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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES
+
+// EXAMPLE : 3.13 :
+// Page number 143-144
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_r = 220.0*10**3 // Receiving end voltage(V)
+Z = complex(20,100) // Impedance(ohm/phase)
+Y = %i*0.0010 // Admittance(mho)
+I_r = 300.0 // Receiving end current(A)
+PF_r = 0.9 // Lagging power factor
+
+// Calculations
+V_2 = V_r/3**0.5 // Receiving end phase voltage(V)
+I_2 = I_r*exp(%i*-acos(PF_r)) // Receiving end current(A)
+I_C2 = (Y/2)*V_2 // Capacitive current at receiving end(A)
+I = I_2+I_C2
+V_1 = V_2+I*Z // Voltage across shunt admittance at sending end(V)
+V_1kV = V_1/1000.0 // Voltage across shunt admittance at sending end(kV)
+I_C1 = (Y/2)*V_1 // Capacitive current at sending end(A)
+I_1 = I_C1+I_2 // Sending end current(A)
+
+// Results
+disp("PART II - EXAMPLE : 3.13 : SOLUTION :-")
+printf("\nSending end voltage, V_1 = %.2f∠%.2f° kV", abs(V_1kV),phasemag(V_1kV))
+printf("\nSending end current, I_1 = %.3f∠%.4f° A", abs(I_1),phasemag(I_1))