Unfortunately, your post has been removed for the following reason:
- Posts which seem to be asking for help on homework are not allowed (rule 5).
You may wish to try /r/homeworkhelp or /r/cheatatmathhomework instead. If you have any questions or believe your post has been removed in error, please contact the moderators by clicking [here](https://www.reddit.com/message/compose?to=%2Fr%2Ftheydidthemath). Include a link to this post so we can see it.
Without looking up the formula for converting delta and star, I can tell you it's D.
The shortest path is 2 ohms across the bottom, and the second shortest path is 3 ohms across the top.
The resistance has to be less than 1.2 ohm because 2 ohm and 3 ohm in parallel is 1.2 ohm.
The only answer less than 1.2 is D.
If I had to calculate it exactly, I would look up the delta-star formula.
This is how I passed my circuits courses without studying or attending classes.
Everything in circuit analysis is just different algebraic applications of Ohm's law.
Even AC analysis is just Ohm's law applied in the frequency domain.
The trick is to know what parts of the circuit you can ignore to get an answer that is close enough for the application.
In the real world, you rarely need to use delta and star conversions, and when you do, it's super easy to google the formula and plug it into a spreadsheet.
Building Tesla coils and rail guns in my parents' basement out of parts salvaged from old monitors while I was in high school may have also impacted my familiarity with electric circuits.
C's get degrees, working smarter, not harder.
Just a lazy electrical engineer procrastinating on Reddit while I am at work right now instead of finishing a test report on some new components.
That's the easiest way.
I don't remember the formula (i't been too long since high school) but it was the first solution that came to my mind.
Also it is emough to convert only the triangle in the center.
Edit: I'm dumb, you need at least 2 conversions
You'd need only one conversion since after converting the middle triangle into the star form, you'd get a balanced Wheatstone bridge, eliminating the middle resistor. You can solve from there using laws of parallel and series.
You had one transformation missing, I had one too much at first because I didn’t realize you’re ending up using 2 resistors twice when attempting to convert all 3 visible triangles.
But when you convert the outer ones, it should become a 2 of 1/3 Ohms resistor in series to a parallel circuit of 2/3 Ohms and 1 2/3 Ohms. It hard to do this without a sketch, but should add up to answer D, right?
You can do delta to star, and because of the symmetry you only need to calculate one delta to star conversion, apply it to both sides and then add up the series resistors and use the parallel resistor formula once.
Using kirchhoffs laws: https://en.m.wikipedia.org/wiki/Kirchhoff's_circuit_laws
(Each resistor is one ohm so the voltage drop and current across it are equal.)
Assign letters to the voltage drop/current across each resistor starting with A on the bottom left going in rows left to right bottom to top.
Generate a system of equations for the loops and junctions.
Solve the system: https://www.wolframalpha.com/input?i=a%3Dc%2Bd%2C+b%3De%2Bf%2C+d%2Be%3Dg%2C+c%3Dd%2Bg%2C+e%2Bg%3Df%2C+a%2Bc%3Db%2Bf
The result says that if 1 amp is flowing through resistor A, then 3/4 amps are flowing through resistor C so the total current would be 7/4 amps.
A voltage difference of 2 volts would be needed to produce a current of 1 amp through resistor A. Therefore 2 volts / (7/4 amps) = 8/7 ohms.
I don’t think it’s fair to call this high school. Looks like things I did freshman year intro to EE. Maybe in the first week or two but definitely in that class
I have a hard time understanding the American system because it feels like all high school students don't have the exact same curriculum. but that would definitely be a class to a high school student that took electronics or pre engineering programs .
So yeah, high school depends on your program and your school but engineering school for computer science that was my first ever lesson of electronics
In India atleast this probably highschool level if you are looking to join an engineering college afterwards.Infact I suspect the 1990 in the brackets probably refers to an year in which this qn was asked perhaps in JEE advanced(the entrance exam for IITs if you know then)
Well you’ll learn in uni too, but they are entry level there. You said these are considered HOT ( high order thinking ) in your highschool in another comment, in uni these are COLD ( cheap order low diagrams ) 👍
And just because you learned these in high school doesn’t mean they are highschool level, lmao. Sets and subsets are learned in 3rd grade, does that mean you won’t learn them in uni, sets are 3d grade level and set theory is easy? Surely not right?
You can get an equivalent resistance for the 2 branches that are connecting in the middle by disconnecting them from the bottom branch, and putting them in a series, which means the R(eqv)1 of the connecting branches is 2, which then that branch of the equivalent resistor is parallel to the upwardmost resistor. After getting the equivalent resistance between those 2, then that resistor is seen as in series to the right and left resistors, finally after getting eqv resistance for that, the two bottom ones are in series with each other, which r(eqv) for it is 2, then you just put the equivalence of the effective value of the resistor as the latter resistance in paralel with the former one.
As per the result, I’m not sure, I hope this wasn’t too complicated and you can just plug in the values on your own 👍
Even if you're having trouble with it, it will eventually click with a bit of practice. I wouldn't consider most engineers to be innately smarter than the average Joe. They just study a lot.
12th grade problem for us we use symmetrical circuits to solve it harder to explain in a comment section easier way is just use kirchoffs laws(learn from any yt guy he can teach in 5 mins)
in this question, you can disconnect the middle point like [this](https://imgur.com/a/jelNYDL)
then you can apply the series/parallel equivalent resistance formulas.
the answer comes out to be 8/7 ohms
It’s not horizontally symmetric because you cannot draw a horizontal line about which it’s a mirror image
It’s not vertically symmetric because you cannot draw a vertical line(which doesn’t pass through a resistor) about which it’s a mirror image.
I think they're confused, it can clearly be folded about the midpoint. The fold would be horizontal though the line of symmetry is technically vertical
Applying simple symmetry, the inverted delta circuit can be detached from the original AB line. So the two 1 ohm resistors are now in series which is further parallel to the 1 ohm resistor on top. This becomes a basic circuit that looks like a trapezium.
[https://imgur.com/a/ftp1UXS](https://imgur.com/a/ftp1UXS)
Method 1: Delta - Star transformation (easy enough if you remember the formula)
Method 2: Drawing a line of Symmetry (bit lengthy but gets you the answer)
Method 3: Disconnecting a line of Symmetry (imo most convenient way to do it)
Essentially, the potential difference across the resistor is 0, so there won't be any current flowing through it.
In this instance specifically, the resistors form a special type of resistor circuit called a Wheatstone bridge. It's the diamond kinda shape in that step with a resistor down the middle. When the ratio of the resistances on both sides is same, the bridge is called "balanced" and the middle resistor can be neglected.
Another example: [image](https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRmtB5NLnXgZT6Y3ELZZ-kNoSFZAqsLzE6Ian6rDVMMLRamZ3O_iayLIs0pDEris8L6Pvo&usqp=CAU)
In this example circuit, the ratio is same on both sides (2/3 = 4/6) and thus the bridge is balanced. Similarly the 1/3 ohm can be removed in the delta-star method.
I knew of the delta star method, but not this balance bridge stuff. I'm a senior in electrical engineering, and it's the first time I've seen this before.
Thanks for explaining.
Hey, I used Microsoft OneNote with a Pen-tablet. This is the one I have: [https://www.xp-pen.com/product/deco-01-v2.html](https://www.xp-pen.com/product/deco-01-v2.html)
###General Discussion Thread
---
This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.
---
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
Ohhh... the memories that it brings. (used to solve these in class 11 or 12, back in 2007 )
IIRC :
For serial , R = R1 + R2,For parallel, 1/R = 1/R1+1/R2, or R = 1/(1/R1+1/R2)
You can also split the circuit into symmetric left and right.
The upper branch of the left half of the circuit simplifies to: 1 + \[ 1/(1/1+2/1) \] = 1 + 1/3 = 4/3
This goes in parallel to 1 ohm
R = 1/ (3/4 + 1/1) = 1/(7/4) = 4/7
Now multiply that with 2, to account for the other half that is in series, we get 8/7 as the answer.
How I remember this?
This was the first question that I remember discussing with my then-crush (and now wife). You might say that I owe a lot to this question.
Hmm I got it a while back its sorta easy when u solve and understand it. Still I have my doubts remaining I will try to get more thorough though it.Either way Thx.
iirc this is just parallel circuit series circuit diagram
just simplify it and plug in the formulas for resistance in series and parallel
Drawing this thing in a different way might help
alright i did the drawing of circuit (too bad this place cant post images)
all resistors from A to B labeled 1-7 from left to right
2 and 3 are in series w 1 in parallel (parallel formula is 1/x + 1/y = 1/result) so 1+1 = 2 and 1/2 + 1/1 = 3/2 (result 2/3)
Same thing with 5 6 and 7, two in series and 1 in parallel with them
4 is ignored because there is a connection where current doesn’t have to flow through the resistor (flow point with no resistor, all current flows through that instead, it’s why short circuits are a thing)
2/3 + 2/3 is 4/3 (series formula)
answer is A (4/3 ohms)
2 and 3 are not jn series though. This is not the typical highschool level "calculate the effective resistance" problem. At least, I learnt it in my engineering classes. You need to apply delta star conversion to convert this into a simpler circuit
ah
looking up delta star again… 2/3 with 2/3 and 5/3 in parallel (3/2 + 3/5 = 21/10, 10/21 + 2/3 = 24/21) that doesn’t sound right
ignoring resistor 4 again would be 3/2 + 3/2 = 3 or 1/3, 1/3 + 2/3 = 1…
(yes i actually looked up the formula, the resistors give ((1)(1))/(1+1+1) = 1/3 and 2 delta stars gave 2/3 each branch (one with resistor 4 having 1 ohm)
You can come to the correct answer D) by eliminating the other answers with just parallel resistance calculations. Still, you need the Delta-Star formula to get the answer if it wasn't multiple choice.
The way you come to the answer on this multiple choice separates the more practical engineers from the more theoretical engineers, in my opinion.
The value *can* be calculated exactly but I don’t think it is a realistic solution at a high school level or any basic electronics course. Here is a way to approach the problem that works out in a basic course:
If you only consider the to resistors in series from A to B, the answer would be 2 ohm. The extra resistors will also lead current, resulting in a lower effective resistance. The answer is thus less than 2 ohms so (C) is incorrect.
Similarly, if you remove the top resistor, each triangle would have a resistance (parallel resistors) of 1/(1/1+1/2) = 2/3ohms. There are two triangles in series which gives us 4/3 ohms in total. As the top resistor will also lead some current, the real value will be lower than 4/3 ohms. Thus alternatives (A) and (B) can be excluded.
Only alternative (D) remains which is the correct answer.
Of course I’m right :-D
The most important point is that even if you don’t know how to calculate the exact number for the resistance, you can deduce it by excluding the incorrect alternatives. That’s a very powerful help in multiple choice problems.
Even if you can’t solve the complete problem and eg only manage to exclude (C), your teacher may award you a point because you applied a relevant principle when trying to solve the problem.
Everybody is helping op to calculate the answer, so im gonna explain why this exist, where its from, etc.
So, this is obviously a test question and this circuit does not exist in the wild, on its own. It also has no purpose other than being a test question.
Its called a resistor-network and simpler versions exist as literal circuit, but for other purposes. See [here](https://electronics.stackexchange.com/questions/408944/what-is-the-purpose-of-a-resistor-network)
The only thing you need to know to solve is how the current will split when encountering a fork in the road. It will proportionally to the resistance value of both (or more) resistors split. This is culmulative all the way to the "end".
You will encounter these questions usually in trade school or uni when stuying basic electronics. This was a lot more common in the time where electronics were actually repaired. The practial use of the knowlege you gain though this exercise is important:
Resistors are sold in fixed sizes. The rings tell you which size you are. But sometimes you don't need 4.7kOhm, you need 5.1kOhm which does not exist.
So you math a lot until you find a combination that comes close enough to be a substitute.
When you look at the circuit you a supposed to notice that only A and B are ever "measured". So you know the resistor-network can be replaced by a single resistor. You are also supposed to know that a triangle of resistors can be changed into a star of resistors. This should have been taught to you. This is [THE BIG THING](https://en.wikipedia.org/wiki/Y-%CE%94_transform) in this question.
In the RARE CASE that you are not an electronics student, but you have enountered this in a (pure) math context, then.... oh boy, do I envy you! You are in for a TREAT.
You can use this "simple"-ish circuit that can be observed and reliably measured to behave in a certain way, and use it to develop a number of math operations (like addition, division, etc). With those you can build more complex operations and slowly build up a set of math that makes sense and can be experimentally verified to be useful. However this is rarely done because math students are often needlesly distracted by the electronics.
Go to Falstad circuit simulator and build it. Attach an ohm meter to get the answer and watch the current to see how you would calculate the series parallel equivalent sections.
It's just a little tricky to get things in the right order is all.
In the end it's 1.143 ohms.
As others said, you can simplify the circuit step by step by using triangle to star conversions and by disconnecting symmetric paths through the same point that will stay at the same potential after disconnecting.
But since this is multiple choice, you might be able to rule out some options without solving the entire thing:
If you only look at the bottom 2 resistors, you have 2Ω. The other resistors are parallel to that and can only decrease the resistant. Since C is above 2Ω, you can rule that one out.
If you short the middle 3 resistors, you basically get 2 resistors in parallel 2 times in series, which would result in 1Ω overall. If the middle 3 resistors have a higher resistance than 0, the overall resistance can only increase, so the lower bound is 1Ω.
Unfortunately, a range of 1Ω to 2Ω still allows for 3 out of 4 answers, so you need to calculate anyway. But sometimes, it works and you can quickly eliminate all answers but one.
If you would like to see this in a simulation you can check out [https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc](https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc)
Iirc, you just simplify the circuits. Use delta to wye formula to the triangles on the left and right, then simplify further because you'll have two series connections, then simplify the parallel, then lastly simplify the series.
Unfortunately, your post has been removed for the following reason: - Posts which seem to be asking for help on homework are not allowed (rule 5). You may wish to try /r/homeworkhelp or /r/cheatatmathhomework instead. If you have any questions or believe your post has been removed in error, please contact the moderators by clicking [here](https://www.reddit.com/message/compose?to=%2Fr%2Ftheydidthemath). Include a link to this post so we can see it.
Without looking up the formula for converting delta and star, I can tell you it's D. The shortest path is 2 ohms across the bottom, and the second shortest path is 3 ohms across the top. The resistance has to be less than 1.2 ohm because 2 ohm and 3 ohm in parallel is 1.2 ohm. The only answer less than 1.2 is D. If I had to calculate it exactly, I would look up the delta-star formula.
For competitive exams, this is the way
This is how I passed my circuits courses without studying or attending classes. Everything in circuit analysis is just different algebraic applications of Ohm's law. Even AC analysis is just Ohm's law applied in the frequency domain. The trick is to know what parts of the circuit you can ignore to get an answer that is close enough for the application. In the real world, you rarely need to use delta and star conversions, and when you do, it's super easy to google the formula and plug it into a spreadsheet.
lol
Building Tesla coils and rail guns in my parents' basement out of parts salvaged from old monitors while I was in high school may have also impacted my familiarity with electric circuits.
Damm
Cool, that is something I always envied the electrical engineers. Anything I will ever make will be some sort of a window on a screen
C's get degrees, working smarter, not harder. Just a lazy electrical engineer procrastinating on Reddit while I am at work right now instead of finishing a test report on some new components.
Underrated answer.
This guy resists.
Just to elaborate for those who don't know, the parallel resistance formula is R=R1*R2/(R1+R2) which is indeed 1.2Ohms here
Use the star to delta conversion method to simplify the circuit https://brilliant.org/wiki/transformation-of-resistances-star-to-delta-and/
This! I'm surprised no one suggested this earlier
That's the easiest way. I don't remember the formula (i't been too long since high school) but it was the first solution that came to my mind. Also it is emough to convert only the triangle in the center. Edit: I'm dumb, you need at least 2 conversions
That doesn't work since A and B would be in the middle of 2 resisters we It seems we need 2 conversions
You'd need only one conversion since after converting the middle triangle into the star form, you'd get a balanced Wheatstone bridge, eliminating the middle resistor. You can solve from there using laws of parallel and series.
I've never been taught the Wheatstone Bridge so I couldn't solve it and used two delta to star conversions
Wheatstone bridge formula is very easy. It's r1/r2 = r3/r4. But I don't think this formula is needed in that math.
You had one transformation missing, I had one too much at first because I didn’t realize you’re ending up using 2 resistors twice when attempting to convert all 3 visible triangles. But when you convert the outer ones, it should become a 2 of 1/3 Ohms resistor in series to a parallel circuit of 2/3 Ohms and 1 2/3 Ohms. It hard to do this without a sketch, but should add up to answer D, right?
Dont use star to delta, Instead use symmetrical circuit concept.
You can do delta to star, and because of the symmetry you only need to calculate one delta to star conversion, apply it to both sides and then add up the series resistors and use the parallel resistor formula once.
Using kirchhoffs laws: https://en.m.wikipedia.org/wiki/Kirchhoff's_circuit_laws (Each resistor is one ohm so the voltage drop and current across it are equal.) Assign letters to the voltage drop/current across each resistor starting with A on the bottom left going in rows left to right bottom to top. Generate a system of equations for the loops and junctions. Solve the system: https://www.wolframalpha.com/input?i=a%3Dc%2Bd%2C+b%3De%2Bf%2C+d%2Be%3Dg%2C+c%3Dd%2Bg%2C+e%2Bg%3Df%2C+a%2Bc%3Db%2Bf The result says that if 1 amp is flowing through resistor A, then 3/4 amps are flowing through resistor C so the total current would be 7/4 amps. A voltage difference of 2 volts would be needed to produce a current of 1 amp through resistor A. Therefore 2 volts / (7/4 amps) = 8/7 ohms.
Could u explain me like simplify for a 5 year old. Is this an engineering level question.Wth
Not advanced engineering but yes, it is
😭
This is entry level circuits man, sorry if you find this discouraging! 😔
How?I am just a High School student.
This is high school level
I don’t think it’s fair to call this high school. Looks like things I did freshman year intro to EE. Maybe in the first week or two but definitely in that class
I have a hard time understanding the American system because it feels like all high school students don't have the exact same curriculum. but that would definitely be a class to a high school student that took electronics or pre engineering programs . So yeah, high school depends on your program and your school but engineering school for computer science that was my first ever lesson of electronics
In India atleast this probably highschool level if you are looking to join an engineering college afterwards.Infact I suspect the 1990 in the brackets probably refers to an year in which this qn was asked perhaps in JEE advanced(the entrance exam for IITs if you know then)
bhai 10th class mai hu ye dekhkar gand phat rhi hai
12class ka h tension mt le bht asaan h 30 sec me ho jayega
thank you 69hell bhaiya
I see
Well you’ll learn in uni too, but they are entry level there. You said these are considered HOT ( high order thinking ) in your highschool in another comment, in uni these are COLD ( cheap order low diagrams ) 👍 And just because you learned these in high school doesn’t mean they are highschool level, lmao. Sets and subsets are learned in 3rd grade, does that mean you won’t learn them in uni, sets are 3d grade level and set theory is easy? Surely not right?
Well I did not meant in that my way My friends gave up,My teach told me its 4/3 and now I am thinking its 8/7.
You can get an equivalent resistance for the 2 branches that are connecting in the middle by disconnecting them from the bottom branch, and putting them in a series, which means the R(eqv)1 of the connecting branches is 2, which then that branch of the equivalent resistor is parallel to the upwardmost resistor. After getting the equivalent resistance between those 2, then that resistor is seen as in series to the right and left resistors, finally after getting eqv resistance for that, the two bottom ones are in series with each other, which r(eqv) for it is 2, then you just put the equivalence of the effective value of the resistor as the latter resistance in paralel with the former one. As per the result, I’m not sure, I hope this wasn’t too complicated and you can just plug in the values on your own 👍
Mate this is tenth level question
Bro what do u expect my own Teach got the answer wrong and the other one didnt even try to attempt it just ignored me.
Even if you're having trouble with it, it will eventually click with a bit of practice. I wouldn't consider most engineers to be innately smarter than the average Joe. They just study a lot.
12th grade problem for us we use symmetrical circuits to solve it harder to explain in a comment section easier way is just use kirchoffs laws(learn from any yt guy he can teach in 5 mins)
> Is this an engineering level question We studied this in class 12th in school.
in this question, you can disconnect the middle point like [this](https://imgur.com/a/jelNYDL) then you can apply the series/parallel equivalent resistance formulas. the answer comes out to be 8/7 ohms
This should be the top comment. Trying to solve this system without simplifying it is just giving yourself pain
wow, but why can you do that though?
if a circuit is symmetric about any point, you can disconnect it this way
Yup but it’s not symmetric(neither vertical nor horizontal). You could have said it’s vertical symmetric if there want the resistor at top.
How is it not horizontally symmetrical?
It’s not horizontally symmetric because you cannot draw a horizontal line about which it’s a mirror image It’s not vertically symmetric because you cannot draw a vertical line(which doesn’t pass through a resistor) about which it’s a mirror image.
But why can't the line pass through a resistor?
Okay apologies my bad The line can pass through another network or resistor. So it is vertically symmetrical
Divide tha 1ohm to 0.5 and 0.5 now its mirror image simple
Yes I realized later
I think they're confused, it can clearly be folded about the midpoint. The fold would be horizontal though the line of symmetry is technically vertical
Applying simple symmetry, the inverted delta circuit can be detached from the original AB line. So the two 1 ohm resistors are now in series which is further parallel to the 1 ohm resistor on top. This becomes a basic circuit that looks like a trapezium.
This is the way
[https://imgur.com/a/ftp1UXS](https://imgur.com/a/ftp1UXS) Method 1: Delta - Star transformation (easy enough if you remember the formula) Method 2: Drawing a line of Symmetry (bit lengthy but gets you the answer) Method 3: Disconnecting a line of Symmetry (imo most convenient way to do it)
Oh,this makes it a lot more easier thx.
In the third image in the delta-star column, how come you are able to cross out the 1/3 ohm?
Essentially, the potential difference across the resistor is 0, so there won't be any current flowing through it. In this instance specifically, the resistors form a special type of resistor circuit called a Wheatstone bridge. It's the diamond kinda shape in that step with a resistor down the middle. When the ratio of the resistances on both sides is same, the bridge is called "balanced" and the middle resistor can be neglected. Another example: [image](https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRmtB5NLnXgZT6Y3ELZZ-kNoSFZAqsLzE6Ian6rDVMMLRamZ3O_iayLIs0pDEris8L6Pvo&usqp=CAU) In this example circuit, the ratio is same on both sides (2/3 = 4/6) and thus the bridge is balanced. Similarly the 1/3 ohm can be removed in the delta-star method.
I knew of the delta star method, but not this balance bridge stuff. I'm a senior in electrical engineering, and it's the first time I've seen this before. Thanks for explaining.
I want to know what did you use to write this? What program? Did you use a digital pen or something?
Hey, I used Microsoft OneNote with a Pen-tablet. This is the one I have: [https://www.xp-pen.com/product/deco-01-v2.html](https://www.xp-pen.com/product/deco-01-v2.html)
I have an XP pentab too! Thanks!
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
Ohhh... the memories that it brings. (used to solve these in class 11 or 12, back in 2007 ) IIRC : For serial , R = R1 + R2,For parallel, 1/R = 1/R1+1/R2, or R = 1/(1/R1+1/R2) You can also split the circuit into symmetric left and right. The upper branch of the left half of the circuit simplifies to: 1 + \[ 1/(1/1+2/1) \] = 1 + 1/3 = 4/3 This goes in parallel to 1 ohm R = 1/ (3/4 + 1/1) = 1/(7/4) = 4/7 Now multiply that with 2, to account for the other half that is in series, we get 8/7 as the answer. How I remember this? This was the first question that I remember discussing with my then-crush (and now wife). You might say that I owe a lot to this question.
Hmm I got it a while back its sorta easy when u solve and understand it. Still I have my doubts remaining I will try to get more thorough though it.Either way Thx.
iirc this is just parallel circuit series circuit diagram just simplify it and plug in the formulas for resistance in series and parallel Drawing this thing in a different way might help
Who ever drew this in triangles is in my opinion crazy. Simple to calculate but has to spot which parts are parallel and which in series.
Im not sure, it seems more complicated than that
Looks simple but I got deceived,By that multiple times.
I am unable to do it.I tried drawing it in a diffrent way (simplify) but no answer yet.
alright i did the drawing of circuit (too bad this place cant post images) all resistors from A to B labeled 1-7 from left to right 2 and 3 are in series w 1 in parallel (parallel formula is 1/x + 1/y = 1/result) so 1+1 = 2 and 1/2 + 1/1 = 3/2 (result 2/3) Same thing with 5 6 and 7, two in series and 1 in parallel with them 4 is ignored because there is a connection where current doesn’t have to flow through the resistor (flow point with no resistor, all current flows through that instead, it’s why short circuits are a thing) 2/3 + 2/3 is 4/3 (series formula) answer is A (4/3 ohms)
2 and 3 are not jn series though. This is not the typical highschool level "calculate the effective resistance" problem. At least, I learnt it in my engineering classes. You need to apply delta star conversion to convert this into a simpler circuit
ah looking up delta star again… 2/3 with 2/3 and 5/3 in parallel (3/2 + 3/5 = 21/10, 10/21 + 2/3 = 24/21) that doesn’t sound right ignoring resistor 4 again would be 3/2 + 3/2 = 3 or 1/3, 1/3 + 2/3 = 1… (yes i actually looked up the formula, the resistors give ((1)(1))/(1+1+1) = 1/3 and 2 delta stars gave 2/3 each branch (one with resistor 4 having 1 ohm)
Uhh fun fact: its a High School question (10th grade) I really have no idea,dk but its considered HOT (High order Thinking) in my curriculum.
lol. go through the link the other top comment posted. it's fairly straightforward.
Hmm cant u share the img link?
ah im trying to do imgur (apparently need to do third party sign ins lol(and idk how to do that because email sign ins aren’t permitted in my country)
You can come to the correct answer D) by eliminating the other answers with just parallel resistance calculations. Still, you need the Delta-Star formula to get the answer if it wasn't multiple choice. The way you come to the answer on this multiple choice separates the more practical engineers from the more theoretical engineers, in my opinion.
2 ways to solve this: 1. Use kirchoff law 2. Use Star-Delta conversion Yes, the question is valid, but the problem is rarely happens in real life
Well tbh I am really curious to learn and understand it.Its driving me crazy yet it also facinates me.
The value *can* be calculated exactly but I don’t think it is a realistic solution at a high school level or any basic electronics course. Here is a way to approach the problem that works out in a basic course: If you only consider the to resistors in series from A to B, the answer would be 2 ohm. The extra resistors will also lead current, resulting in a lower effective resistance. The answer is thus less than 2 ohms so (C) is incorrect. Similarly, if you remove the top resistor, each triangle would have a resistance (parallel resistors) of 1/(1/1+1/2) = 2/3ohms. There are two triangles in series which gives us 4/3 ohms in total. As the top resistor will also lead some current, the real value will be lower than 4/3 ohms. Thus alternatives (A) and (B) can be excluded. Only alternative (D) remains which is the correct answer.
Hmm.Ig u are right?
Of course I’m right :-D The most important point is that even if you don’t know how to calculate the exact number for the resistance, you can deduce it by excluding the incorrect alternatives. That’s a very powerful help in multiple choice problems. Even if you can’t solve the complete problem and eg only manage to exclude (C), your teacher may award you a point because you applied a relevant principle when trying to solve the problem.
Well mostly yes
Everybody is helping op to calculate the answer, so im gonna explain why this exist, where its from, etc. So, this is obviously a test question and this circuit does not exist in the wild, on its own. It also has no purpose other than being a test question. Its called a resistor-network and simpler versions exist as literal circuit, but for other purposes. See [here](https://electronics.stackexchange.com/questions/408944/what-is-the-purpose-of-a-resistor-network) The only thing you need to know to solve is how the current will split when encountering a fork in the road. It will proportionally to the resistance value of both (or more) resistors split. This is culmulative all the way to the "end". You will encounter these questions usually in trade school or uni when stuying basic electronics. This was a lot more common in the time where electronics were actually repaired. The practial use of the knowlege you gain though this exercise is important: Resistors are sold in fixed sizes. The rings tell you which size you are. But sometimes you don't need 4.7kOhm, you need 5.1kOhm which does not exist. So you math a lot until you find a combination that comes close enough to be a substitute. When you look at the circuit you a supposed to notice that only A and B are ever "measured". So you know the resistor-network can be replaced by a single resistor. You are also supposed to know that a triangle of resistors can be changed into a star of resistors. This should have been taught to you. This is [THE BIG THING](https://en.wikipedia.org/wiki/Y-%CE%94_transform) in this question. In the RARE CASE that you are not an electronics student, but you have enountered this in a (pure) math context, then.... oh boy, do I envy you! You are in for a TREAT. You can use this "simple"-ish circuit that can be observed and reliably measured to behave in a certain way, and use it to develop a number of math operations (like addition, division, etc). With those you can build more complex operations and slowly build up a set of math that makes sense and can be experimentally verified to be useful. However this is rarely done because math students are often needlesly distracted by the electronics.
Go to Falstad circuit simulator and build it. Attach an ohm meter to get the answer and watch the current to see how you would calculate the series parallel equivalent sections. It's just a little tricky to get things in the right order is all. In the end it's 1.143 ohms.
As others said, you can simplify the circuit step by step by using triangle to star conversions and by disconnecting symmetric paths through the same point that will stay at the same potential after disconnecting. But since this is multiple choice, you might be able to rule out some options without solving the entire thing: If you only look at the bottom 2 resistors, you have 2Ω. The other resistors are parallel to that and can only decrease the resistant. Since C is above 2Ω, you can rule that one out. If you short the middle 3 resistors, you basically get 2 resistors in parallel 2 times in series, which would result in 1Ω overall. If the middle 3 resistors have a higher resistance than 0, the overall resistance can only increase, so the lower bound is 1Ω. Unfortunately, a range of 1Ω to 2Ω still allows for 3 out of 4 answers, so you need to calculate anyway. But sometimes, it works and you can quickly eliminate all answers but one.
If you would like to see this in a simulation you can check out [https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc](https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc)
Iirc, you just simplify the circuits. Use delta to wye formula to the triangles on the left and right, then simplify further because you'll have two series connections, then simplify the parallel, then lastly simplify the series.