PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 161

MATLAB Solution

>> R = [3 -2 0; -2 9 -4; 0 -4 10];

>> V = [20; -15; -10];

>> I = inv(R)*V;

>> Result = [I(1) I(2) I(3)];

>> disp(‘***********************************************’)

>> disp(‘The loop currents I1, I2 and I3(in amp) are:’);

>> disp(Result’);

>> disp(‘***********************************************’)

*************************************************

The loop currents I1, I2 and I3 (in amp) are:

6.0440

-0.9341

-1.3736

*************************************************

Example 2.18

For the circuit diagram shown in Figure 2.61

1. Write the two node equations (for nodes X and Y)


2. Arrange the result of part 1 into a matrix equation


3. Use MATLAB to solve for the two voltages (VX and VY)

ANALYTICAL Solution

1. The node equations are

For node X: I 1 = (^) ( ___^1
R 1

• ^1
  R 2
  (^) ) VX – VY or 2 = (1 + 2) VX − VY
  For node Y: I 2 = – VX + (^) ( ^1
  R 3


• ___^1
  R 1
  (^) ) VY or 3 = −VX + (3 + 1)VY
  The two simplifi ed and rearranged node equations are
  2 = 3 VX − VY
  3 = −VX = + 4 VY
  FIGURE 2.61
  Network of Example 2.18.
  XYR^1 =^1 Ω
  I 2 = 3 A
  I 1 = 2 A
  R 3 =
  3
  1


• 
  
  
  
  • −−
    R 2 = Ω
    2
    (^1) Ω
    VX VY