How Do You Solve Kirchhoff's Loop Equations for Currents I1, I2, and I3?

[StephenDoty]

My professor gave us the three equations for a circuit and said to find I1, I2, and I3 However, I cannot figure out how to solve it.

The equations:

I1 + I2 = I3

5 - 10*I1 -10 - 5*I3 = 0

5 - 5*I2 - 5*I3 - 10 = 0

Every time I try to solve it I end up with two unknown variables. How would you solve them? I am having difficulty solving systems of equations. A step by step guide would be a great help.
Thanks.
Stephen

 

[Hootenanny]

Try substituting the the first equation into the second and third equations to eliminate I3. You will then have two equations with two unknowns, which you can solve my eliminating a further variable.

Alternatively if you are comfortable with matrices you can form a matrix equation and put the coefficient matrix into REF.

 

[thomate1]

First, it is important to understand that when solving systems of equations, it is necessary to have the same number of equations as unknown variables. In this case, we have three equations and three unknown variables (I1, I2, and I3), so we should be able to solve the system.

To begin, let's rearrange the equations to put all the terms with the unknown variables on one side and all the constant terms on the other side.

I1 + I2 - I3 = 0

-10*I1 - 5*I3 = -15

-5*I2 - 5*I3 = 5

Now, we can use any method to solve the system, such as substitution or elimination. Let's use substitution by solving the first equation for I2:

I2 = I3 - I1

Now, we can substitute this expression for I2 into the second and third equations:

-10*I1 - 5*I3 = -15 becomes -10*I1 - 5*(I3 - I1) = -15

-5*I3 - 5*(I3 - I1) = 5 becomes -5*I3 - 5*(I3 - I1) = 5

Simplifying these equations, we get:

-15*I1 - 5*I3 = -15

-10*I3 + 5*I1 = 5

Now, we have two equations with two unknown variables, I1 and I3. We can solve this system using any method, such as elimination or substitution. Let's use elimination by multiplying the first equation by 2 and adding it to the second equation:

-30*I1 - 10*I3 = -30

-10*I3 + 5*I1 = 5

Adding these equations, we get:

-25*I1 = -25

Solving for I1, we get I1 = 1. Now, we can substitute this value for I1 into any of the original equations to solve for I2 and I3.

I2 = I3 - I1 = I3 - 1

-5*I3 - 5*(I3 - 1) = 5 becomes -10*I3 = 10

Solving for I3, we get I3 = -1. Substituting this value into I2 = I3 - 1, we get I2 = -2.