Calculate Equivalent Resistance of a 5 Resistor Bridge Circuit | Kirchhoff's Loop & Junction Rules
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- Опубліковано 21 лип 2024
- This circuit can NOT be reduced using basic series and parallel reductions. Instead this problem must be solved using loop rule and junction rules.
The physics of this problem is not overly difficult, however the algebra causes most people to get lost in solving the problem. Here we take a look at using Ohm's Law to set up a system of equations that can be used to solve for the current and voltage at each resistor as well as the equivalent resistance of the entire circuit.
Yes you can use a Delta-Y transformation here, but often people encounter this problem before learning that technique.
The topic of Wheatstone Bridges often comes up in introductory physics and engineering courses such as AP Physics C E&M, Electricity and Magnetism, Physics 202, Physics 8.02 and appears on A Level physics as well as the JEE.
I was having a bit of trouble grasping the concept of Kirchhoff's rules, so thank you for your explanation! :)
Any time!
This was extremely helpful even for cementing the basics. I got stuck every time when making my equations but your explanation helped me understand how to start combining them.
Glad it helped!
Thank you for this video! may I ask why I1=1.92 A?
That is an absolutely great explanation! Keep up the great work.
Thanks, will do!
WOW! appreciate your efforts. Thank you so much for explaining the way of doing this math.
Glad it was helpful!
how did you get the 21 I^3/7 ???
Thank you so much for your wonderful explanation. Take love & respect from Bangladesh....🥰🥰🥰🥰🥰🥰🥰🥰
Thanks and welcome
thank you best explanation yet!
You're welcome!
Thank you mr❤
I have a question, in 14:55, where did 1.92 come from for I1? I have already done the previous steps before that, yet I couldn't seem to get the desired answer in the end. Thank you in advance!
I think I get it now:
0 = (-183 / 28)I1 + 351 / 28
(183 / 28)I1 = 351 / 28
I1 = (351 / 28) / (183 / 28)
I1 = 351 / 183 = 1.92A
Is this correct?
What a superb explanation
Thanks! This was one of my favorite problems when learning circuits.
Only question, on rearranging. Why in the first equation the I3 equal to I3=(5I1-9)/4 while for I2 it is equal to I2=(9-5I3)/7? Why isn't it I2=(5I3-9)/7?
In the I3 equation I moved everything else to the other side of the equals sign. (so the 9 and the 5I1 changed sign) In the I2 equation I moved the I2 across the equals sign so it changed sign.
In 12:38 where did 4I1 and 4I3 came from? More specifically where did that "4" came from? Since I1 and I3 have 1Ω and 3Ω resistors, then what that "4" does there? Is that the substitute of so called 4I4??
Oooh i see. Now I understand how it came. Just to be sure nobody will fail on that:
0 = 9 - 1I1 - 4I4 (we replace 1I1 with I1 as there is no point to write like 1I1. We replace 4I4 by the substitution of I4 which is I4=I1-I3)
0 = 9 - I1 - 4(I1-I3) (we multiply 4 with each part in the parentheses. Be careful with signs)
0 = 9 - I1 - 4I1 + 4I3
You got it! Thanks for clarifying the algebra.
love from bangladesh!
thank you sooooo much you really saved me
Glad I could help!
Pls, any practical result out of this except a practice in a simple algebra? The bridge itself is used to measure voltage only between the middle points. And is widely used because any input voltage noise is eliminated and does not affect a measured result.
I'm sure someone smarter than I uses it effectively in some application. But the point I think you are making is largely correct; I as a student then teacher really have only ever seen this bridge circuit as a common way to start roughing up kids in introductory circuits courses.
You can use the online HTML5 "PhET Circuit Construction kit (DC)" to "build" this circuit and check the result. I found that the battery current was 3.098 A to a couple of extra significant figures.
I check my work on a few of these bridge circuits using PhET. =)
Thank you so so so much
No problem
Thanks brother
Gotta love a good bridge circuit.
Book name please?
This type of bridge circuit shows up in lots of books. However the numbers used in this problem are not out of a book. I just sat down and made up, then worked this one out.
Use the D-Y transformation and you solve it in less than 2 lines.
Not quite two lines but inarguably easier. The issue is, the Delta-Y conversion is something that virtually no person seeing a bridge circuit for the first time is going to understand. They may utilize it as a magical tool, but not comprehend what the transformation is doing.
The real point of the video, regardless of D-Y or lots of algebra, is the diagram @35 seconds. The circuit cant be solved using typical series and parallel combination methods.
@@INTEGRALPHYSICS Sure, I get that. I don't understand the D-Y conversion either, but it's a cool tool to use when you can't think of a faster way to solve a problem on a test. I was searching the web the last 2 days up and down for circuits, cause my teacher is a total a$$h0le and can't teach. This video was helpful on some fronts.
Please provide your 2-line solution using D-Y.
Why is I3 1.4 instead of 0.147 I did ur calculation in my calculator for I3 and I got 0.147
Not sure what you are referencing. @15:58 I solved I3=.147A then rambled about sig figs.
@@INTEGRALPHYSICS I agree on the question, if you use the calculator to solve the equation it ends up with 0.15
(5*1,92 - 9 ) / 4 = 0,15!
I get what you're saying.
"Don't Get What You Are Saying. 😿"
Delta to Y simple
I agree, but the Delta-Y transformation always seems a bit like a band-aid or cheating.