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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 IV : UTILIZATION AND TRACTION
// CHAPTER 2: HEATING AND WELDING
// EXAMPLE : 2.6 :
// Page number 732
clear ; clc ; close ; // Clear the work space and console
// Given data
l = 10.0 // Length of material(cm)
b = 10.0 // Breadth of material(cm)
t = 3.0 // Thickness of material(cm)
f = 20.0*10**6 // Frequency(Hz)
P = 400.0 // Power absorbed(W)
e_r = 5.0 // Relative permittivity
PF = 0.05 // Power factor
// Calculations
e_0 = 8.854*10**-12 // Absolute permittivity
A = l*b*10**-4 // Area(Sq.m)
C = e_0*e_r*A/(t/100) // Capacitace of parallel plate condenser(F)
X_c = 1.0/(2*%pi*f*C) // Reactance of condenser(ohm)
phi = acosd(PF) // Φ(°)
R = X_c*tand(phi) // Resistance of condenser(ohm)
V = (P*R)**0.5 // Voltage necessary for heating(V)
I_c = V/X_c // Current flowing in the material(A)
// Results
disp("PART IV - EXAMPLE : 2.6 : SOLUTION :-")
printf("\nVoltage necessary for heating, V = %.f V", V)
printf("\nCurrent flowing in the material, I_c = %.2f A\n", I_c)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook")
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