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+//chapter12
+//example12.15
+//page252
+
+R_C=2.2 // kilo ohm
+V_CC=9 // V
+gain_beta=50
+V_BE=0.3 // V
+I_C=2 // mA
+V_CE=3 // V
+
+I_B=I_C/gain_beta
+I1=10*I_B
+
+// I1=V_CC/(R1+R2) so let Rt=R1+R2 thus we get
+Rt=V_CC/I1
+
+// by Kirchoff voltage law to collector side we get
+// V_CC=I_C*R_C+V_CE+I_E*R_E and also we have I_C=I_E so
+// V_CC=I_C*R_C+V_CE+I_C*R_E so making R_E as subject we get
+R_E=((V_CC-V_CE)/I_C)-R_C // in kilo ohm
+
+V2=V_BE+I_C*R_E // since V_E=I_C*R_E
+R2=V2/I1
+R1=Rt-R2
+
+printf("emitter resistance = %.3f ohm \n",R_E*1000)
+printf("R1 = %3f kilo ohm \n",R1)
+printf("R2 = %3f kilo ohm \n",R2)