An electric circuit problem involving 5 current variables

[vcsharp2003]

Homework Statement

An electric circuit is as shown in diagram below. Determine the current through the 6 V battery.

Relevant Equations

Kirchhoff's Junction rule and Kirchhoff's Loop rule

I came up with the following set of equations. I try to express all currents in terms of $I_1$ and then substitute in the first equation to get the value of $I_1$. Since $I_4$ needs to be determined and I have already expressed it in terms of $I_1$, so it can be calculated. The problem is that I get a value of $I_4= \dfrac {62}{25}$ A when it's given in the book as 2 A. I am not sure if my approach of expressing all currents in terms of $I_1$ and then substitute in any equation is the correct way of solving such system of linear equations in 5 variables. We're not allowed to use online equation solvers, but we need to solve them manually.

Another problem is that how can I know that all equations I get by applying Kirchhoff's laws are independent, If I look at the first 3 equations I got then I can clearly see that when second and third equations are combined they give the first equation. I used to think that each application of Kirchhoff's rules to a different loop or junction will yield an independent equation, but clearly its not the case with the above set of equations.

UPDATE 1

I checked my solution using an online system of equations solver and the value of $I_4= \dfrac {62}{25}$ is correct. This can be seen at Solution from online equations solver. So, this means my initial approach of finding all current in terms of $I_1$ and then substitute in an equation to get the value of $I_1$ was correct.

UPDATE 2

I checked the solution in the book and it seems to be wrong when applying loop rules to loop 2 and loop 3. So, the answer of $I_4= \dfrac {62}{25}$ obtained from online equations solver seems to be correct, provided my initial set of equations obtained by applying Kirchhoff's laws were correct.

 

[nasu]

If you have N junctions you only get N-1 independent equations from junction rule.

 

[kuruman]

Also, you have 5 unknown currents and 6 equations. One of the 6 equaitons cannot be independent.

 

[vcsharp2003]

If you have N junctions you only get N-1 independent equations from junction rule.

Ok, thanks. So, for loops also can we say the same thing i.e. if there are N loops that give us N equations then we should only take any N-1 of the N loop equations?

 

[vcsharp2003]

Also, you have 5 unknown currents and 6 equations. One of the 6 equaitons cannot be independent.

So, I should remove any one of the junction rule equations. That would give me 5 independent equations in 5 variables.

 

[kuruman]

Yes.

 

[vcsharp2003]

Yes.

Thanks. Can I say that each different loop will yield an independent equation or there will be some dependent equations using different loops?

 

[kuruman]

Thanks. Can I say that each different loop will yield an independent equation or there will be some dependent equations using different loops?

The latter. Look at the drawing. For example, the equation you get from loop AKMLA is the sum of the two equations you get from ALKA and KLMK.

 

[Steve4Physics]

Ok, thanks. So, for loops also can we say the same thing i.e. if there are N loops that give us N equations then we should only take any N-1 of the N loop equations?

That doesn't sound right.

Note, in this problem there are six possible ‘loops’ I can see. Apart from the three you have used, there are AEBCDA, AECDA, and BCDEB.

One strategy is this:

With N=3 nodes you set up N-1 = 3-1 = 2 equations using Kirchhoff's 1st law (junction rule).

Since you have U= 5 unknowns (currents) you need to setup a further U - (N-1) = 5 - 2 = 3 equations using Kirchhoff's 2nd law (loop rule). You must select (3) loops which together cover the whole circuit.

I'd guess that strategy should always work. (But if I'm wrong I am confident someone here will correct me!)

Your equations look OK to me. But when I put the equations into an online 'solver' I get $I_4=2.48A$ which disagrees with both the official answer and your answer!

 

[vela]

UPDATE 1

I checked my solution using an online system of equations solver and the value of $I_4= \dfrac {62}{25}$ is correct. This can be seen at Solution from online equations solver. So, this means my initial approach of finding all current in terms of $I_1$ and then substitute in an equation to get the value of $I_1$ was correct.

I get the same answer you did.

UPDATE 2

I checked the solution in the book and it seems to be wrong when applying loop rules to loop 2 and loop 3. So, the answer of $I_4= \dfrac {62}{25}$ obtained from online equations solver seems to be correct, provided my initial set of equations obtained by applying Kirchhoff's laws were correct.

Actually, there's a sign mistake is in the equation for loop 1 in the book's solution. Because of the direction of $I_1-I_2$ as shown, the voltage drop across the 1-ohm resistor should be $-1~\Omega(I_1-I_2)$ in the loop equation.

There's also a mistake in the circuit diagram. The current through the 2-ohm resistor should be $I_1+I_3$, not $I_2+I_3$. The book's solution, however, uses the correct expression for the current in the loop equations.

 

[vcsharp2003]

But when I put the equations into an online 'solver' I get I4=2.48A which disagrees with both the official answer and your answer!

I also got the same answer as you i.e. $I_4= \dfrac {62}{25}=2.48$ as mentioned in OP under UPDATE 1.

 

[vcsharp2003]

I get the same answer you did.

Actually, there's a sign mistake is in the equation for loop 1 in the book's solution. Because of the direction of $I_1-I_2$ as shown, the voltage drop across the 1-ohm resistor should be $-1~\Omega(I_1-I_2)$ in the loop equation.

There's also a mistake in the circuit diagram. The current through the 2-ohm resistor should be $I_1+I_3$, not $I_2+I_3$. The book's solution, however, uses the correct expression for the current in the loop equations.

Thanks for recognizing the mistakes in the book's solution. I was not able to see that current in branch LM was incorrect.

By incorporating your suggestions, the same answer (as from the original approach) is obtained for current through 6 V battery as can be seen at the following link: Online equation solver.

 

[vcsharp2003]

That doesn't sound right.

Note, in this problem there are six possible ‘loops’ I can see. Apart from the three you have used, there are AEBCDA, AECDA, and BCDEB.

One strategy is this:

With N=3 nodes you set up N-1 = 3-1 = 2 equations using Kirchhoff's 1st law (junction rule).

Since you have U= 5 unknowns (currents) you need to setup a further U - (N-1) = 5 - 2 = 3 equations using Kirchhoff's 2nd law (loop rule). You must select (3) loops which together cover the whole circuit.

I'd guess that strategy should always work. (But if I'm wrong I am confident someone here will correct me!)

Your equations look OK to me. But when I put the equations into an online 'solver' I get $I_4=2.48A$ which disagrees with both the official answer and your answer!

I think the following will always be true when it comes to getting independent equations from applying Kirchhoff's Laws to an electric circuit.

1. If there are N ( > 1) junctions in an electric circuit then apply Kirchhoff's Junction rule to only N-1 junctions.
2. If there are N ( > 1) loops in an electric circuit then apply Kirchhoff's Loop rule to only N-1 junctions.

By following the above points, one will always get independent equations.

 

[Steve4Physics]

1. If there are N ( > 1) junctions in an electric circuit then apply Kirchhoff's Junction rule to only N-1 junctions.
2. If there are N ( > 1) loops in an electric circuit then apply Kirchhoff's Loop rule to only N-1 junctions.

By following the above points, one will always get independent equations.

Agree with point 1. But uncertain about point 2. It depends how you are defining 'loop'. For the original question, N=6 (as noted in Post #9) but you don't use 6-1 = 5 loop-equations. What about a messy network such as this: https://upload.wikimedia.org/wikipe...clique_K_3_3.svg/150px-Biclique_K_3_3.svg.png with supplies and resistors at random places in this network - how do you count the loops?

It's an interesting topological-ish problem. It reminds me of Euler's formula for graphs.

 

[vcsharp2003]

QBut uncertain about point 2. It depends how you are defining 'loop'. For the original question, N=6 (as noted in Post #9) but you don't use 6-1 = 5 loop-equations.

Yes, you're right. Point 2 is not true. As @kuruman pointed out in post#8 that if 2 smaller loops add up to another bigger loop, then there would be only two independent equations coming from applying Loop rule to the 2 smaller loops and the 1 bigger loop.

 

[vcsharp2003]

What about a messy network such as this: https://upload.wikimedia.org/wikipe...clique_K_3_3.svg/150px-Biclique_K_3_3.svg.png with supplies and resistors at random places in this network - how do you count the loops?

Yes, it would be impossible to count the number of loops in this circuit.

 

[kuruman]

Yes, it would be impossible to count the number of loops in this circuit.

Not that impossible. First you unpack the circuit diagram by moving nodes around assuming that the blue circles represent nodes and the black lines represent circuit elements.

Then you count nodes, $n=6$ and branches $b=9$ and apply the formula found here. The number of independent loops is $$l=b-n+1=9-6+1=4.$$That makes sense just by looking at the unpacked circuit. It has 4 loops that are not sums of smaller loops therefore 4 independent loop equations.

On edit:
The figure above was carelessly drawn. See post #21 for the corrected version.

 

[hutchphd]

If you write down too many equations (correctly) you will not get an incorrect answer. You will simply do some extra work. If you do it correctly you will just show the same result twice.....so don't fret. I always try to write the smallest equations first. However if you limit the number of equations to a fixed number and you inadvertantly duplicate one you will not be able to solve. This is a skill you will acquire quickly with practice.

 

[vcsharp2003]

Not that impossible. First you unpack the circuit diagram by moving nodes around assuming that the blue circles represent nodes and the black lines represent circuit elements.

View attachment 327433
Then you count nodes, $n=6$ and branches $b=9$ and apply the formula found here. The number of independent loops is $$l=b-n+1=9-6+1=4.$$That makes sense just by looking at the unpacked circuit. It has 4 loops that are not sums of smaller loops therefore 4 independent loop equations.

Wow, that is great knowledge. I never knew these facts. Thanks for clearing such an important and often missed concept.

So, the key is to apply Kirchhoff's Loop rule to loops such that no considered loop is fully contained in another considered loop, in order to get independent equations.

One question regarding independent equations obtained by applying Kirchhoff's Junction rule. Can we just say that if there are N (>1) junctions, then consider only N-1 junctions to get independent equations? Here a junction is defined as a point of intersection of three or more wires (so an intersection of two wires is not a junction according to this definition).

 

[vcsharp2003]

If you write down too many equations (correctly) you will not get an incorrect answer. You will simply do some extra work. If you do it correctly you will just show the same result twice.....so don't fret. I always try to write the smallest equations first. However if you limit the number of equations to a fixed number and you inadvertantly duplicate one you will not be able to solve. This is a skill you will acquire quickly with practice.

Would you say that the approach followed by the book's solution (that is mentioned under UPDATE 2 in OP) is a good one for limiting the number of unknowns? In the book's solution, only 3 unknowns are there, whereas in my approach I had 5 unknowns. It would be quicker to solve for 3 unknowns rather than 5 unknowns when solving equations manually.

 

[kuruman]

I believe that the unpacking was done carelessly in post #17. If the nodes are labeled so that the branches can be correctly identified in the unpacked circuit, the figure below is what one would get. This correction does not change the number of independent loops which is still $4$ because the number of nodes and the number of branches are the same.

So, the key is to apply Kirchhoff's Loop rule to loops such that no considered loop is fully contained in another considered loop, in order to get independent equations.

I am not sure what you mean by fully contained. Look at the circuit above right. You have 5 clockwise loops
16251
52435
15341
16351
62436
They have 4 nodes each and you need to pick 4 out of these 5 loops. Which one is "fully contained" and must, therefore, be thrown out?

 

[vcsharp2003]

I believe that the unpacking was done carelessly in post #17. If the nodes are labeled so that the branches can be correctly identified in the unpacked circuit, the figure below is what one would get. This correction does not change the number of independent loops which is still $4$ because the number of nodes and the number of branches are the same.

View attachment 327435

I am not sure what you mean by fully contained. Look at the circuit above right. You have 5 clockwise loops
16251
52435
15341
16351
62436
They have 4 nodes each and you need to pick 4 out of these 5 loops. Which one is "fully contained" and must, therefore, be thrown out?

I meant the following when I said "fully contained".

If you take the loop 1624351, then it will contain the loops 16351 and 62436. So to get independent equations, either take the loop 1624351 or take the two loops 16351 and 62436 which are fully contained in the loop 1624351.

In your list of loops, I don't know why you haven't taken the loop 1624351.

 

[vcsharp2003]

I believe that the unpacking was done carelessly in post #17. If the nodes are labeled so that the branches can be correctly identified in the unpacked circuit, the figure below is what one would get. This correction does not change the number of independent loops which is still $4$ because the number of nodes and the number of branches are the same.

View attachment 327435

I am not sure what you mean by fully contained. Look at the circuit above right. You have 5 clockwise loops
16251
52435
15341
16351
62436
They have 4 nodes each and you need to pick 4 out of these 5 loops. Which one is "fully contained" and must, therefore, be thrown out?

When you add the first two loops you get the same loop as when you add the last two loops. So clearly, there are overlapping loops which are going to yield dependent equations. I'm not sure if that's a valid way of looking at it.

So, I would either ignore any one of the first two loops or any one of the last two loops from the list of loops you provided, in order to get independent equations.

 

[kuruman]

When you add the first two loops you get the same loop as when you add the last two loops. So clearly, there are overlapping loops which are going to yield dependent equations. I'm not sure if that's a valid way of looking at it.

You missed my point. I agree that when you add any of the five loops listed in post #21, you will not get a new equation. My point is that in this case you only need 4 loops out of the five, however none of the five loops is the sum of two other loops. Which 4 are you going to choose and what justification do you have to throw out the other one?

 

[vcsharp2003]

You missed my point. I agree that when you add any of the five loops listed in post #21, you will not get a new equation. My point is that in this case you only need 4 loops out of the five, however none of the five loops is the sum of two other loops. Which 4 are you going to choose and what justification do you have to throw out the other one?

I would either ignore any one of the first two loops or any one of the last two loops from the list of loops you provided, in order to get 4 independent equations. And the justification would be the first paragraph in post#23 .i.e. "When you add the first two loops you get the same loop as when you add the last two loops. So clearly, there are overlapping loops which are going to yield dependent equations.".

So it's like saying we don't want loops such that any one loop can be formed by multiple other loops OR we don't want loops such that adding some of the loops results in the same loop as obtained by adding some other loops. We need to eliminate loop(s) from our considered loops such that the above fact is not true

 

[kuruman]

I would either ignore any one of the first two loops or any one of the last two loops from the list of loops you provided, in order to get 4 independent equations. And the justification would be the first paragraph in post#23 .i.e. "When you add the first two loops you get the same loop as when you add the last two loops. So clearly, there are overlapping loops which are going to yield dependent equations.".

So it's like saying we don't want loops such that any one loop can be formed by multiple other loops OR we don't want loops such that adding some of the loops results in the same loop as obtained by adding some other loops. We need to eliminate loop(s) from our considered loops such that the above fact is not true

That is not incorrect thinking. It is important, however, to also keep in mind that your choice of loops must include all elements. In this example, element 1-4 appears in only one loop and that loop must be included. Doing so automatically eliminates the possibility of including all four of the remaining loops.

 

[vcsharp2003]

That is not incorrect thinking. It is important, however, to also keep in mind that your choice of loops must include all elements. In this example, element 1-4 appears in only one loop and that loop must be included. Doing so automatically eliminates the possibility of including all four of the remaining loops.

Yes. The loops considered must be independent loops to get independent equations i.e. no considered loop can be formed from a combination of loops being considered.

 

[kuruman]

I think you got it now!

 

[vcsharp2003]

I think you got it now!

Thankyou for your great help.