Equation Solving Worksheet for EE 188

General Description: Use of various techniques for solving equations. Other skills involved include plotting, putting equations into matrix form and manipulation of equations. Applications in EE188 include solving circuit equations, especially nodal and mesh equations.

1. Given two equations in two unknowns obtained from KCL or KVL, review various techniques for grouping terms and simplifying these equations.

Graphical method – plot the two equations by putting them both into the slope intercept form and estimate the solution from the graph. For example, using nodal analysis on the following circuit:

The nodal equations at V1 and V2 are the following:

(40-V1)/10000 = V1/15000 + (V1-V2)/18000 and (V1-V2)/18000 + .02 = V2/12000

Manipulate both equations until you get the slope intercept form, V2 = M V1+B. ∙

Graph the two equations on the graph below:

Check your plot. Did you label each axis? Did you title the plot? Did you remember that the axes are not at the edge of the graph, but go through (0,0)? What were the coordinates of the intersection of the two graphs (the solution)? (V1,V2) =(__16.4___,__6.54_______)

Substitution method – reduce one equation so that V2=function(V1). This is the same equation you would get by manipulating the equation into the slope intercept form. Substitute the function into the second equation for V2 and then solve the resulting equation for V1.

V1 = _______16.3_____________

Then substitute that value in either original equation to get V2. To check your work, you should substitute into both equations to be sure you get the same value and that it is close to the value obtained graphically.

6.4

V2 = ___________________

Linear combination method – multiply each side of one equation by a factor so that when that modified equation is added to the second equation, one of the variable terms subtract out. For instance, if you have 20 V1-5 V2=360 volts and -2 V1+5 V2=720 volts, you could multiply the ∙ ∙ ∙ ∙second equation by 10, to get -20 V1+50 V2=7200 volts to add to the first equation. When ∙ ∙added together, the two equations become one with one variable. They would add to be (20-20) V1+(50-5) V2=7200+360 volts, or 45 V2=7560 volts, so V2=168 volts. As before, substitute ∙ ∙ ∙V2=168 volts into either original equation to get V1. To check your work, you should substitute into both equations to be sure you get the same value and that it is close to the value obtained graphically.

2. Given three or more equations obtained from KCL or KVL.

Substitution method - reduce one equation so that Variable1=function(Variable2,Variable3). Substitute that function in for Variable1 in the other two original equations. Simplify those equations and then follow the same methods given in #1 above, since you now have two equations in two unknowns. The three mesh equations would be:

-12 V+i1 4Ω+(i1-i2) 2Ω = 0 V∙ ∙

(i2-i1) 2Ω+i2 8Ω+(i2-i3) 2Ω+8 V=0 V∙ ∙ ∙

and (i3-i2) 2Ω+i3 5Ω+20 V = 0 V∙ ∙

Hint: Solve for i2 in the first equation and plug that expression in for i2 in the other equations.

Linear combination method – using the method described above, eliminate one variable from equations 1 and 2 and the same variable from equations 1 and 3. Then use the method above since you now have 2 equations in 2 unknowns.

If you multiply the entire 1st equation by 6 and add it to the second equation, the i2 terms should subtract out. What is the resulting equation?

24i1-12i3=64

If you multiply the entire 1st equation by -1 and add it to the third equation, the i2 terms should again subtract out. What is the resulting equation?

32i1-2i3=64

Now solve the two new equations by the methods discussed earlier.

1i=1.9 and i3=1.52

Matrix method – You can put the three equations into matrix form and then solve by using matrix inversion, premultiplying both sides by the inverse. To get the equations ready to put into matrix form, they must be of the form: i1 A + i2 B + i3 C = D∙ ∙ ∙

Manipulate the first equation into the above form. (Group terms and put constants on right side)

Manipulate the second equation into the above form.

Manipulate the third equation into the above form.

The three equations can be expressed in matrix form as follows:

A1 B1 C1 i1 D1

A2 B2 C2 i2 = D2 ∙ or R I = D∙

A3 B3 C3 i3 D3

This can be solved if you can find the inverse of R, or R-1, since R-1 R = Identity Matrix∙

So premultiplying both sides by the inverse, R-1 R I = R∙ ∙ -1 D∙ yields I = R-1 D∙

For this example, you should have obtained the following three equations:

i1 6Ω + i2 (-2Ω) + i3 0Ω = 12 V∙ ∙ ∙

i1 (-2Ω) + i2 12Ω + i3 (-2Ω) =-8 V and∙ ∙ ∙

i1 0Ω + i2 (-2Ω) + i3 7Ω = -20 V∙ ∙ ∙

Expressed in matrix form as:

6 -2 0 i1 12

-2 12 -2 i2 = -8 ∙ or R I = D∙

0 -2 7 i3 -20

Finding the inverse of R

.1770 .0310 .0088 1 0 0 i1

R-1 = .0310 .0929 .0265 and R-1 R∙ = 0 1 0 and R-1 R I = I = ∙ ∙ i2

.0088 .0265 .1504 0 0 1 i3

So I = R-1 D∙

Multiplying the inverse of R times D yields each of the currents in amps

.1770 .0310 .0088 12 1.6991

I = R-1 ∙ D= .0310 .0929 .0265 -8 = -.9027 Amps∙

.0088 .0265 .1504 -20 -3.115