Engineering Thevenin Equivalent for circuit with diagonal resistors

AI Thread Summary
The discussion focuses on finding the Thevenin equivalent of a circuit with diagonal resistors. The user initially removes part B to analyze the open circuit and identifies series connections among the resistors. Suggestions are made to redraw the circuit for clarity, emphasizing that nodes can be repositioned without altering the circuit's topology. The user successfully rewrites the circuit and calculates the voltage across the open terminals. The conversation highlights the importance of visualizing circuit components and their interconnections for easier analysis.
John Malenay
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Homework Statement


Hi everyone! I'm trying to find the thevenin equivalent for part A of this circuit: http://imgur.com/4Jdg8w8

Homework Equations

The Attempt at a Solution


I first remove part B, that is, open circuit it entirely. Then I try to find Voc across the two terminals that are open. I can see that the 47 Kohm is in series with the 18 Kohm since no circuit flows to the open circuit, and that the 15,000 Kohm is in series with the 33,00 Kohm. However I'm having a lot of trouble redrawing the circuit. Not really sure about how to make it look simpler.
 
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Hi John, Welcome to Physics Forums!

John Malenay said:

Homework Statement


Hi everyone! I'm trying to find the thevenin equivalent for part A of this circuit: http://imgur.com/4Jdg8w8

Homework Equations

The Attempt at a Solution


I first remove part B, that is, open circuit it entirely. Then I try to find Voc across the two terminals that are open. I can see that the 47 Kohm is in series with the 18 Kohm since no circuit flows to the open circuit, and that the 15,000 Kohm is in series with the 33,00 Kohm. However I'm having a lot of trouble redrawing the circuit. Not really sure about how to make it look simpler.

Yes, redrawing the figure is a good idea. Don't be too concerned about preserving the location of nodes b and c at the right-hand end of the figure. They might be right at home in the middle... Start by migrating b so that it lies vertically between a and d. Does that help you picture what else can be done?
 
gneill said:
Yes, redrawing the figure is a good idea. Don't be too concerned about preserving the location of nodes b and c at the right-hand end of the figure. They might be right at home in the middle... Start by migrating b so that it lies vertically between a and d. Does that help you picture what else can be done?
Hi gneill! I can see that d is directly connected to ground to I rewrote the circuit like this: http://imgur.com/nhnpDNJ

What do you think? The expression I get for Voc is = (18,000/65,000)*Vi
 
That's the voltage at the + terminal of Voc. Now you have to subtract the voltage at the - terminal of Voc from the voltage at the + terminal of Voc. That difference is the voltage Voc.
 
John Malenay said:
Hi gneill! I can see that d is directly connected to ground to I rewrote the circuit like this: http://imgur.com/nhnpDNJ

What do you think? The expression I get for Voc is = (18,000/65,000)*Vi
Okay, I was sort of hoping you'd abandon forcing the b and c nodes to the right of the circuit :smile: You're almost there, but not quite embracing the idea that you can move any component anywhere on the page so long as their interconnections are retained; The circuit topology remains the same so long as all the interconnections are the same.

Here's a re-arrangement of the components on the page that may give you some ideas:

Fig2.gif

Note that all the component connections are the same as in the original.

I'd expect the potential from terminal b to terminal c to end up negative, based on the voltage dividers in evidence.

Clever bit: When a problem is presented to you with diagonal components or crossovers of some type, usually it's meant to confuse your sense of symmetry so you don't recognize a simpler interpretation. Try to practice picturing (in your mind) "moving" components around on the page, keeping their interconnections intact but allowing the wires to stretch and move as required. Eventually you'll be able to "unfold" these circuits instinctively. Or at at least with a small bit of scratchpad doodling :smile:
 
gneill said:
Okay, I was sort of hoping you'd abandon forcing the b and c nodes to the right of the circuit :) You're almost there, but not quite embracing the idea that you can move any component anywhere on the page so long as their interconnections are retained; The circuit topology remains the same so long as all the interconnections are the same.

Here's a re-arrangement of the components on the page that may give you some ideas:

View attachment 73898
Note that all the component connections are the same as in the original.

I'd expect the potential from terminal b to terminal c to end up negative, based on the voltage dividers in evidence.

Clever bit: When a problem is presented to you with diagonal components or crossovers of some type, usually it's meant to confuse your sense of symmetry so you don't recognize a simpler interpretation. Try to practice picturing (in your mind) "moving" components around on the page, keeping their interconnections intact but allowing the wires to stretch and move as required. Eventually you'll be able to "unfold" these circuits instinctively. Or at at least with a small bit of scratchpad doodling :)
Thank you so much for your clear explanation gneill! I'm going to keep practicing!
 

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