# Signs when analysing simple circuits

#### EobardThawne

Joined Apr 30, 2017
8
I'm having trouble finding out which sign to use when analysing circuits. I try to be very methodical, but if I analyse the exact same circuits that were analysed in class, but in another direction (for instance, analysing clockwise what was analysed counter-clockwise in class) I arrive to different results. I'm uploading an example: Here, I tried to analyse the right-hand mesh clockwise, and arrived to two different equations than my teacher, who analysed it counter-clockwise.

R1 = 4 Ohm
R2 = 5 Ohm
R3 = 16 Ohm
E1 = 6 V
E2 = 2 V

The upper diagram and equations are mine; the lower are my teacher's. All signs and loops (mine in red, those of my teacher in blue) weren't given in the letter of the exercise.

It obviously has something to do with the resistance in the middle. I don't see how it's possible, as they say, to analyse a circuit in different directions and arrive at the same results. This example I'm showing isn't even the only one where I had the exact same problem.

#### Attachments

• 415.6 KB Views: 9

#### WBahn

Joined Mar 31, 2012
26,148
What different results do you get?

You won't get identical equations because each set of equations is fundamentally tied to the directions assigned to the currents. But at the end of the day when if the first approach says that the current in R3 is 10 mA (just making that up) flowing downward, then the other approach must also say that the current in R3 is 10 mA flowing downward.

You are probably getting hung up because you have the same variable names, i1 and i2, in both sets of equations. But they are not the same variables.

Replace the second equations with the variables i3 and i4.

Then see if the resulting equations are compatible with the first by noting that i1 = i3 while i2 = -i4. Also keep in mind that you can always multiply both sides of an equation by -1 without affecting equality.

• EobardThawne

#### EobardThawne

Joined Apr 30, 2017
8
What different results do you get?

You won't get identical equations because each set of equations is fundamentally tied to the directions assigned to the currents. But at the end of the day when if the first approach says that the current in R3 is 10 mA (just making that up) flowing downward, then the other approach must also say that the current in R3 is 10 mA flowing downward.

You are probably getting hung up because you have the same variable names, i1 and i2, in both sets of equations. But they are not the same variables.

Replace the second equations with the variables i3 and i4.

Then see if the resulting equations are compatible with the first by noting that i1 = i3 while i2 = -i4. Also keep in mind that you can always multiply both sides of an equation by -1 without affecting equality.
Thanks for answering. When I say they're different equations, I mean that some signs are the same, others aren't. I know if all signs were opposite with the values being equal, then the equations would actually be the same. Besides, I just found out what I was doing wrong in that exercise, so it isn't a good example of my confusion anymore, unfortunately. I was not only taking a clockwise direction on both meshes, but I was also taking a different convention for each one of them! (As in, in the first one I was considering the conventional electric current, while in the second one I was considering the current flowing from negative to positive terminal).

This other exercise has me super puzzled, though. If you could help me telling me which one of the two equations for mesh [M1] is correct, I'd really appreciate it. I post the document with the details of the letter and the two "possible" M1 equations. One of those equations is the one I arrived at, even after analysing the circuit with different approaches, while the other one is the one the teacher arrived at in class. I truly think he made a mistake, but before I move on to other exercises, I'd like to be sure I wasn't the one who made the mistake. Notice that both equations are essentially the same (signs are opposite for E1, E3 and voltage in R1, but, as we've talked about before, that doesn't make for different equations), but the real difference is in R3 (i1 + i2) vs R3 (i2 - i1).

#### Attachments

• 136.4 KB Views: 5

#### RBR1317

Joined Nov 13, 2010
563
Without knowing the direction of the mesh currents, one can only guess at the correctness of your mesh equation. Firstly, the direction of mesh current is unconcerned with the direction of conventional electric current. Secondly, the simplicity and consistency of having all mesh currents circulate in the same direction is the overriding consideration. That way, the mesh current for any mesh will always be positive in the equation for that mesh, while all other mesh currents will be negative. This makes it easy to know whether the current signs are correct. Otherwise there will be confusion and errors. Of course, you can make any set of random mesh current directions give the correct answer, but who really wants to work that hard?

• EobardThawne

#### WBahn

Joined Mar 31, 2012
26,148
Thanks for answering. When I say they're different equations, I mean that some signs are the same, others aren't. I know if all signs were opposite with the values being equal, then the equations would actually be the same.
You are clearly still looking at both sets of equations insisting that i1 and i2 mean the same thing in both equations. They don't!

Use i1 and i2 as you have defined them in the top attempt. You have

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
-$$16 \Omega$$ i_1 \; + \; $$21 \Omega$$ i_2 \; = \; -2V
\)

Now, in the second set of equations, use i3 and i4 instead of i1 and i2, keeping everything else the same. You have

$$\( 20 \Omega$$ i_3 \; + \; $$16 \Omega$$ i_4 \; = \; 6V
$$16 \Omega$$ i_3 \; + \; $$21 \Omega$$ i_4 \; = \; 2V
\)

Now see what happens when you convert the second set from i3,i4 to i1,i2 using the relationships

$$i_3 \; = \; i_1 i_4 \; = \; -i_2$$

You get

$$\( 20 \Omega$$ $$i_1$$ \; + \; $$16 \Omega$$ $$-i_2$$ \; = \; 6V
$$16 \Omega$$ $$i_1$$ \; + \; $$21 \Omega$$ $$-i_2$$\; = \; 2V
\)

which becomes

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
$$16 \Omega$$ i_1 \; - \; $$21 \Omega$$ i_2 \; = \; 2V
\)

Now multiply both sides of the bottom equation by -1 and you get

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
-$$16 \Omega$$ i_1 \; + \; $$21 \Omega$$ i_2 \; = \; -2V
\)

Look familiar?

Besides, I just found out what I was doing wrong in that exercise, so it isn't a good example of my confusion anymore, unfortunately. I was not only taking a clockwise direction on both meshes, but I was also taking a different convention for each one of them! (As in, in the first one I was considering the conventional electric current, while in the second one I was considering the current flowing from negative to positive terminal).
I think you are confusing yourself. Look at your work. In the top one you defined both mesh currents as flowing clockwise. Because you marked the voltage across each resistor as being positive on the side that the current enters, you are using conventional current. You did EXACTLY the same thing in the bottom, except you happened to define i2 as going counter-clockwise. But you can flip a coin as to which direction each mesh current is flowing. You can even (but I don't suggest you do) flip a coin and define some of the currents to be conventional and some to be electron flow. As long as you define the voltage drops properly according to each current's definition, everything will work out properly. But in your case, you defined all of your currents as being conventional current.

This other exercise has me super puzzled, though. If you could help me telling me which one of the two equations for mesh [M1] is correct, I'd really appreciate it. I post the document with the details of the letter and the two "possible" M1 equations. One of those equations is the one I arrived at, even after analysing the circuit with different approaches, while the other one is the one the teacher arrived at in class. I truly think he made a mistake, but before I move on to other exercises, I'd like to be sure I wasn't the one who made the mistake. Notice that both equations are essentially the same (signs are opposite for E1, E3 and voltage in R1, but, as we've talked about before, that doesn't make for different equations), but the real difference is in R3 (i1 + i2) vs R3 (i2 - i1).
It's impossible to tell which, if either, is correct. Not only do you not define which direction each current is going, you don't even define which mesh current is i1 and which is i2. You MUST properly define all of the components used in all of your equations, otherwise your work becomes meaningless.[/tex][/tex]

• EobardThawne

#### EobardThawne

Joined Apr 30, 2017
8
Without knowing the direction of the mesh currents, one can only guess at the correctness of your mesh equation. Firstly, the direction of mesh current is unconcerned with the direction of conventional electric current. Secondly, the simplicity and consistency of having all mesh currents circulate in the same direction is the overriding consideration. That way, the mesh current for any mesh will always be positive in the equation for that mesh, while all other mesh currents will be negative. This makes it easy to know whether the current signs are correct. Otherwise there will be confusion and errors. Of course, you can make any set of random mesh current directions give the correct answer, but who really wants to work that hard?
Thanks for the answer. From now on, I think I'll go with conventional current and clockwise direction every time

You are clearly still looking at both sets of equations insisting that i1 and i2 mean the same thing in both equations. They don't!

Use i1 and i2 as you have defined them in the top attempt. You have

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
-$$16 \Omega$$ i_1 \; + \; $$21 \Omega$$ i_2 \; = \; -2V
\)

Now, in the second set of equations, use i3 and i4 instead of i1 and i2, keeping everything else the same. You have

$$\( 20 \Omega$$ i_3 \; + \; $$16 \Omega$$ i_4 \; = \; 6V
$$16 \Omega$$ i_3 \; + \; $$21 \Omega$$ i_4 \; = \; 2V
\)

Now see what happens when you convert the second set from i3,i4 to i1,i2 using the relationships

$$i_3 \; = \; i_1 i_4 \; = \; -i_2$$

You get

$$\( 20 \Omega$$ $$i_1$$ \; + \; $$16 \Omega$$ $$-i_2$$ \; = \; 6V
$$16 \Omega$$ $$i_1$$ \; + \; $$21 \Omega$$ $$-i_2$$\; = \; 2V
\)

which becomes

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
$$16 \Omega$$ i_1 \; - \; $$21 \Omega$$ i_2 \; = \; 2V
\)

Now multiply both sides of the bottom equation by -1 and you get

$$\( 20 \Omega$$ i_1 \; - \; $$16 \Omega$$ i_2 \; = \; 6V
-$$16 \Omega$$ i_1 \; + \; $$21 \Omega$$ i_2 \; = \; -2V
\)

Look familiar?

I think you are confusing yourself. Look at your work. In the top one you defined both mesh currents as flowing clockwise. Because you marked the voltage across each resistor as being positive on the side that the current enters, you are using conventional current. You did EXACTLY the same thing in the bottom, except you happened to define i2 as going counter-clockwise. But you can flip a coin as to which direction each mesh current is flowing. You can even (but I don't suggest you do) flip a coin and define some of the currents to be conventional and some to be electron flow. As long as you define the voltage drops properly according to each current's definition, everything will work out properly. But in your case, you defined all of your currents as being conventional current.

It's impossible to tell which, if either, is correct. Not only do you not define which direction each current is going, you don't even define which mesh current is i1 and which is i2. You MUST properly define all of the components used in all of your equations, otherwise your work becomes meaningless.[/tex][/tex]
Wow, thanks a lot for the thorough answer! Now I understand. In my second example, the different thing was the same, now I notice, so your answer also relieved me of that doubt. Thank you so much!!!