          • #14
                            Re: Question as requested by Woussko

                            Thevenin's Theorem

                            1. remove the load
                            2. determine voltage as seen by the load Vth
                            3. replace the voltage source with a short (not like shorting the power supply)
                            4. determine the resistance seen by the source Rth


                            With the load removed current only flows in the first half of the circuit
                            Vth = voltage across the 40 ohm resistor
                            Vth=(40/(10+40+5))*30= 21.8

                            Now replace the source with a short and solve Rth, resistance seen by the load



                            So 40 in parallel with 10 +5 which is then in series with 20 and 30

                            Rth=((40*15)/(40*15))+20+30 = 60.9 ohm

                            The Thevenin equivalent circuit would be ....



                            Vo = (40/(40+60.9)*21.8= 8.6 V
                            Attached Files

                            Comment


                            • #15
                              Re: Question as requested by Woussko

                              Thanks for the information. I need to print this post and keep it with my notes here. An engineer I am not and never will be. Brain it too old and has too many bum cells .

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