What is the equivalent resistance through circuit?

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SUMMARY

The discussion centers on calculating the equivalent resistance and current through a circuit with multiple resistors. The equivalent resistance is determined to be 1.638 ohms, while the current through the 5-ohm resistor is calculated to be 1.911 amps downward. Participants emphasize the importance of correctly identifying series and parallel configurations and applying Kirchhoff's laws for accurate circuit analysis. The conversation highlights common challenges faced by learners when simplifying complex circuits and setting up current equations.

PREREQUISITES
  • Understanding of series and parallel resistor configurations
  • Familiarity with Kirchhoff's laws (loop rule and junction rule)
  • Basic circuit analysis techniques
  • Ability to set up and solve simultaneous equations
NEXT STEPS
  • Study advanced circuit analysis techniques using Kirchhoff's laws
  • Learn how to simplify complex circuits with multiple loops
  • Practice calculating equivalent resistance in mixed resistor networks
  • Explore tools for circuit simulation, such as LTspice or Multisim
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Students studying electrical engineering, educators teaching circuit analysis, and hobbyists working on electronics projects who need to understand complex resistor networks.

tMANn
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Homework Statement


uploaded image
529d4c4b-d58d-44ac-a5d8-18ebd9a3ef12.png


a) What is the equivalent resistance through circuit? (Answer is 1.638 ohms)

B) what is the current through the 5ohm resistor? (Answer is 1.911 amps downward)

The main part I am confuse about in part A) is what is in parallel and what part is in series because I haven't really worked with a circuit like this before.(Am I even suppose to simplify it?) For part B) I am confused about how to set up the current(like i1, i2, i3,etc). I know you can set up the current anyway you like, but if you can give me an example it will help a lot. You don't have to get the answer for me, just explain those 2 things will be fine. thanks!

Homework Equations


Parallel: 1/Req = 1/R1 + 1/R2 + ...
Series: Req = R1 + R2 + ...

The Attempt at a Solution



I've tried doing it several times but it failed. I just want an explanation as stated above.
 
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Not all circuits can be simplified using the rules for resistors in parallel or series. Have you learned about Kirchoff's laws? (also known as the loop rule and junction rule)
 
yes I have learned about Kirchoff's laws, but I thought it would be easier to simplify it first (if possible). Like I can apply kirchhoffs laws pretty easy in 2 or 3 loop circuits, but i am having trouble in this one. Sometimes I get confused in saying where the current (I) goes, especially in a more complex circuit as shown above. I'll show you how I set up the current (I am sure this is where my mistake comes in), and hopefully you can lead me to the right direction. Ill post in about 5mins. thanks though for quick response
 
cb067ae8-4cb7-4f47-90ab-b476ff8005ec.png


I don't know if you can read it
 
3OHM = I-3 downward
1OHM = I-1 to the right
4OHM = I-2 downward
2OHM = I-4 downward
6OHM = I-4 + I-2 to the left
5OHM = (I-1) - (I-4) downward
 
Got it. What will be your label for the current flowing through the battery? (Remember that when applying Kirchoff's laws, you need to label all segments of wire with their currents)

Once you figure that out, try writing the loop laws. Since this is a slightly complicated circuit, you might first want to think about how many loops (and thus how many equations) there will be, before you spend time figuring out what they are.
 
flowing through the battery looks like (I-1) + (I-2) + (I-3). And I think I will need 4 equations. I think I've tried it doing it like this before, but I'll try it again (maybe I made a mistake somewhere)
thanks!

edit: also is the 5OHM resistor (I-1) - (I-4) downward or (I-1) - (I-4) + (I-2)downward, cause it looks like I-2 gets added in? not sure about this

and will i have to use the flow through the battery anywhere in the equations? (I've never really used the I's without a resistor before that's why i left it blank)
 
Last edited:
yep, wrong answer again =\
 
tMANn said:
flowing through the battery looks like (I-1) + (I-2) + (I-3). And I think I will need 4 equations. I think I've tried it doing it like this before, but I'll try it again (maybe I made a mistake somewhere)
thanks!
Makes sense. You might not need the current flowing through the battery, but it's good to label it for consistency.
tMANn said:
edit: also is the 5OHM resistor (I-1) - (I-4) downward or (I-1) - (I-4) + (I-2)downward, cause it looks like I-2 gets added in? not sure about this
I can't know anything more than you do on that count.
tMANn said:
yep, wrong answer again =\
If you show your work, we can help you see if you've done something wrong.
 
  • #10
figured it out, tried a different way, thanks diazona
 

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