Thevenin Equivalent Voltage: why ignore the 3-kΩ resistor?












1














enter image description here



In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)






passive-networks thevenin







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    The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
    – thece
    58 mins ago






  • 1




    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    – Tony EE rocketscientist
    53 mins ago
















1














enter image description here



In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)






passive-networks thevenin







share|improve this question




















  • 1




    The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
    – thece
    58 mins ago






  • 1




    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    – Tony EE rocketscientist
    53 mins ago














1












1








1







enter image description here



In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)






passive-networks thevenin







share|improve this question















enter image description here



In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)






passive-networks thevenin




passive-networks thevenin






share|improve this question















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edited 40 mins ago









Verbal Kint

3,1941312




3,1941312










asked 1 hour ago









fred

9218




9218








  • 1




    The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
    – thece
    58 mins ago






  • 1




    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    – Tony EE rocketscientist
    53 mins ago














  • 1




    The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
    – thece
    58 mins ago






  • 1




    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    – Tony EE rocketscientist
    53 mins ago








1




1




The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
58 mins ago




The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
58 mins ago




1




1




The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
53 mins ago




The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
53 mins ago










3 Answers
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The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).



When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.






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    4














    The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.






    share|improve this answer





























      2














      Computing a Thevenin Equivalent requires two steps:




      1. Obtain the Thevenin Impedance $Z_{TH}$

      2. Obtain the Thevenin Voltage $V_{TH}$


      In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.



      In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!



      Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
      $$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
      No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
      $$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
      Being I = 8mA and R = 7KOhms.



      This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.





      schematic





      simulate this circuit – Schematic created using CircuitLab






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      P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        3 Answers
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        The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).



        When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.






        share|improve this answer




























          4














          The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).



          When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.






          share|improve this answer


























            4












            4








            4






            The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).



            When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.






            share|improve this answer














            The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).



            When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 14 mins ago

























            answered 1 hour ago









            Spehro Pefhany

            203k4150408




            203k4150408

























                4














                The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.






                share|improve this answer


























                  4














                  The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.






                  share|improve this answer
























                    4












                    4








                    4






                    The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.






                    share|improve this answer












                    The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 1 hour ago









                    Shamtam

                    2,3831022




                    2,3831022























                        2














                        Computing a Thevenin Equivalent requires two steps:




                        1. Obtain the Thevenin Impedance $Z_{TH}$

                        2. Obtain the Thevenin Voltage $V_{TH}$


                        In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.



                        In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!



                        Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
                        $$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
                        No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
                        $$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
                        Being I = 8mA and R = 7KOhms.



                        This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.





                        schematic





                        simulate this circuit – Schematic created using CircuitLab






                        share|improve this answer








                        New contributor




                        P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.























                          2














                          Computing a Thevenin Equivalent requires two steps:




                          1. Obtain the Thevenin Impedance $Z_{TH}$

                          2. Obtain the Thevenin Voltage $V_{TH}$


                          In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.



                          In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!



                          Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
                          $$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
                          No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
                          $$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
                          Being I = 8mA and R = 7KOhms.



                          This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.





                          schematic





                          simulate this circuit – Schematic created using CircuitLab






                          share|improve this answer








                          New contributor




                          P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





















                            2












                            2








                            2






                            Computing a Thevenin Equivalent requires two steps:




                            1. Obtain the Thevenin Impedance $Z_{TH}$

                            2. Obtain the Thevenin Voltage $V_{TH}$


                            In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.



                            In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!



                            Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
                            $$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
                            No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
                            $$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
                            Being I = 8mA and R = 7KOhms.



                            This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.





                            schematic





                            simulate this circuit – Schematic created using CircuitLab






                            share|improve this answer








                            New contributor




                            P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            Computing a Thevenin Equivalent requires two steps:




                            1. Obtain the Thevenin Impedance $Z_{TH}$

                            2. Obtain the Thevenin Voltage $V_{TH}$


                            In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.



                            In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!



                            Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
                            $$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
                            No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
                            $$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
                            Being I = 8mA and R = 7KOhms.



                            This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.





                            schematic





                            simulate this circuit – Schematic created using CircuitLab







                            share|improve this answer








                            New contributor




                            P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|improve this answer



                            share|improve this answer






                            New contributor




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                            answered 36 mins ago









                            P. Collado

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                            212




                            New contributor




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                            P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            Check out our Code of Conduct.






























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