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+//chapter1
+//example1.8
+//page21
+
+R=8 // ohm
+R1=10 // ohm
+R2=20 // ohm
+R3=12 // ohm
+//removing 100 ohm resistance, we form linear equations by assuming currents I1 through loop1 and I2 through loop2
+
+//100=10*I1+20*(I1-I2)
+//0=(12+8)*I2+20*(I2-I1)
+
+//thus we get the following linear equations
+
+//30*I1-20*I2=100
+//-20*I1+40*I2=0
+//solving these equations
+
+a=[30 -20;-20 40]
+b=[100;0]
+x=inv(a)*b // matrix of I1 and I2
+
+I2=x(2,1) // current through 8 ohm resistor
+
+E0=I2*R
+printf("voltage across AB with 100 ohm resistance not connected = %.3f V \n",E0)
+
+R_equi=(R1*R2/(R1+R2))+R3
+R0=R_equi*R/(R_equi+R)
+printf("resistance between AB with 100 ohm removed and voltage source shorted = %.3f ohm \n",R0)
+
+I=E0/(R0+Rl)
+printf("current through 100 ohm resistor = %.3f A",I)
+