Kirchoff's Voltage Law and Current Law --

[shibuzgeorge]

Homework Statement

Homework Equations

Using Loop 1, 2 and 3

The Attempt at a Solution

 

[shibuzgeorge]

 

[ehild]

You have to give I1, I2, I3 in terms of the resistances R1,R2,R3 and the Emf -s ε1 and ε2. The equations for loops 1 and 2 are correct, and adding the third equation I1=I2+I3 is enough to solve for the three currents.

 

[berkeman]

Welcome to the PF.

Things look okay so far to me. You still need to plug the equation for I2 into your equation for I1, and then the equation for I3 also gets combined. The result they want you to show only has the values of the voltage sources and resistances in it.

 

[shibuzgeorge]

Welcome to the PF.

Things look okay so far to me. You still need to plug the equation for I2 into your equation for I1, and then the equation for I3 also gets combined. The result they want you to show only has the values of the voltage sources and resistances in it.

I tried so many times but just can't get it with voltages and resistances. Can you show me step by step?

 

[gneill]

You need only two of the KVL loops and the one KCL equation to solve the circuit.

Your circuit has only two independent loops. Once you've written KVL for loops 1 and 2, every component has been "touched" by one or the other or both equations, and no new information can be gained by writing another loop equation; the equation set will simply become overdetermined.

To see that this is so you might cast your three loop equations into matrix form:

If you take the determinant of the impedance matrix (the square matrix), you'll find that it is zero, and hence no unique solution can be found from them.

 

[vela]

Only two of the loop equations are useful. The third one isn't independent; if you add the equations for loop 1 and loop 2 together, you get the equation for loop 3. So solve the system consisting of two of the loop equations and the KCL equation.

At this point, it's algebra. You've likely solved this kind of problem before: 3 equations, 3 unknowns. It just doesn't look familiar because you're not using the variables x, y, and z.

 

[shibuzgeorge]

Only two of the loop equations are useful. The third one isn't independent; if you add the equations for loop 1 and loop 2 together, you get the equation for loop 3. So solve the system consisting of two of the loop equations and the KCL equation.

At this point, it's algebra. You've likely solved this kind of problem before: 3 equations, 3 unknowns. It just doesn't look familiar because you're not using the variables x, y, and z.

I can't do the algebra its just not working for me! I just need a full method step by step in how you get the final answer with voltages and resistances

 

[vela]

Can you solve a problem like the following? (Not asking you to solve it, but do you know what to do?)
\begin{align*}
2x + y &= 6 \\
-x + 2y &= 0
\end{align*}
How about this?
\begin{align*}
2x + 3y &= 2 \\
4y-2z &= 0 \\
-3x + 5z &= 10
\end{align*}

 

[shibuzgeorge]

Can you solve a problem like the following? (Not asking you to solve it, but do you know what to do?)

\begin{align*}
2x + y &= 6 \\
-x + 2y &= 0
\end{align*}

How about this?
$$\begin{align*}
2x + 3y &= 2 \\
4y-2z &= 0 \\
-3x + 5z &= 10
\end{align*}$$

WHAT?? NO IDEA! Thought simultaneous equations but no

 

[vela]

Yes, they're simultaneous equations. You've solved them before. If you don't remember how, you need to go back and review the methods.

No one here is going to work it out step by step for you because that's against forum rules. Berkeman already gave you a suggestion on how to finish the problem above, but you need to do the actual work. Don't be afraid to make mistakes. Be brave.

 

[kuruman]

Here is another strategy. You have three equations. One of them is KCL $I_1=I_2+I_3$. OK, replace all occurrences of $I_1$ with $(I_2 + I_3)$ in the two KVL equations. You will get two equations having unknowns $I_2$, $I_3$. Solve one of them to find $I_2$ in terms of $I_3$. Replace the expression for $I_2$ that you get in the other equation. This will give you a single equation with $I_3$ as the only unknown. Solve it. Then go back and find $I_2$ and finally go back and find $I_1$. There are less cumbersome ways of getting to the answer as other have suggested, but this is straightforward and easy to understand.

 

[shibuzgeorge]

Here is another strategy. You have three equations. One of them is KCL $I_1=I_2+I_3$. OK, replace all occurrences of $I_1$ with $(I_2 + I_3)$ in the two KVL equations. You will get two equations having unknowns $I_2$, $I_3$. Solve one of them to find $I_2$ in terms of $I_3$. Replace the expression for $I_2$ that you get in the other equation. This will give you a single equation with $I_3$ as the only unknown. Solve it. Then go back and find $I_2$ and finally go back and find $I_1$. There are less cumbersome ways of getting to the answer as other have suggested, but this is straightforward and easy to understand.

That is really long and I want to know a quicker way please

 

[kuruman]

You can try using the method of determinants, but I am not sure that it is quicker. In fact it will be rougher going if you are not familiar with the method.
https://www.cliffsnotes.com/study-g...tions-using-determinants-with-three-variables