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+//Problem 35.10: An ac. source of 30/_0° V and internal resistance 20 kohm is matched to a load by a 20:1 ideal transformer. Determine for maximum power transfer (a) the value of the load resistance, and (b) the power dissipated in the load.
+
+//initializing the variables:
+rv = 30; // in volts
+thetav = 0; // in degrees
+r = 20000; // in ohms
+tr = 20; // turn ratio
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180) 
+//The network diagram is shown in Figure 35.13.
+//For maximum power transfer, r1 must be equal to
+r1 = r
+//load resistance RL
+RL = r1/tr^2
+//The total input resistance when the source is connected to the matching transformer is
+RT = r + r1
+//Primary current
+I1 = V/RT
+//N1/N2 = I2/I1
+I2 = I1*tr
+//Power dissipated in load resistance RL is given by
+P = RL*I2^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the value of the load resistance is %.0f ohm",RL)
+printf("\n (b) Power dissipated in the load resistance is %.2E W",P)
+\ No newline at end of file