How do I deal with “crossing resistances”? [closed]












9












$begingroup$


I'm doing some physics homework calculating the total resistance between points A and B and while some circuits are comprehensible, there are two that I can't understand how they work at all.



So, it is my pleasure to present to you these two monstrosities:



The divided squareThe hellborn triangles of death



How do I approach the crossing resistances in each one? Here's where I am as of writing this post:



The divided square:




  • The two 10k resistances are parallel to each other

  • The 20k and 9k resistances are parallel to each other

  • I don't know what to do with the 5k one


The hellborn triangles of death:




  • The first and last pair of 5k resistances are parallel to each other

  • The bottom center 5k resistance is in series with the previous two couples

  • I don't know what to do with the 15k ones


So, long story short, how should I handle these situations when the resistances are connected by both sides and the circuit looks like the voltage should converge inside them and start burning?










share|cite|improve this question











$endgroup$



closed as off-topic by John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos Jan 8 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 5




    $begingroup$
    Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
    $endgroup$
    – N. Steinle
    Jan 7 at 17:18






  • 4




    $begingroup$
    Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
    $endgroup$
    – Hal Hollis
    Jan 7 at 19:23










  • $begingroup$
    For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
    $endgroup$
    – Massimo Ortolano
    Jan 8 at 10:51










  • $begingroup$
    Questions like this should probably go on Electronics.SE.
    $endgroup$
    – knzhou
    Jan 8 at 11:10










  • $begingroup$
    As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
    $endgroup$
    – Groctel
    Jan 8 at 11:17
















9












$begingroup$


I'm doing some physics homework calculating the total resistance between points A and B and while some circuits are comprehensible, there are two that I can't understand how they work at all.



So, it is my pleasure to present to you these two monstrosities:



The divided squareThe hellborn triangles of death



How do I approach the crossing resistances in each one? Here's where I am as of writing this post:



The divided square:




  • The two 10k resistances are parallel to each other

  • The 20k and 9k resistances are parallel to each other

  • I don't know what to do with the 5k one


The hellborn triangles of death:




  • The first and last pair of 5k resistances are parallel to each other

  • The bottom center 5k resistance is in series with the previous two couples

  • I don't know what to do with the 15k ones


So, long story short, how should I handle these situations when the resistances are connected by both sides and the circuit looks like the voltage should converge inside them and start burning?










share|cite|improve this question











$endgroup$



closed as off-topic by John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos Jan 8 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 5




    $begingroup$
    Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
    $endgroup$
    – N. Steinle
    Jan 7 at 17:18






  • 4




    $begingroup$
    Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
    $endgroup$
    – Hal Hollis
    Jan 7 at 19:23










  • $begingroup$
    For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
    $endgroup$
    – Massimo Ortolano
    Jan 8 at 10:51










  • $begingroup$
    Questions like this should probably go on Electronics.SE.
    $endgroup$
    – knzhou
    Jan 8 at 11:10










  • $begingroup$
    As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
    $endgroup$
    – Groctel
    Jan 8 at 11:17














9












9








9


2



$begingroup$


I'm doing some physics homework calculating the total resistance between points A and B and while some circuits are comprehensible, there are two that I can't understand how they work at all.



So, it is my pleasure to present to you these two monstrosities:



The divided squareThe hellborn triangles of death



How do I approach the crossing resistances in each one? Here's where I am as of writing this post:



The divided square:




  • The two 10k resistances are parallel to each other

  • The 20k and 9k resistances are parallel to each other

  • I don't know what to do with the 5k one


The hellborn triangles of death:




  • The first and last pair of 5k resistances are parallel to each other

  • The bottom center 5k resistance is in series with the previous two couples

  • I don't know what to do with the 15k ones


So, long story short, how should I handle these situations when the resistances are connected by both sides and the circuit looks like the voltage should converge inside them and start burning?










share|cite|improve this question











$endgroup$




I'm doing some physics homework calculating the total resistance between points A and B and while some circuits are comprehensible, there are two that I can't understand how they work at all.



So, it is my pleasure to present to you these two monstrosities:



The divided squareThe hellborn triangles of death



How do I approach the crossing resistances in each one? Here's where I am as of writing this post:



The divided square:




  • The two 10k resistances are parallel to each other

  • The 20k and 9k resistances are parallel to each other

  • I don't know what to do with the 5k one


The hellborn triangles of death:




  • The first and last pair of 5k resistances are parallel to each other

  • The bottom center 5k resistance is in series with the previous two couples

  • I don't know what to do with the 15k ones


So, long story short, how should I handle these situations when the resistances are connected by both sides and the circuit looks like the voltage should converge inside them and start burning?







homework-and-exercises electric-circuits electrical-resistance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 19:07









David Z

63.4k23136252




63.4k23136252










asked Jan 7 at 17:13









GroctelGroctel

518




518




closed as off-topic by John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos Jan 8 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos Jan 8 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ZeroTheHero, Norbert Schuch, Chair, Kyle Kanos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    $begingroup$
    Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
    $endgroup$
    – N. Steinle
    Jan 7 at 17:18






  • 4




    $begingroup$
    Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
    $endgroup$
    – Hal Hollis
    Jan 7 at 19:23










  • $begingroup$
    For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
    $endgroup$
    – Massimo Ortolano
    Jan 8 at 10:51










  • $begingroup$
    Questions like this should probably go on Electronics.SE.
    $endgroup$
    – knzhou
    Jan 8 at 11:10










  • $begingroup$
    As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
    $endgroup$
    – Groctel
    Jan 8 at 11:17














  • 5




    $begingroup$
    Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
    $endgroup$
    – N. Steinle
    Jan 7 at 17:18






  • 4




    $begingroup$
    Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
    $endgroup$
    – Hal Hollis
    Jan 7 at 19:23










  • $begingroup$
    For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
    $endgroup$
    – Massimo Ortolano
    Jan 8 at 10:51










  • $begingroup$
    Questions like this should probably go on Electronics.SE.
    $endgroup$
    – knzhou
    Jan 8 at 11:10










  • $begingroup$
    As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
    $endgroup$
    – Groctel
    Jan 8 at 11:17








5




5




$begingroup$
Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
$endgroup$
– N. Steinle
Jan 7 at 17:18




$begingroup$
Have you tried redrawing the diagrams? You'd be surprised how helpful it can be to draw them in the way your brain wants to see them :)
$endgroup$
– N. Steinle
Jan 7 at 17:18




4




4




$begingroup$
Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
$endgroup$
– Hal Hollis
Jan 7 at 19:23




$begingroup$
Since you've accepted an answer, I'll just leave this comment for completeness: there are no resistor pairs in series or in parallel in either network. This is why students find both networks perplexing at first.
$endgroup$
– Hal Hollis
Jan 7 at 19:23












$begingroup$
For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
$endgroup$
– Massimo Ortolano
Jan 8 at 10:51




$begingroup$
For the first circuit, you can have a look at this answer of mine, where I present a less-known method, for the other at this one, where there is a very similar circuit.
$endgroup$
– Massimo Ortolano
Jan 8 at 10:51












$begingroup$
Questions like this should probably go on Electronics.SE.
$endgroup$
– knzhou
Jan 8 at 11:10




$begingroup$
Questions like this should probably go on Electronics.SE.
$endgroup$
– knzhou
Jan 8 at 11:10












$begingroup$
As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
$endgroup$
– Groctel
Jan 8 at 11:17




$begingroup$
As a new contributor I just new there was a physics community and that it's the subject I'm studing right now. I'll keep it in mind for future questions!
$endgroup$
– Groctel
Jan 8 at 11:17










4 Answers
4






active

oldest

votes


















11












$begingroup$

Try then to apply the Y-$Delta$ tranformations! I am glad you found the light! If you cannot carry calculation drop again here a line and i'll try to work more on the answer!






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    For the top one, you can separate out the $9 kOmega$ resistor at the top, since it's in series with the rest. Now you have a graph with four nodes and five edges. You can assign the variables $V_1$ through $V_4$ to the nodes, but WLOG you can consider the top node to be 0 and the bottom to be 1, so that leaves just the left and right voltages. You then have currents $I_1$ through $I_5$ through the edges. You can then start applying the circuit laws to these variables: for each edge, the current is equal to change in voltage divided by the resistance. The net current through each edge is zero (if you take current flowing in to be positive and current flowing out to be negative, that is). You'll then get a bunch of equations, and solving them will give the effective resistances.



    It's probably best to work out what the effective resistances are from basic principles and algebra first, and then try to work through applying Y-Δ to get a sense of how that works.



    Also, any symmetry in the diagram should be preserved in the values. The second circuit has left-right symmetries, so the currents should be symmetrical: the two top resistors should have the same currents as each other, the two outsides diagonal resistors should have the same as each other, and the two inside diagonal should have the same as each other. So you can use just four variables to represent the nine different currents.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The second circuit can be solved easily by perpendicular axis of symmetry.enter image description here






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        I would apply an abritrary voltage and calculate current in each loop using mesh currents. Then R = Vab/I_total. Then the circuit can be as messy as it wants, but it won't really matter. You'll always be able to solve the matrix for unknowns.






        share|cite|improve this answer









        $endgroup$




















          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          11












          $begingroup$

          Try then to apply the Y-$Delta$ tranformations! I am glad you found the light! If you cannot carry calculation drop again here a line and i'll try to work more on the answer!






          share|cite|improve this answer









          $endgroup$


















            11












            $begingroup$

            Try then to apply the Y-$Delta$ tranformations! I am glad you found the light! If you cannot carry calculation drop again here a line and i'll try to work more on the answer!






            share|cite|improve this answer









            $endgroup$
















              11












              11








              11





              $begingroup$

              Try then to apply the Y-$Delta$ tranformations! I am glad you found the light! If you cannot carry calculation drop again here a line and i'll try to work more on the answer!






              share|cite|improve this answer









              $endgroup$



              Try then to apply the Y-$Delta$ tranformations! I am glad you found the light! If you cannot carry calculation drop again here a line and i'll try to work more on the answer!







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 7 at 17:25









              Pietro OlivaPietro Oliva

              261111




              261111























                  4












                  $begingroup$

                  For the top one, you can separate out the $9 kOmega$ resistor at the top, since it's in series with the rest. Now you have a graph with four nodes and five edges. You can assign the variables $V_1$ through $V_4$ to the nodes, but WLOG you can consider the top node to be 0 and the bottom to be 1, so that leaves just the left and right voltages. You then have currents $I_1$ through $I_5$ through the edges. You can then start applying the circuit laws to these variables: for each edge, the current is equal to change in voltage divided by the resistance. The net current through each edge is zero (if you take current flowing in to be positive and current flowing out to be negative, that is). You'll then get a bunch of equations, and solving them will give the effective resistances.



                  It's probably best to work out what the effective resistances are from basic principles and algebra first, and then try to work through applying Y-Δ to get a sense of how that works.



                  Also, any symmetry in the diagram should be preserved in the values. The second circuit has left-right symmetries, so the currents should be symmetrical: the two top resistors should have the same currents as each other, the two outsides diagonal resistors should have the same as each other, and the two inside diagonal should have the same as each other. So you can use just four variables to represent the nine different currents.






                  share|cite|improve this answer









                  $endgroup$


















                    4












                    $begingroup$

                    For the top one, you can separate out the $9 kOmega$ resistor at the top, since it's in series with the rest. Now you have a graph with four nodes and five edges. You can assign the variables $V_1$ through $V_4$ to the nodes, but WLOG you can consider the top node to be 0 and the bottom to be 1, so that leaves just the left and right voltages. You then have currents $I_1$ through $I_5$ through the edges. You can then start applying the circuit laws to these variables: for each edge, the current is equal to change in voltage divided by the resistance. The net current through each edge is zero (if you take current flowing in to be positive and current flowing out to be negative, that is). You'll then get a bunch of equations, and solving them will give the effective resistances.



                    It's probably best to work out what the effective resistances are from basic principles and algebra first, and then try to work through applying Y-Δ to get a sense of how that works.



                    Also, any symmetry in the diagram should be preserved in the values. The second circuit has left-right symmetries, so the currents should be symmetrical: the two top resistors should have the same currents as each other, the two outsides diagonal resistors should have the same as each other, and the two inside diagonal should have the same as each other. So you can use just four variables to represent the nine different currents.






                    share|cite|improve this answer









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      For the top one, you can separate out the $9 kOmega$ resistor at the top, since it's in series with the rest. Now you have a graph with four nodes and five edges. You can assign the variables $V_1$ through $V_4$ to the nodes, but WLOG you can consider the top node to be 0 and the bottom to be 1, so that leaves just the left and right voltages. You then have currents $I_1$ through $I_5$ through the edges. You can then start applying the circuit laws to these variables: for each edge, the current is equal to change in voltage divided by the resistance. The net current through each edge is zero (if you take current flowing in to be positive and current flowing out to be negative, that is). You'll then get a bunch of equations, and solving them will give the effective resistances.



                      It's probably best to work out what the effective resistances are from basic principles and algebra first, and then try to work through applying Y-Δ to get a sense of how that works.



                      Also, any symmetry in the diagram should be preserved in the values. The second circuit has left-right symmetries, so the currents should be symmetrical: the two top resistors should have the same currents as each other, the two outsides diagonal resistors should have the same as each other, and the two inside diagonal should have the same as each other. So you can use just four variables to represent the nine different currents.






                      share|cite|improve this answer









                      $endgroup$



                      For the top one, you can separate out the $9 kOmega$ resistor at the top, since it's in series with the rest. Now you have a graph with four nodes and five edges. You can assign the variables $V_1$ through $V_4$ to the nodes, but WLOG you can consider the top node to be 0 and the bottom to be 1, so that leaves just the left and right voltages. You then have currents $I_1$ through $I_5$ through the edges. You can then start applying the circuit laws to these variables: for each edge, the current is equal to change in voltage divided by the resistance. The net current through each edge is zero (if you take current flowing in to be positive and current flowing out to be negative, that is). You'll then get a bunch of equations, and solving them will give the effective resistances.



                      It's probably best to work out what the effective resistances are from basic principles and algebra first, and then try to work through applying Y-Δ to get a sense of how that works.



                      Also, any symmetry in the diagram should be preserved in the values. The second circuit has left-right symmetries, so the currents should be symmetrical: the two top resistors should have the same currents as each other, the two outsides diagonal resistors should have the same as each other, and the two inside diagonal should have the same as each other. So you can use just four variables to represent the nine different currents.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 7 at 18:18









                      AcccumulationAcccumulation

                      2,486312




                      2,486312























                          2












                          $begingroup$

                          The second circuit can be solved easily by perpendicular axis of symmetry.enter image description here






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            The second circuit can be solved easily by perpendicular axis of symmetry.enter image description here






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              The second circuit can be solved easily by perpendicular axis of symmetry.enter image description here






                              share|cite|improve this answer









                              $endgroup$



                              The second circuit can be solved easily by perpendicular axis of symmetry.enter image description here







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 7 at 20:33







                              user212727






























                                  0












                                  $begingroup$

                                  I would apply an abritrary voltage and calculate current in each loop using mesh currents. Then R = Vab/I_total. Then the circuit can be as messy as it wants, but it won't really matter. You'll always be able to solve the matrix for unknowns.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    I would apply an abritrary voltage and calculate current in each loop using mesh currents. Then R = Vab/I_total. Then the circuit can be as messy as it wants, but it won't really matter. You'll always be able to solve the matrix for unknowns.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      I would apply an abritrary voltage and calculate current in each loop using mesh currents. Then R = Vab/I_total. Then the circuit can be as messy as it wants, but it won't really matter. You'll always be able to solve the matrix for unknowns.






                                      share|cite|improve this answer









                                      $endgroup$



                                      I would apply an abritrary voltage and calculate current in each loop using mesh currents. Then R = Vab/I_total. Then the circuit can be as messy as it wants, but it won't really matter. You'll always be able to solve the matrix for unknowns.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 7 at 23:40









                                      AdvancedNewbieAdvancedNewbie

                                      1




                                      1















                                          Popular posts from this blog

                                          "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                                          Alcedinidae

                                          RAC Tourist Trophy