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I have this system of equations: \begin{cases} I_1 = I_2 + I_3 \\ \epsilon_1 - I_1(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - I_1(R_1 + R_2) - I_3(R_4 + R_5) + \epsilon_2 = 0 \end{cases} I want to solve it for $I_1, I_2$ and $I_3$ (so that neither is expressed in function of the other).

This is what I did so far:

From the first equation we have that $I_2 = I_1 - I_3$. Substitute that into the second equation for $I_2$ to get \begin{align*} \epsilon_1 - I_1(R_1 + R_2) - (I_1 - I_3) R_3 = 0. \end{align*} Substracting the third equation from that one gives \begin{align*} -(I_1 - I_3) R_3 + I_3(R_4 + R_5) - \epsilon_2 = 0. \end{align*} Then I'm not sure what to do, I still have that $I_1$ which I want to remove.

Any help?

Edit: Another try, making use of a given hint. Substituting $I_1 = I_2 + I_3$ into equation two and three gives \begin{cases} \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_3(R_4+R_5) + \epsilon_2 = 0. \end{cases} After distribution this becomes \begin{cases} \epsilon_1 - I_2 R_1 - I_2 R_2 - I_3 R_1 - I_3 R_2 - I_2 R_3 = 0 \\ \epsilon_1 - I_2 R_1 - I_2 R_2 - I_3 R_1 - I_3 R_2 - I_3 R_4 - I_3 R_5 + \epsilon_2 = 0. \end{cases} Substracting the second from the first gives \begin{align*} -I_2 R_3 + I_3 R_4 + I_3 R_5 + \epsilon_2 = 0, \end{align*} which has still the $I_2$ factor in it =(

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  • $\begingroup$ Are you allowed to use a calculator, like a TI83/84? With matrices, this will go a lot easier. $\endgroup$
    – imranfat
    Mar 5, 2015 at 18:25
  • $\begingroup$ No, it's for my electromagnetism class. We can't use calculators during the exam, and I might get a similar problem so I need to know how to solve it. $\endgroup$
    – Kamil
    Mar 5, 2015 at 18:27
  • $\begingroup$ Oh, ok, yeah in that case, the answer below may give a clue. $\endgroup$
    – imranfat
    Mar 5, 2015 at 18:28

2 Answers 2

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If you substitue $I_2+I_3$ for $I_1$ in the second and third equations, then you have two equations in two variables, $I_2$ and $I_3$. Do the simplifications, then do a similar thing to reduce it to one unknown quantity.

Later update in response to comments: You have $$ \begin{cases} \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_3(R_4+R_5) + \epsilon_2 = 0. \end{cases} $$ Collecting all of the instances of $I_2$ into one term and all of those of $I_3$ into one term, we get $$ \begin{cases} \epsilon_1 - I_2(R_1+R_2+R_3) - I_3(R_1+R_2) = 0 \\ \epsilon_1 - I_2(R_1+R_2) - I_3(R_1+R_2+R_4+R_5) +\epsilon_2 = 0 \end{cases} $$ If you multiply both sides of the first of these by $R_1+R_2$ and both sides of the second by $R_1+R_2+R_3$ and then subtract, then you eliminate $I_2$ from the system, and you're left with only one variable to solve for, namely $I_3$. When you've done that, you can plug in the solution for $R_3$ in place of $I_3$ in one of the two equations above and solve it for $I_2$. Then you can find $I_1$ since you've shown it's the sum $I_1+I_2$. Notice that $(R_1+R_2)\epsilon_1 - (R_1+R_2+R_3)\epsilon_1$, which you get when you subtract, simplifies to $-R_3\epsilon_1$.

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  • $\begingroup$ Ok, I just tried that, and it seems like that complicates the matters even more. You want me to edit my post and write what I found? $\endgroup$
    – Kamil
    Mar 5, 2015 at 18:34
  • $\begingroup$ Perhaps. After you do that substitution you should be able to do routine simplifications so that each of the two equations has just one occurrence of the variable $I_2$ and just one occurrence of $I_3$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Mar 5, 2015 at 18:35
  • $\begingroup$ What do you mean with routine simplifications? Can you check my post? I edited it at the bottom. $\endgroup$
    – Kamil
    Mar 5, 2015 at 18:45
  • $\begingroup$ Did you mean, multiply first side by $(R_1 + R_2)$? Because you said $(R_2 + R_2)$. $\endgroup$
    – Kamil
    Mar 6, 2015 at 11:06
  • $\begingroup$ ^ That was a typo; I've fixed it. $\endgroup$
    – Michael Hardy
    Mar 6, 2015 at 13:15
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This is a circuit system of equations. Be sure to utilize Kirchhoff's Current Law and Ohm's Law to obtain additional equations to allow you to solve more easily. If you want to solve for just the I's in terms of R and E, then organize equations with I's as the coefficients.

Then solve system of 3 equations for the 3 unknowns.

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