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+//Problem 43.16:The mutual inductor of problem 43.15 is connected to the circuit of Figure 43.19. Determine the source and load currents for (a) the windings as shown (i.e. with the dots adjacent), and (b) with one winding reversed (i.e. with the dots at opposite ends). 
+
+//initializing the variables:
+E1 = 50; // in Volts
+thetae1 = 0; // in degrees
+r = 5; // in ohm
+R1 = 20; // in ohm
+L1 = 0.2; // in Henry
+R = 8; // in ohm
+L = 0.1; // in Henry
+L2 = 0.4; // in Henry
+R2 = 25; // in ohm
+RL = 20; // in ohm
+M = 0.1; // in Henry
+f = 75/%pi; // in Hz
+
+//calculation:
+w = 2*%pi*f
+//voltage
+E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180)
+//Applying Kirchhoff’s voltage law to the primary circuit gives
+//(r + R1 + %i*w*L1 + R + %i*w*L)*I1 - (%i*w*M + R + %i*w*L)*I2 = E1
+//Applying Kirchhoff’s voltage law to the secondary circuit gives
+//-1*(%i*w*M + R + %i*w*L)*I1 + (R2 + RL + %i*w*L2 + R + %i*w*L)*I2 = 0
+//solving these two
+I2 = E1/((r + R1 + %i*w*L1 + R + %i*w*L)*(R2 + RL + %i*w*L2 + R + %i*w*L)/((%i*w*M + R + %i*w*L)) + (-1*(%i*w*M + R + %i*w*L)))
+I1 = I2*(R2 + RL + %i*w*L2 + R + %i*w*L)/(%i*w*M + R + %i*w*L)
+//reversing
+//Applying Kirchhoff’s voltage law to the primary circuit gives
+//(r + R1 + %i*w*L1 + R + %i*w*L)*I1r - (-1*%i*w*M + R + %i*w*L)*I2r = E1
+//Applying Kirchhoff’s voltage law to the secondary circuit gives
+//-1*(-1*%i*w*M + R + %i*w*L)*I1r + (R2 + RL + %i*w*L2 + R + %i*w*L)*I2r = 0
+//solving these two
+I2r = E1/((r + R1 + %i*w*L1 + R + %i*w*L)*(R2 + RL + %i*w*L2 + R + %i*w*L)/((-1*%i*w*M + R + %i*w*L)) + (-1*(-1*%i*w*M + R + %i*w*L)))
+I1r = I2r*(R2 + RL + %i*w*L2 + R + %i*w*L)/((-1*%i*w*M + R + %i*w*L))
+
+printf("\n\n Result \n\n")
+printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1))
+printf("\n load current I2 is %.2f +(%.2f)i A",real(I2), imag(I2))
+printf("\n reversed primary current I1r is %.2f +(%.2f)i A",real(I1r), imag(I1r))
+printf("\n reversed load current I2r is %.2f +(%.2f)i A",real(I2r), imag(I2r))