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












5












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    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
    $endgroup$
    – thece
    Jan 4 at 16:36






  • 1




    $begingroup$
    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    $endgroup$
    – Sunnyskyguy EE75
    Jan 4 at 16:41
















5












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    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
    $endgroup$
    – thece
    Jan 4 at 16:36






  • 1




    $begingroup$
    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    $endgroup$
    – Sunnyskyguy EE75
    Jan 4 at 16:41














5












5








5





$begingroup$


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











$endgroup$




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















share|improve this question













share|improve this question




share|improve this question








edited Jan 4 at 16:55









Verbal Kint

3,3331412




3,3331412










asked Jan 4 at 16:23









fredfred

123110




123110








  • 2




    $begingroup$
    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
    $endgroup$
    – thece
    Jan 4 at 16:36






  • 1




    $begingroup$
    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    $endgroup$
    – Sunnyskyguy EE75
    Jan 4 at 16:41














  • 2




    $begingroup$
    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
    $endgroup$
    – thece
    Jan 4 at 16:36






  • 1




    $begingroup$
    The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
    $endgroup$
    – Sunnyskyguy EE75
    Jan 4 at 16:41








2




2




$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36




$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36




1




1




$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41




$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41










3 Answers
3






active

oldest

votes


















12












$begingroup$

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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.



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











$endgroup$













  • $begingroup$
    Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
    $endgroup$
    – Massimo Ortolano
    Jan 5 at 17:33












  • $begingroup$
    Thanks a lot for the corrections. Is it ok now? ;)
    $endgroup$
    – P. Collado
    Jan 5 at 23:21



















10












$begingroup$

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









$endgroup$





















    8












    $begingroup$

    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











    $endgroup$













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      3 Answers
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      3 Answers
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      12












      $begingroup$

      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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.



      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











      $endgroup$













      • $begingroup$
        Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        $endgroup$
        – Massimo Ortolano
        Jan 5 at 17:33












      • $begingroup$
        Thanks a lot for the corrections. Is it ok now? ;)
        $endgroup$
        – P. Collado
        Jan 5 at 23:21
















      12












      $begingroup$

      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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.



      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











      $endgroup$













      • $begingroup$
        Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        $endgroup$
        – Massimo Ortolano
        Jan 5 at 17:33












      • $begingroup$
        Thanks a lot for the corrections. Is it ok now? ;)
        $endgroup$
        – P. Collado
        Jan 5 at 23:21














      12












      12








      12





      $begingroup$

      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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.



      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











      $endgroup$



      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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.



      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














      share|improve this answer



      share|improve this answer








      edited Jan 5 at 23:20

























      answered Jan 4 at 16:58









      P. ColladoP. Collado

      1365




      1365












      • $begingroup$
        Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        $endgroup$
        – Massimo Ortolano
        Jan 5 at 17:33












      • $begingroup$
        Thanks a lot for the corrections. Is it ok now? ;)
        $endgroup$
        – P. Collado
        Jan 5 at 23:21


















      • $begingroup$
        Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        $endgroup$
        – Massimo Ortolano
        Jan 5 at 17:33












      • $begingroup$
        Thanks a lot for the corrections. Is it ok now? ;)
        $endgroup$
        – P. Collado
        Jan 5 at 23:21
















      $begingroup$
      Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
      $endgroup$
      – Massimo Ortolano
      Jan 5 at 17:33






      $begingroup$
      Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
      $endgroup$
      – Massimo Ortolano
      Jan 5 at 17:33














      $begingroup$
      Thanks a lot for the corrections. Is it ok now? ;)
      $endgroup$
      – P. Collado
      Jan 5 at 23:21




      $begingroup$
      Thanks a lot for the corrections. Is it ok now? ;)
      $endgroup$
      – P. Collado
      Jan 5 at 23:21













      10












      $begingroup$

      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









      $endgroup$


















        10












        $begingroup$

        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









        $endgroup$
















          10












          10








          10





          $begingroup$

          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









          $endgroup$



          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 Jan 4 at 16:31









          ShamtamShamtam

          2,5331023




          2,5331023























              8












              $begingroup$

              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











              $endgroup$


















                8












                $begingroup$

                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











                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  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











                  $endgroup$



                  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 Jan 4 at 17:21

























                  answered Jan 4 at 16:32









                  Spehro PefhanySpehro Pefhany

                  210k5161425




                  210k5161425






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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