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












5














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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  • 2




    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
    yesterday






  • 1




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
















5














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




















  • 2




    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
    yesterday






  • 1




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














5












5








5







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















share|improve this question













share|improve this question




share|improve this question








edited yesterday









Verbal Kint

3,2291312




3,2291312










asked yesterday









fred

11219




11219








  • 2




    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
    yesterday






  • 1




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














  • 2




    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
    yesterday






  • 1




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








2




2




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
yesterday




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
yesterday




1




1




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




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










3 Answers
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11














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






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New contributor




P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
    – Massimo Ortolano
    12 hours ago












  • Thanks a lot for the corrections. Is it ok now? ;)
    – P. Collado
    6 hours ago



















10














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





























    8














    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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      3 Answers
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      3 Answers
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      active

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      11














      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










      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.


















      • Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        – Massimo Ortolano
        12 hours ago












      • Thanks a lot for the corrections. Is it ok now? ;)
        – P. Collado
        6 hours ago
















      11














      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










      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.


















      • Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        – Massimo Ortolano
        12 hours ago












      • Thanks a lot for the corrections. Is it ok now? ;)
        – P. Collado
        6 hours ago














      11












      11








      11






      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










      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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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










      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








      edited 6 hours ago





















      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.









      answered yesterday









      P. Collado

      1265




      1265




      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.





      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.






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












      • Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        – Massimo Ortolano
        12 hours ago












      • Thanks a lot for the corrections. Is it ok now? ;)
        – P. Collado
        6 hours ago


















      • Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
        – Massimo Ortolano
        12 hours ago












      • Thanks a lot for the corrections. Is it ok now? ;)
        – P. Collado
        6 hours ago
















      Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
      – Massimo Ortolano
      12 hours ago






      Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
      – Massimo Ortolano
      12 hours ago














      Thanks a lot for the corrections. Is it ok now? ;)
      – P. Collado
      6 hours ago




      Thanks a lot for the corrections. Is it ok now? ;)
      – P. Collado
      6 hours ago













      10














      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


























        10














        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
























          10












          10








          10






          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 yesterday









          Shamtam

          2,4431023




          2,4431023























              8














              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




























                8














                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


























                  8












                  8








                  8






                  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 yesterday

























                  answered yesterday









                  Spehro Pefhany

                  203k4150408




                  203k4150408






























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                      Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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                      Display a custom attribute below product name in the front-end Magento 1.9.3.8

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                      0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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