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+//Problem 36.11: A supply voltage v given by
+// v = 25 + 100sinwt + 40sin(3wt + pi/6) + 20sin(5wt + pi/12) Volts
+//where w = 10000 rad/s. The voltage is applied to a series circuit comprising a 5.0 ohm resistance and a 500 μH inductance. Determine (a) an expression to represent the current flowing in the circuit, (b) the rms value of current, correct to two decimal places, and (c) the power dissipated in the circuit, correct to three significant figures.
+
+//initializing the variables:
+Vom = 25; // in volts
+V1m = 100; // in volts
+V3m = 40; // in volts
+V5m = 20; // in volts
+w1 = 10000; // fundamental
+R = 5; // in ohm
+L = 500E-6; // in Henry
+phiv1 = 0; // in rad
+phiv3 = %pi/6; // in rad
+phiv5 = %pi/12; // in rad
+
+//calculation:
+//voltage
+V1 = V1m*cos(phiv1) + %i*V1m*sin(phiv1)
+V3 = V3m*cos(phiv3) + %i*V3m*sin(phiv3)
+V5 = V5m*cos(phiv5) + %i*V5m*sin(phiv5)
+//Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by
+Iom = Vom/R
+//fundamental or first harmonic
+//inductive reactance,
+XL1 = w1*L
+//impedance at the fundamental frequency,
+Z1 = R + %i*XL1
+//Maximum current at fundamental frequency
+I1m = V1/Z1
+I1mag = (real(I1m)^2 + imag(I1m)^2)^0.5
+phii1 = atan(imag(I1m)/real(I1m))
+//Third harmonic
+XL3 = 3*XL1
+//impedance at the third harmonic frequency,
+Z3 = R + %i*XL3
+//Maximum current at third harmonic frequency
+I3m = V3/Z3
+I3mag = (real(I3m)^2 + imag(I3m)^2)^0.5
+phii3 = atan(imag(I3m)/real(I3m))
+//fifth harmonic
+XL5 = 5*XL1
+//impedance at the third harmonic frequency,
+Z5 = R + %i*XL5
+//Maximum current at third harmonic frequency
+I5m = V5/Z5
+I5mag = (real(I5m)^2 + imag(I5m)^2)^0.5
+phii5 = atan(imag(I5m)/real(I5m))
+//rms current
+Irms = (Iom^2 + (I1mag^2 + I3mag^2 + I5mag^2)/2)^0.5
+//power dissipated
+P = R*Irms^2
+
+printf("\n\n Result \n\n")
+printf("\n(b)the rms value of current is %.2f A",Irms)
+printf("\n(c)the total power dissipated %.0f W",P)
+\ No newline at end of file