1 files changed, 31 insertions, 0 deletions
diff --git a/608/CH33/EX33.12/33_12.sce b/608/CH33/EX33.12/33_12.sce
new file mode 100755
index 000000000..fbeff8302
--- /dev/null
+++ b/608/CH33/EX33.12/33_12.sce
@@ -0,0 +1,31 @@
+//Problem 33.12:For the network shown in Figure 33.60, obtain the Norton equivalent network at terminals AB. Hence determine the power dissipated in a 5 ohm resistor connected between A and B.
+
+//initializing the variables:
+rv = 20; // in volts
+thetav = 0; // in degrees
+R1 = 2; // in ohm
+R2 = 4; // in ohm
+R3 = %i*3; // in ohm
+R4 = -1*%i*3; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Terminals AB are initially short-circuited, as shown in Figure 33.61.
+//The circuit impedance Z presented to the voltage source is given by
+Z = R1 + R4*(R2 + R3)/(R2 + R3 + R4)
+//Thus current I in Figure 33.61 is given by
+I = V/Z
+Isc = ((R2 + R3)/(R2 + R3 + R4))*I
+//Removing the voltage source of Figure 33.60 gives the network Figure 33.62 of Figure 33.62. Impedance, z, ‘looking in’ at terminals AB is given by
+z = R4 + R1*(R2 + R3)/(R2 + R3 + R1)
+//The Norton equivalent network is shown in Figure 33.63.
+R = 5; // in ohms
+//Current IL
+IL = (z/(z + R))*Isc
+ILmag = (real(IL)^2 + imag(IL)^2)^0.5
+//the power dissipated in the 5 ohm resistor is
+Pr5 = R*ILmag^2
+
+printf("\n\n Result \n\n")
+printf("\n the power dissipated in the 5 ohm resistor is %.2f W", Pr5)
+\ No newline at end of file