Ok, so using those equations I get I1=1.74A, I2=-1.42A, I3=0.32A Which are not the correct values. I officially quit this question. It is impossible.
Finding Current with Kirchoffs Laws
- Thread starter hogesjzz30
- Start date
Yes, 1.74 + (-1.42) = 0.32....What are you talking about?
if ....I3 = I1 + I2and....
I1=1.74A, I2=-1.42A, I3=0.32AIs that equation true or not?
I'm going to kill you; you made me do the whole exercise... the initial equations I just gave you are right!!!
Your result is wrong, even though it adds up. It's what you have on the last pic of your OP (I1 = -2.56, I2 = 0.53, and I3 = -2.03). You made a mistake somewhere along the way....
These are the initial equations including the internal resistors and their currents:
Your result is wrong, even though it adds up. It's what you have on the last pic of your OP (I1 = -2.56, I2 = 0.53, and I3 = -2.03). You made a mistake somewhere along the way....
These are the initial equations including the internal resistors and their currents:
E1 + E2 = - (I1*r1) - (I1*R4) + (I2*R2) + (I2*r2) - (I1*R1)
-E2 + E3 = - (I2*r2) - (I2*R2) - (I3*R3) - (I3*r3)
Good luck... Now I'm definitely going to sleep.-E2 + E3 = - (I2*r2) - (I2*R2) - (I3*R3) - (I3*r3)
Last edited:
Thanks very much for your help with that, sorry to be such a pain/dumbass, I went back and started from scratch again and it finally came out to the correct answer! I hope I never have to do another one of those stupid questions ever again (I'm doing mechanical engineering so here's hoping I dont!)
Glad you finally got the answer, though it seemed to be much more of a struggle than it should have been. Whether you go on with any EE or not, rest assured that you will have the same math in mechanics since mechanical systems are governed by the same types of conservation laws and constituitive equations.
Here is a piece of advice I think you are in a very good position to recognize the value of, having gone through what you just did. Consider the amount of time and effort represented by just the work you included in your initial post (and thank you very, very much for including YOUR effort to solve YOUR problem -- that is something we see all too rarely around here). Now what if I told you that you should have caught that your very first equation was wrong and would have caught it if you had adopted a philosophy of always asking if the answer makes sense as you proceed through your work? Because you didn't, everything you did after that point was wasted time and effort, because you were guaranteed not to get the correct answer.
Here are two checks you should have done after writing your first loop equation.
(1) Does it make sense that the voltage sources have different signs?
A quick glance at the circuit shows that as you go around the loop you are crossing each source in the same direction, therefore how can they possibly have different signs in the resulting loop equation?
(2) Does it make sense that I1 and I2 should have the same sign?
As you go around the loop, you are going in the direction of one of these currents and in the opposite direction of the other, therefore they have to have opposite signs.
So your very first equation violates BOTH common sense sanity checks and thus has at least two errors.
We humans tend to be pretty bad at math. We make lots of mistakes. I know I certainly do. So simple prudence dictates that we adopt effective error detection strategies that we can employ as we work through a problem to catch errors as soon after they are made as we can so that we can correct them when doing so is simple and so that we don't invest a lot of time and effort going down blind alleys.
Let me walk you through two ways to tackle the problem. The first isn't going to be the most efficient, but it is perhaps the easier to understand the physics of.
The first one that popped into my head is to look to the fact that we have three paths that can take us from Pt E to Pt C. We know that the voltage between two points is independent of the path taken, so that gives us a way to get three equations in the three unknowns -- we simply write the equations for V_ec along each path.
The key is to be sure that you understand when you have a voltage rise and when you have a voltage drop. As you go from one side of a resistor to the other, you have a voltage rise if the current is flowing toward you and a voltage drop if it is flowing with you. As you go from one side of a voltage souirce, you have a voltage rise if you go from the negative terminal to the positive terminal and a voltage drop if you go from positive to negative.
Finally, you have to decide if you are going to sum up voltage rises or voltage drops along the path. If you want V_ec, which is the voltage at Pt E relative to the voltage at Pt C, then you can either start from Pt C and sum up the voltage rises until you get to Pt E, or you can start at Pt E and sum up the voltage drops until you get to Pt C.
Let's do the latter and go from Pt E to Pt C summing up the voltage drops. So for the top, middle, and bottom branches respectively, we get:
\(
V_{ec} \, = \, -(I_1R_1)-(E_1)-(I_1r_1)-(I_1)(R_4) \, = \, -I_1(r_1+R_1+R_4)-E_1
\
V_{ec} \, = \, (E_2)-(I_2r_2)-(I_2R_2) \, = \, -I_2(r_2+R_2)+E_2
\
V_{ec} \, = \, (E_3)+(I_3r_3)+(I_3R_3) \, = \, I_3(r_3+R_3)+E_3
\)
When I did my little sanity checks I discovered that I had put R2 instead of R4 in the left side of the first equation and that I had used R1 throughout the third equation. Part of that was due to the fact that I am working with the equations via the tex equation editor and not writing anything down on paper. But it would have messed up everything up from this point on if I hadn't asked if it made sense for an equation to include components not on that equation's path.
We have three equations but four unknowns (V_ec is the fourth). So we need another equation, which is provided by KCL:
\(
I_3 \, = \, I_1+I_2
\)
At this point the physics and electronics is done and all we have is an algebra problem of the type you will have to deal with in mechanical engineering, chemistry, even economics.
We can proceed to solve these a number of ways.
Let's eliminate V_ec from the system by setting the first two branch equations equal and then the second two branch equations equal:
\(
-I_1(r_1+R_1+R_4)-E_1 \, = \, -I_2(r_2+R_2)+E_2
\
-I_2(r_2+R_2)+E_2 \, = \, I_3(r_3+R_3)+E_3
\)
These are what your Equation 1 and Equation 2 should have been. Written in the form you have yours so that we can compare them:
\(
-E_2-E_1 \, = \, I_1(r_1+R_1+R_4)-I_2(r_2+R_2)
\
E_2-E_3 \, = \, I_2(r_2+R_2)+I_3(r_3+R_3)
\)
So your second equation is correct and the first equation just as the two sign errors that the sanity checks caught. Hopefully you can start to see how powerful this simple tool is!
Now let's substitute our KCL equation into the second equation:
\(
-E_2-E_1 \, = \, I_1(r_1+R_1+R_4)-I_2(r_2+R_2)
\
E_2-E_3 \, = \, I_2(r_2+R_2)+(I_1+I_2)(r_3+R_3) \, = \, I_1(r_3+R_3)+I_2(r_2+r_3+R_2+R_3)
\)
Since I2 has the opposite sign in the two equations, let's eliminate it by multiplying both sides of the first equation by (r2+r3+R2+R3) and both sides of the second by (r2+R2) and then adding the two equations together:
\(
(-E_2-E_1)(r_2+r_3+R_2+R_3)+(E_2-E_3)(r_2+R_2) \, = \, I_1[(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)]
\)
We can now solve for I_1 directly:
\(
I_1 \, = \, \frac{(-E_2-E_1)(r_2+r_3+R_2+R_3)+(E_2-E_3)(r_2+R_2)}{(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)}
\)
Multiplying out the numerator and simplifying gives us:
\(
I_1 \, = \, -\frac{E_1(r_2+r_3+R_2+R_3)+E_2(r_3+R_3)+E_3(r_2+R_2)}{(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)}
\)
Having solved for I1, it is now a simple matter to go back to our first three equations and find Vec from the first one and then to use the second and third to find I2 and I3.
\(
V_{ec} \, = \, -I_1(r_1+R_1+R_4)-E_1
\
I_2 \, = \, \frac{E_2-V_{ec}}{r_2+R_2}
\
I_3 \, = \, \frac{V_{ec}-E_3}{r_3+R_3}
\)
Let's go ahead and solve for I1:
\(
I_1 = -\frac{4V(0.1\Omega + 0.3\Omega + 3\Omega + 1\Omega)+2V(0.3\Omega + 1\Omega)+3V(0.1\Omega + 3\Omega)}{(0.2\Omega + 1\Omega +0.5\Omega)(0.1\Omega + 0.3\Omega + 3\Omega + 1\Omega)+(0.3\Omega + 1\Omega)(0.1\Omega + 3\Omega)} \, = \, -2.56A
\)
Which agrees with your simulation (depending on the polarity of the ammeter, which I can't discern).
Notice that the work up to that last step was done completely symbolically. This has several advantages. First, it makes it a lot easier to do effective sanity checks, which we've already seen the benefit of in both your work and my work. Second, it makes it trivial to solve the problem again using different component values, which you know you have to do in the second part of the problem.
Another advantage is that it let's us get a glimpse into the structure of the solution. But that is going beyond any reasonable scope of this discussion. If you are interested, we can show how this leads to a form of the solution that you can actually write down almost by inspection.
Here is a piece of advice I think you are in a very good position to recognize the value of, having gone through what you just did. Consider the amount of time and effort represented by just the work you included in your initial post (and thank you very, very much for including YOUR effort to solve YOUR problem -- that is something we see all too rarely around here). Now what if I told you that you should have caught that your very first equation was wrong and would have caught it if you had adopted a philosophy of always asking if the answer makes sense as you proceed through your work? Because you didn't, everything you did after that point was wasted time and effort, because you were guaranteed not to get the correct answer.
Here are two checks you should have done after writing your first loop equation.
(1) Does it make sense that the voltage sources have different signs?
A quick glance at the circuit shows that as you go around the loop you are crossing each source in the same direction, therefore how can they possibly have different signs in the resulting loop equation?
(2) Does it make sense that I1 and I2 should have the same sign?
As you go around the loop, you are going in the direction of one of these currents and in the opposite direction of the other, therefore they have to have opposite signs.
So your very first equation violates BOTH common sense sanity checks and thus has at least two errors.
We humans tend to be pretty bad at math. We make lots of mistakes. I know I certainly do. So simple prudence dictates that we adopt effective error detection strategies that we can employ as we work through a problem to catch errors as soon after they are made as we can so that we can correct them when doing so is simple and so that we don't invest a lot of time and effort going down blind alleys.
Let me walk you through two ways to tackle the problem. The first isn't going to be the most efficient, but it is perhaps the easier to understand the physics of.
The first one that popped into my head is to look to the fact that we have three paths that can take us from Pt E to Pt C. We know that the voltage between two points is independent of the path taken, so that gives us a way to get three equations in the three unknowns -- we simply write the equations for V_ec along each path.
The key is to be sure that you understand when you have a voltage rise and when you have a voltage drop. As you go from one side of a resistor to the other, you have a voltage rise if the current is flowing toward you and a voltage drop if it is flowing with you. As you go from one side of a voltage souirce, you have a voltage rise if you go from the negative terminal to the positive terminal and a voltage drop if you go from positive to negative.
Finally, you have to decide if you are going to sum up voltage rises or voltage drops along the path. If you want V_ec, which is the voltage at Pt E relative to the voltage at Pt C, then you can either start from Pt C and sum up the voltage rises until you get to Pt E, or you can start at Pt E and sum up the voltage drops until you get to Pt C.
Let's do the latter and go from Pt E to Pt C summing up the voltage drops. So for the top, middle, and bottom branches respectively, we get:
\(
V_{ec} \, = \, -(I_1R_1)-(E_1)-(I_1r_1)-(I_1)(R_4) \, = \, -I_1(r_1+R_1+R_4)-E_1
\
V_{ec} \, = \, (E_2)-(I_2r_2)-(I_2R_2) \, = \, -I_2(r_2+R_2)+E_2
\
V_{ec} \, = \, (E_3)+(I_3r_3)+(I_3R_3) \, = \, I_3(r_3+R_3)+E_3
\)
When I did my little sanity checks I discovered that I had put R2 instead of R4 in the left side of the first equation and that I had used R1 throughout the third equation. Part of that was due to the fact that I am working with the equations via the tex equation editor and not writing anything down on paper. But it would have messed up everything up from this point on if I hadn't asked if it made sense for an equation to include components not on that equation's path.
We have three equations but four unknowns (V_ec is the fourth). So we need another equation, which is provided by KCL:
\(
I_3 \, = \, I_1+I_2
\)
At this point the physics and electronics is done and all we have is an algebra problem of the type you will have to deal with in mechanical engineering, chemistry, even economics.
We can proceed to solve these a number of ways.
Let's eliminate V_ec from the system by setting the first two branch equations equal and then the second two branch equations equal:
\(
-I_1(r_1+R_1+R_4)-E_1 \, = \, -I_2(r_2+R_2)+E_2
\
-I_2(r_2+R_2)+E_2 \, = \, I_3(r_3+R_3)+E_3
\)
These are what your Equation 1 and Equation 2 should have been. Written in the form you have yours so that we can compare them:
\(
-E_2-E_1 \, = \, I_1(r_1+R_1+R_4)-I_2(r_2+R_2)
\
E_2-E_3 \, = \, I_2(r_2+R_2)+I_3(r_3+R_3)
\)
So your second equation is correct and the first equation just as the two sign errors that the sanity checks caught. Hopefully you can start to see how powerful this simple tool is!
Now let's substitute our KCL equation into the second equation:
\(
-E_2-E_1 \, = \, I_1(r_1+R_1+R_4)-I_2(r_2+R_2)
\
E_2-E_3 \, = \, I_2(r_2+R_2)+(I_1+I_2)(r_3+R_3) \, = \, I_1(r_3+R_3)+I_2(r_2+r_3+R_2+R_3)
\)
Since I2 has the opposite sign in the two equations, let's eliminate it by multiplying both sides of the first equation by (r2+r3+R2+R3) and both sides of the second by (r2+R2) and then adding the two equations together:
\(
(-E_2-E_1)(r_2+r_3+R_2+R_3)+(E_2-E_3)(r_2+R_2) \, = \, I_1[(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)]
\)
We can now solve for I_1 directly:
\(
I_1 \, = \, \frac{(-E_2-E_1)(r_2+r_3+R_2+R_3)+(E_2-E_3)(r_2+R_2)}{(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)}
\)
Multiplying out the numerator and simplifying gives us:
\(
I_1 \, = \, -\frac{E_1(r_2+r_3+R_2+R_3)+E_2(r_3+R_3)+E_3(r_2+R_2)}{(r_1+R_1+R_4)(r_2+r_3+R_2+R_3)+(r_3+R_3)(r_2+R_2)}
\)
Having solved for I1, it is now a simple matter to go back to our first three equations and find Vec from the first one and then to use the second and third to find I2 and I3.
\(
V_{ec} \, = \, -I_1(r_1+R_1+R_4)-E_1
\
I_2 \, = \, \frac{E_2-V_{ec}}{r_2+R_2}
\
I_3 \, = \, \frac{V_{ec}-E_3}{r_3+R_3}
\)
Let's go ahead and solve for I1:
\(
I_1 = -\frac{4V(0.1\Omega + 0.3\Omega + 3\Omega + 1\Omega)+2V(0.3\Omega + 1\Omega)+3V(0.1\Omega + 3\Omega)}{(0.2\Omega + 1\Omega +0.5\Omega)(0.1\Omega + 0.3\Omega + 3\Omega + 1\Omega)+(0.3\Omega + 1\Omega)(0.1\Omega + 3\Omega)} \, = \, -2.56A
\)
Which agrees with your simulation (depending on the polarity of the ammeter, which I can't discern).
Notice that the work up to that last step was done completely symbolically. This has several advantages. First, it makes it a lot easier to do effective sanity checks, which we've already seen the benefit of in both your work and my work. Second, it makes it trivial to solve the problem again using different component values, which you know you have to do in the second part of the problem.
Another advantage is that it let's us get a glimpse into the structure of the solution. But that is going beyond any reasonable scope of this discussion. If you are interested, we can show how this leads to a form of the solution that you can actually write down almost by inspection.
It's alright. If I tell you the truth, and as you may have noticed, in my first posts I also got myself a bit confused; so this was a good practice for me too.
There are some useful tricks around that are worthwhile reminding ourselves of occasionally - such as Millman's Theorem.
http://www.allaboutcircuits.com/vol_1/chpt_10/6.html
http://www.allaboutcircuits.com/vol_1/chpt_10/6.html
