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+//example1.32
+clc
+disp("Use the loop analysis")
+disp("From the current source branch,")
+disp("I3=1 A")
+disp("Applying KVL to the loops without current source we get,")
+disp("-6(I1)-4-5(I1)+5(I2)=0    i.e.    -11(I1)+5(I2)=4  ..(1)")
+disp("-5(I2)+5(I1)-6-4(I2)-4(I3)=0  i.e.    5(I1)-9(I2)=10  (2)")
+disp("Solving, we get:")
+disp("-11(I1)+5((5I1-10)/9)=4")
+disp("Therefore,    -99(I1)+25(I1)-50=36")
+i=86/(-74)
+format(7)
+disp(i,"Therefore,      I1(in A)=")
+i=((5*(-1.1621))-10)/9
+disp(i,"and,    I2(in A)=")
+disp("Current through 5ohm in specified direction is,")
+i=(-1.7567+1.1621)
+disp(i,"I(5ohm)[in A]=I2-I1= -1.7567-(-1.1621)=")
+disp("As negative, current through 5ohm flows in opposite direction to that specified in the circuit.")