Searched for hours for help on this homework problem only to discover it can't be represented using series or parallel conventions. Thank you so much. I have no idea why this isn't mentioned anywhere else on the internet.
This is definitely one of the harder first or second year circuit problems. There is a much simpler method using what is called a Y-Delta transformation but you need to know the trick to know how to use it. It's not just series or parallel but a rather more complicated method to solve certain type of resistor network. Good luck with your studies! I'll try to make a video on Y-Delta transforms this weekend!
It can be simplified by doubling the bridging impedance into two parallel impedances and transforming each delta (triangle) into star. Then it's a piece of cake to reduce it to a single impedance.
We can also use the DELTA to WYE Formulas for Resistances to find Rt, 'Total Resistance'. However, I would prefer this method you have demonstrated to save time. Find the all the currents flowing and the 'equivalent resistance' comes so easy to calculate in the end. Thank You Sir 'PN' -Junction.... :)
Thank you. Until now I didn't realise some circuits defy classic series & parallel analysis. So you've up levelled my applied knowledge. I'm guessing the 4th loop, at the bottom be used also or instead.
Several years later and I come back to this video and channel for help again! I do have one question about Prime 2 Calculation at 12:45. I don't understand where you got the coefficient 3 from, when all you did was replace i2 rather than i3. (lowercase i for clarity). Since term rearrangement gives 13 = i3 + 2(i2+i3), the substitution of i2 would give us 13 = i3 + 2(i1 + 2i3). This simplifies to 13 = i3 + **2i1** + 4i3 = 2i1 + 5i3 I don't see where the 3 i1 comes from.
I completely agree. I have a video on the same problem using that transformation. My only problem with y-delta is you need to know the trick before applying it. It’s not an obvious transformation.
Great Explanation. Can we also find Potential at Va & Vb by voltage divider rule . Then find Vab = Va-Vb, Then find Current i3 =Vab/1. Is it right way to analyse if we just want to find current i3. Please let me know.
@@PhysicsNinja Unfortunately this is incorrect as exact voltage division requires no bleed current i.e. the bridge would need to be in balance. As this example has an unbalanced bridge arrangement, the use of simple voltage division will result in errors since the current in R3 is not zero. HTH. BTW this was a great video and thanks for sharing.
Thank you, very clear and well explained. Determination of which loops are used is the only thing I'm unsure on. I'm guessing you use the minimum number to cover all lines between each junction.
RNAlh thanks, you can certainly write down more loop equations than you need. I like to use 2 junction equation then determine how many variables are remaining. If there are 5 currents and I use 2 junction rules I know I will need 3 loop rules. Happy learning!
This is really awesome but I have one question. If you slightly complicate this problem by changing the value of the resistor that i2 flows through, to anything other than 1 you end up with a fraction in the prime of three. Basically I tried this method with a range of resistor values and was significantly stumped at the point where I end up with a fraction representing i2. What are the options in such a case?
Can also solve these circuits by Node Analysis as then you only have to solve two equations with two unknowns (Va and Vb) versus Mesh Analysis which has three equations and three unknowns: (Va - 13)/1 + (Va - Vb)/1 + (Va - 0)/1 = 0 (Vb - 13)/1 + (Vb - Va)/1 + (Vb - 0)/2 = 0 Then all currents can be calculated via Ohm's law.
Andrew Kostiuk I have another video where I solve the circuit using a delta-y transform. It’s super easy if you know the method. The method in this video is by far the most tedious algebraically but still straightforward..
Wonderful work! appreciate your help and effort. I want to add, unfortunately I could not understand why there was only 3 loops instead of 4. I thought (abd) junctions are also a loop. Can you enlighten me on this regard please? Thank you so much.
I would consider a very simple parallel circuit (1 battery and 2 resistors in parallel). WHen you solve this problem you are solving for 2 unknowns (the currents through the resistors). When you set up loop rules for this simple circuit you can write 3 loop rules. You only need 2 equations to solve this problem. One of the loop rules (around both resistors) is actually the difference between the 2 other loop rules (each with battery and 1 resistor). So you gain nothing by writing this additional loop rule. The same applied for the bridge circuit. There are 5 unknown currents and therefore i need 5 equations. I know there are 2 junctions and therefore i need to write 3 independent loop rules. For this problem, you may be also able to solve if you write 4 loops rules and use only 1 junction rule. If you try to solve and can't it's because one of your loops rules is a linear combination of the others.
I was following the circuit perfectly until 14 minutes in as I had 11 I3 in equation 1.1 and 20 I3 in equation 2.1 so can't simply multiply the bottom top one to cancel out I3. What do I do now
I think there is something wrong, I solved the 3 equations by calculator and I found the equivalent R is 1, not 1.2, maybe you can double check this. Thanks.
Is it pssible to do in way,at first remove bridge resistor,calculate Vcd,then put it back and calculate I bridge,we will also get direction from it,but my valuea is little different,left side 6,5A and 7,5A,right side 4,33A and 3,33A
Hi Saif, great question. In general the number of "independent" loops should be equal to the number of unknowns in the problems. If we want to solve for 5 current values then we would require 5 independent loops. The only things that you have to be careful with are that the loop equations are indeed independent. Sometimes if you add 2 loop equations you get another loop equations - these are NOT independent from each other.
Electrical engineers often use Thevenin and Norton equivalent circuit methods to quickly simplify complex circuits. I have another video where i use Thevenin Equivalent to simplify the circuit in this video. Good Luck!
subhasis das this is one of the most basic sums in current electricity. Just eliminate resisitor btween a and b coz the have same potential the solve (wheatstone bridge)
Thanks for the comment. To be perfectly honest i've never heard of a delta conversion so i'm going to have to look it up before i can comment. I'm certain there are probably faster way to solve this problem. The way showed in the video in not the most elegant proof but it's straightforward using simple circuit rule.
It can't be simplified because of the resistor in the middle, right? With open terminals you would have two simple series combinations in parallel and RT would be RA||RB?
Yes, it can't be simplified using standard rules for parallel and series resistors. There are other methods (delta - Y, see my other video) that can be used but these are move advanced thicks.
For a trick question, if the ratio of the resistances between the two resistors on the left is equal to the ratio of the two resistances on the right, then there is no voltage between "a" and "b", and the middle resistor has no effect, and can thus be removed. Then R_total would be R_left || R_right :-)
My guess is that since you have calculated the current (I) and the resistance (R) is given, you can calculate the voltage drop. This is a great question, would like for Online Physics Ninja to reply to this.
Hello sir, I have a problem about finding the equivalent reactance. It’s the most hardest one I’ve ever seen, how can I share it with you? And I wish if you can do help me please
Peter Chan A loop is shorthand for Kirchhoff voltage law. If you go around a closed path (loop) the voltage differences should sum to zero - Conservation of energy. The junction law is also called Kirchhoff current law. The junction rule says whatever current flows into a point must equal to the current flowing out= conservation of charge.
Great question! Yes you can write this in matrix form. Any linear system of equations like this can be written in matrix form. This is essentially applying what is called the voltage node method.
Great question. You can definitely use the smaller loop (3). The equation you get from this loop will be a combination of the loop equations that i obtain so you end up with the same thing.
Great questions. For complicated circuits i first determine the number of unknowns that i'm trying to solve for. In this problem i'd like to find the voltages at the nodes and the currents. Some are dependent on each other but at the end I have 5 unknowns that i'd like to solve for (5 current values) if you know the current values you will also know the voltages. This means i'll need at least 5 equations. I use 3 KVL (loop equations) and 2 (KCL) equations to give me a total of 5 equations.
So there is no way of rationalizing the equivalent resistance apart from adding a test current to the network? How did pi-t conversions were first conceived then?
Zupprezed you can write down on equivalent resistance using delta-y, combined with series and parallel combination. This is something Electrical engineering students would cover in detail and beyond the scope of this video. I’ll add a video on this method but I do have videos on y- delta transforms. Thanks for the comment.
Here's a video a did before using the same problem but with Y-Delta conversion. You can easily write a general equation for Reff! th-cam.com/video/TIVcNNpPn_w/w-d-xo.html
Searched for hours for help on this homework problem only to discover it can't be represented using series or parallel conventions. Thank you so much. I have no idea why this isn't mentioned anywhere else on the internet.
This is definitely one of the harder first or second year circuit problems.
There is a much simpler method using what is called a Y-Delta transformation but you need to know the trick to know how to use it. It's not just series or parallel but a rather more complicated method to solve certain type of resistor network. Good luck with your studies! I'll try to make a video on Y-Delta transforms this weekend!
Here's a video solving the same circuit using the delta-y transform. th-cam.com/video/TIVcNNpPn_w/w-d-xo.html
It can be simplified by doubling the bridging impedance into two parallel impedances and transforming each delta (triangle) into star. Then it's a piece of cake to reduce it to a single impedance.
🎉@@PhysicsNinja
That was the simplest explanation that I've come across. Thank you so much for your help.
This is the exact assignment my professor has given to me. Thank you Sir! I understand it more clearly!
Thank you for this video, not only did I understand the process but also gain new insights on handling this kind of circuit!
Thank you very much man, I really couldn't understand shit from the book because they don't explain anything but you did it explain it very well
5 years later it is still helping out students like me. thanks mate.
Thanks! Ninja is still around creating content.
Bhai bhai. 😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭thank u, i was searching lecture for this type of circuit, today i found, thank you sir...... 😭😭😭😭😭😭
I love your usage of colors in the circuit. Thank you for the educational video.
Thanks for your support
Thanks for the clear and concise explanation!
You are welcome!
im so fukked for electricity tmrw
same bro
same broooooooooo
This man knows how to teach, excellent job!
Thank you so much sir, you have saved me from all the mess I have been going through with this type circuit
I love you sir your explanation was the best
THANK YOU VERY MUCH!!!!
Thank you very much. Now I can face my exam!
when instruction clear, questions like this is kinda fun to solve
Love from India. ,,, Brother ❣️❣️❣️❣️❣️
Thank you very much! This was very helpful and your explained it perfectly!!
well done sir
i need to find resistance across a and b
I have understood thank you mwalimu (teacher)
Great!
Thanks, Good Explanation 🤩
Thanks
10/10 thank you. Why do engineering teachers have to suck so bad and random youtubers be so good
Thank you
really , helped sir! thank you.
Thanks a lot, I understood perfectly !!!
We can also use the DELTA to WYE Formulas for Resistances to find Rt, 'Total Resistance'. However, I would prefer this method you have demonstrated to save time. Find the all the currents flowing and the 'equivalent resistance' comes so easy to calculate in the end. Thank You Sir 'PN' -Junction.... :)
Yes, I have another video where I use delta-Y for the same circuit. Best regards
thanks a lot that was so helpful
Thank you so much
You're most welcome
Thanks thalaiva
thank you very much you are the best teacher
the delta to y conversion were made for these types of weird circuits :)
i have another video on delta-Y for the same circuit. I agree it's easier but you have to know about delta-Y
Please share that video I see
THANKS FROM INDIA.........🙏
At the end of the day you are an effing genius man
Thank you. Until now I didn't realise some circuits defy classic series & parallel analysis. So you've up levelled my applied knowledge. I'm guessing the 4th loop, at the bottom be used also or instead.
Yes, the 4th loop can also be used.
you made it easier
Great video! Thanks
very helpful
it was perfectly explained
damn this was a great explanation
Several years later and I come back to this video and channel for help again! I do have one question about Prime 2 Calculation at 12:45.
I don't understand where you got the coefficient 3 from, when all you did was replace i2 rather than i3. (lowercase i for clarity).
Since term rearrangement gives 13 = i3 + 2(i2+i3), the substitution of i2 would give us 13 = i3 + 2(i1 + 2i3).
This simplifies to 13 = i3 + **2i1** + 4i3 = 2i1 + 5i3
I don't see where the 3 i1 comes from.
true hero ❤
This helped so much thank you
Thanks 🙏 that was so smooth ❤️
Thanks man, great explaining!
You rescued me
The content is really valuable. I just hope that you raise your voice a little bit more in the coming videos. Thanks in advance!
You are simply amazing!
You can convert any one of the delta parts to star and can find Req without applying loops. That will be easy in my opinion.
I completely agree. I have a video on the same problem using that transformation. My only problem with y-delta is you need to know the trick before applying it. It’s not an obvious transformation.
Can you give a formula for the equivalent resistance? ( without numbers, so it can be used elsewhere)
See my other video on delta-y transform. It’s the easiest way to write as an equivalent resistance
Great Explanation. Can we also find Potential at Va & Vb by voltage divider rule . Then find Vab = Va-Vb, Then find Current i3 =Vab/1. Is it right way to analyse if we just want to find current i3. Please let me know.
Yes, that seems ok. Test it out and if you get to the same answer it will give you confidence. There are several ways to solve this circuit!
@@PhysicsNinja Unfortunately this is incorrect as exact voltage division requires no bleed current i.e. the bridge would need to be in balance. As this example has an unbalanced bridge arrangement, the use of simple voltage division will result in errors since the current in R3 is not zero. HTH. BTW this was a great video and thanks for sharing.
Interesting to see a video on things I learned 50 years ago.
Now, pls add some volume control to your recording ;)
Awesome video
Thank you, very clear and well explained. Determination of which loops are used is the only thing I'm unsure on. I'm guessing you use the minimum number to cover all lines between each junction.
RNAlh thanks, you can certainly write down more loop equations than you need. I like to use 2 junction equation then determine how many variables are remaining. If there are 5 currents and I use 2 junction rules I know I will need 3 loop rules. Happy learning!
RNAlh, "I'm guessing you use the minimum number to cover all lines between each junction." Right on!
awesome video
Thanks this was very helpful
Would you be able to make a video on the Maxwell Bridge Circuit? It would be much appreciated
Why didn't you subtract opposing currents where you have resistors common to two loops?
This is really awesome but I have one question. If you slightly complicate this problem by changing the value of the resistor that i2 flows through, to anything other than 1 you end up with a fraction in the prime of three. Basically I tried this method with a range of resistor values and was significantly stumped at the point where I end up with a fraction representing i2. What are the options in such a case?
Can you do the same questions with resistances 2 ,4,8,3,6ohms pls?
At 8:35, when doing loop 1, you do not account for the opposing current in the first resistor--WHY?
Thanks!
thank you
Happy to help! Thanks for watching
Thank you so much appreciated it
Can also solve these circuits by Node Analysis as then you only have to solve two equations with two unknowns (Va and Vb) versus Mesh Analysis which has three equations and three unknowns:
(Va - 13)/1 + (Va - Vb)/1 + (Va - 0)/1 = 0
(Vb - 13)/1 + (Vb - Va)/1 + (Vb - 0)/2 = 0
Then all currents can be calculated via Ohm's law.
Andrew Kostiuk I have another video where I solve the circuit using a delta-y transform. It’s super easy if you know the method. The method in this video is by far the most tedious algebraically but still straightforward..
THANK YOU!!!
Wonderful work! appreciate your help and effort. I want to add, unfortunately I could not understand why there was only 3 loops instead of 4. I thought (abd) junctions are also a loop. Can you enlighten me on this regard please? Thank you so much.
I would consider a very simple parallel circuit (1 battery and 2 resistors in parallel). WHen you solve this problem you are solving for 2 unknowns (the currents through the resistors). When you set up loop rules for this simple circuit you can write 3 loop rules. You only need 2 equations to solve this problem. One of the loop rules (around both resistors) is actually the difference between the 2 other loop rules (each with battery and 1 resistor). So you gain nothing by writing this additional loop rule. The same applied for the bridge circuit. There are 5 unknown currents and therefore i need 5 equations. I know there are 2 junctions and therefore i need to write 3 independent loop rules. For this problem, you may be also able to solve if you write 4 loops rules and use only 1 junction rule. If you try to solve and can't it's because one of your loops rules is a linear combination of the others.
I was following the circuit perfectly until 14 minutes in as I had 11 I3 in equation 1.1 and 20 I3 in equation 2.1 so can't simply multiply the bottom top one to cancel out I3. What do I do now
thanks sir .
If the last two resistors were both 2ohms, would current flow through the 3rd one?
Proteeti Sarkar no current in that case. Check out my new video on Wheatstone bridge.
Directly use the Mesh Current Method will solve this quicker.
where u hava written 2ohm if we take resistor out and there is only a wire then how will we solve it
sorry for bad english
Great
How we can calculate effective resistance without battery
That's my question also
I think there is something wrong, I solved the 3 equations by calculator and I found the equivalent R is 1, not 1.2, maybe you can double check this. Thanks.
I request for the video of the other method that I can use
Awesome.........
if there are no mention voltage, Only hava resistor value. How to find EQ resistor that bridge connection?
I would use a delta Y transform. Check out my other video.
Bridge Circuit Equivalent Resistance using Delta-Y Transform
th-cam.com/video/TIVcNNpPn_w/w-d-xo.html
@@PhysicsNinja Thank You so much..
Is there a way someday in the future that this circuit can be represented using series or parallel resistors?
Is it pssible to do in way,at first remove bridge resistor,calculate Vcd,then put it back and calculate I bridge,we will also get direction from it,but my valuea is little different,left side 6,5A and 7,5A,right side 4,33A and 3,33A
Super ❤️
the four resistors 20 ohms,40ohms, (20+x)ohms , 80 ohms respecrively from a whitestone bridge. find the value of "x".give answer for this please
How would you get the total resistance if you have two wheatstone bridges in one circuit that share a common wire?
How should we consider loops???
Hi Saif, great question. In general the number of "independent" loops should be equal to the number of unknowns in the problems. If we want to solve for 5 current values then we would require 5 independent loops. The only things that you have to be careful with are that the loop equations are indeed independent. Sometimes if you add 2 loop equations you get another loop equations - these are NOT independent from each other.
Are there any method of find equivalent resistance without applying KVL,KCL??
Electrical engineers often use Thevenin and Norton equivalent circuit methods to quickly simplify complex circuits. I have another video where i use Thevenin Equivalent to simplify the circuit in this video. Good Luck!
subhasis das this is one of the most basic sums in current electricity. Just eliminate resisitor btween a and b coz the have same potential the solve (wheatstone bridge)
Sir Vedio is super , but your voice is not clear. ...
Why should the unknown resistance value be 50%
i think it well be more easy by using dalta conversion to find equivalent resistance and than to find current. what are you thinking will it work?
Thanks for the comment. To be perfectly honest i've never heard of a delta conversion so i'm going to have to look it up before i can comment. I'm certain there are probably faster way to solve this problem. The way showed in the video in not the most elegant proof but it's straightforward using simple circuit rule.
i have used delta conversion to solve circuit like this one. and i think that is awesome
It can't be simplified because of the resistor in the middle, right? With open terminals you would have two simple series combinations in parallel and RT would be RA||RB?
Yes, it can't be simplified using standard rules for parallel and series resistors. There are other methods (delta - Y, see my other video) that can be used but these are move advanced thicks.
For a trick question, if the ratio of the resistances between the two resistors on the left is equal to the ratio of the two resistances on the right, then there is no voltage between "a" and "b", and the middle resistor has no effect, and can thus be removed. Then R_total would be R_left || R_right :-)
I understood shit like my brain
Why can't we simplify the circuit with series and parallel combinations??? an explanation would be nice.
Very tedious haha. Thanks though I suck at circuit theory.
Why the current I2 is not divided in ba branch??
Thank you
How to find the voltage between two junctions in circuits like this?
My guess is that since you have calculated the current (I) and the resistance (R) is given, you can calculate the voltage drop. This is a great question, would like for Online Physics Ninja to reply to this.
Walter Montalvo thank you for your help I tried that and it worked 👍
You are welcome, glad I could help.
would i2 flow in the 1 i ohm resistor? the resistor in the middle?
Hello sir, I have a problem about finding the equivalent reactance. It’s the most hardest one I’ve ever seen, how can I share it with you? And I wish if you can do help me please
JEMSHAR please email it to me onlinephysicsninja@gmail.com
Online Physics Ninja
I just sent it to your email
still don't see it in my inbox
Starting from 6:44 . . . what in the world is a 'loop' and and what are the mathematics related to them?
Peter Chan A loop is shorthand for Kirchhoff voltage law. If you go around a closed path (loop) the voltage differences should sum to zero - Conservation of energy. The junction law is also called Kirchhoff current law. The junction rule says whatever current flows into a point must equal to the current flowing out= conservation of charge.
@@PhysicsNinja
Thanks for the clarification.
I already know about the junction thing. It was the loop thing which sort of puzzled me.
might be a dumb question but can we find the values of Is in loop equations by using matrix form?
Great question! Yes you can write this in matrix form. Any linear system of equations like this can be written in matrix form. This is essentially applying what is called the voltage node method.
Why not consider the smaller loop below loop 3?
Great question. You can definitely use the smaller loop (3). The equation you get from this loop will be a combination of the loop equations that i obtain so you end up with the same thing.
why did you only take 3 loops and not 4 or 2?
Great questions. For complicated circuits i first determine the number of unknowns that i'm trying to solve for. In this problem i'd like to find the voltages at the nodes and the currents. Some are dependent on each other but at the end I have 5 unknowns that i'd like to solve for (5 current values) if you know the current values you will also know the voltages. This means i'll need at least 5 equations. I use 3 KVL (loop equations) and 2 (KCL) equations to give me a total of 5 equations.
So there is no way of rationalizing the equivalent resistance apart from adding a test current to the network? How did pi-t conversions were first conceived then?
Zupprezed you can write down on equivalent resistance using delta-y, combined with series and parallel combination. This is something Electrical engineering students would cover in detail and beyond the scope of this video. I’ll add a video on this method but I do have videos on y- delta transforms. Thanks for the comment.
Here's a video a did before using the same problem but with Y-Delta conversion. You can easily write a general equation for Reff! th-cam.com/video/TIVcNNpPn_w/w-d-xo.html
Sir can we calculate the equivalent resistance in this case if the emf is not given?
EMF actual value does not matter for equivalent resistance. Double the voltage will double the current, R = V / I :-)
what do you mean by I3 and I2 will combine? Does it give a null value? As in the current there disappears (changes to zero) or what?
Jamilla Dambo approx what time in the video did I say this