Loop equations for this battery-resistor circuit

[erobz]

Alright, uhhmm.. I'm here again, and... I found no success in understanding how the mesh operation works. So, uhm... Can you guide me? :(

You define a loop currents for each loop that does not enclose another loop $I_1,I_2$. Then go around the resistive elements and give them polarities based on the loop current direction. Voltage source polarity is fixed. Next apply KVL around a loop. I'll do the loop 1:

$$-I_1 R_1 - I_1 R_2 + I_2 R_2 - V = 0 $$

That is the equation for loop one. Notice that the where the loop currents "mesh" the contributions of voltage drop in resistor 2 are "superimposed"

Now you try loop 2 equation.

 

[eleking16]

View attachment 357660

You define a loop currents for each loop that does not enclose another loop $I_1,I_2$. Then go around the resistive elements and give them polarities based on the loop current direction. Voltage source polarity is fixed. Next apply KVL around a loop. I'll do the loop 1:

$$-I_1 R_1 - I_1 R_2 + I_2 R_2 - V = 0 $$

That is the equation for loop one. Notice that the where the loop currents "mesh" the contributions of voltage drop in resistor 2 are "superimposed"

Now you try loop 2 equation.

So it's uhhh...

$$ V - I_2R_2 - I_2R_3 - I_1R_2 = 0 $$

 

[erobz]

So it's uhhh...

$$ V - I_2R_2 - I_3R_3 - I_3R_2 = 0 $$

It should be involving $I_1,I_2$ there is no loop current $I_3$. Think about how I got each of the terms in the equation for loop 1. Can you see where they come from?

 

[eleking16]

It should be involving $I_1,I_2$ there is no loop current $I_3$.

Like that? I edited my previous message
$V - I_2R_2 - I_2R_3 - I_1R_2 = 0$

 

[erobz]

Like that? I edited my previous message
$V - I_2R_2 - I_2R_3 - I_1R_2 = 0$

Very close, but check your signs. Notice the polarities across the resistor in the middle. They are different with respect to the direction of each loop current passing through it.

 

[eleking16]

Oh wait, hm...
$V - I_2R_2 - I_2R_3 + I_1R_2 = 0$

Closer, but check your signs.

 

[erobz]

Oh wait, hm...
$V - I_2R_2 - I_2R_3 + I_1R_2 = 0$

Yes, that's it. So you are left with two equations in two unknowns for loop currents $I_1,I_2$. you solve, and then realize that where in the circuit $I_1$ and $I_2$ mesh ( the branch containing $R_2$), the loop currents must be superimposed.

 

[eleking16]

Yes, that's it. So you are left with two equations in two unknowns for loop currents $I_1,I_2$

And then, I'm just doing the math... That's it.
How about the 3 loop? Is it the same? I think someone suggested me to break the circuit into 2 parts of circuit. One with e1 and the other with the other components such as e2 until e4. The problem is, from your example, it was given the fixated polar of each voltage source and resistance. But my question doesn't even have the polar sign (for resistance). How do we sort it out?

 

[erobz]

And then, I'm just doing the math... That's it.
How about the 3 loop? Is it the same? I think someone suggested me to break the circuit into 2 parts of circuit. One with e1 and the other with the other components such as e2 until e4.

Is there some confusion? Thats a different problem.

You said you weren't understanding "Mesh Analysis", so I was giving you a simple example to explore/ understand before you go back to your original problem with the newly learned method.

The problem is, from your example, it was given the fixated polar of each voltage source and resistance. But my question doesn't even have the polar sign (for resistance). How do we sort it out?

The voltage drops across a resistor by convention So you imagine the currents going around the loop in the directions you chose for loop currents ( clockwise, counterclockwise ) and assign polarities for resistive elements with that in mind.

 

[eleking16]

Is there some confusion? Thats a different problem.

You said you weren't understanding "Mesh Analysis", so I was giving you a simple example to explore/ understand before you go back to your original problem with the newly learned method.

Oh, I see. I miscaught the direct intention you gave me. My bad . But, really though... With mesh analysis, I feel like what I need to search is technically much less than what I was taught. Instead of having 6 different currents, I have 3 currents. That's amazing!

 

[erobz]

Oh, I see. I miscaught the direct intention you gave me. My bad . But, really though... With mesh analysis, I feel like what I need to search is technically much less than what I was taught. Instead of having 6 different currents, I have 3 currents. That's amazing!

Well, just apply with care. Likely less variables to deal with algebraically, but still some thought required. Just don't forget that you directly find "loop currents" - a theoretical construct in this technique. Your question is asking for branch currents, so don't forget to mesh the loop currents if elements share loop currents.

 

[eleking16]

Well, just apply with care. Less variables to deal with algebraically, but still some thought required.

I'll try to implant this new idea with another 3 loop problem tomorrow morning, and I hope it works! :D

 

[haruspex]

View attachment 357657
This?

No, both diagrams should have all the wires and resistors. All you need to do is change selected cells to be just a wire (or, equivalently, set their voltages to zero). In one diagram it is $\epsilon_1$, in the other it is all the others.
In the first of those, identify the symmetry and think about what it implies.

 

[eleking16]

No, both diagrams should have all the wires and resistors. All you need to do is change selected cells to be just a wire (or, equivalently, set their voltages to zero). In one diagram it is $\epsilon_1$, in the other it is all the others.
In the first of those, identify the symmetry and think about what it implies.

Seems like I can't visualize it, can you draw it for me?

 

[SammyS]

Alright, uhm... I actually have sent you the reply, but it's not been sent because it needs the moderator's approval... So, I actually changed the second loop equation because it was wrong. The other 2 loops are correct, tho.


The equation I got was:
Loop 1: $$ 5I_1 + 4I_2 - 4I_3 = 11 $$
Loop 2: $$ -6I_1 + 11I_2 + I_3 = 11 $$
Loop 3: $$ 6I_1 + I_2 + 11I_3 = 11 $$

The problem with both the with both Loop 2 and Loop 3 equations is that $\mathscr E_2= \mathscr E_3= 3$ volts and $\mathscr E_4= 7$ volts .

You should have the following.

Loop 2: $\displaystyle \quad -6I_1 + 11I_2 + I_3 = 10$

Loop 3: $\displaystyle \quad 6I_1 + I_2 + 11I_3 = 10$

 

[eleking16]

The problem with both the with both Loop 2 and Loop 3 equations is that $\mathscr E_2= \mathscr E_3= 3$ volts and $\mathscr E_4= 7$ volts .

You should have the following.

Loop 2: $\displaystyle \quad -6I_1 + 11I_2 + I_3 = 10$

Loop 3: $\displaystyle \quad 6I_1 + I_2 + 11I_3 = 10$

Thanks for the correction, I realized and fixed it after I posted it. Then, I did the math

 

[haruspex]

Since you have solved it, I'll post my solution.

First, collapse $R_5, R_6$ into one resistance of 6Ω. Likewise 7 and 8.

Just taking the $\epsilon_1$ source, there is a left-right symmetry in that $R_2=R_3$ and $R_5+R_6=R_7+R_8$. This means there is no current through $R_4$. This leaves us with $R_2+R_3$ in parallel with $R_5+R_6+R_7+R_8$, and all of that in series with $R_1$. We can easily solve that to find the currents due to the $\epsilon_1$ source.

Now dropping the $\epsilon_1$ source and instead including the rest, we can use the symmetry to say that there are equal and opposite currents around the left and right lower loops, with no current through $R_1$.

Having found the currents due to the various sources we can just sum them in each resistor.

 

[eleking16]

Since you have solved it, I'll post my solution.

First, collapse $R_5, R_6$ into one resistance of 6Ω. Likewise 7 and 8.

Just taking the $\epsilon_1$ source, there is a left-right symmetry in that $R_2=R_3$ and $R_5+R_6=R_7+R_8$. This means there is no current through $R_4$. This leaves us with $R_2+R_3$ in parallel with $R_5+R_6+R_7+R_8$, and all of that in series with $R_1$. We can easily solve that to find the currents due to the $\epsilon_1$ source.

Now dropping the $\epsilon_1$ source and instead including the rest, we can use the symmetry to say that there are equal and opposite currents around the left and right lower loops, with no current through $R_1$.

Having found the currents due to the various sources we can just sum them in each resistor.

So, I need to;

1. Make the equation from one circuit which only has epsilon_1 as the voltage source while R4 doesn't have any current going through.

2. Make the equation from one circuit which only has R2 until R8, and the remaining epsilons (e2, e3, and e4).

Is that it?

 

[haruspex]

So, I need to;

1. Make the equation from one circuit which only has epsilon_1 as the voltage source while R4 doesn't have any current going through.

Yes, but do you see why $R_4$ would have no current?

2. Make the equation from one circuit which only has R2 until R8, and the remaining epsilons (e2, e3, and e4).

Yes, but because of the symmetry you can split it into two identical circuits, just mirror images. One has $\epsilon_2, \epsilon_4, R_2, R_5, R_6$ and a half of $R_4$, while the other has $\epsilon_3, \epsilon_4, R_3, R_7, R_8$ and the other half of $R_4$.
When I write that $R_4$ is split into two halves, I mean each will have twice the resistance of$R_4$. This is just the parallel resistors formula applied in reverse.
Note that each copy of $\epsilon_4$ has the same voltage as the original, not halved or doubled. Again, do you see why?