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diff --git a/2459/CH12/EX12.11/Ex12_11.sce b/2459/CH12/EX12.11/Ex12_11.sce
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+//chapter12
+//example12.11
+//page248
+
+Vcc=15 // V
+Re=2 // kilo ohm
+Rc=1 // kilo ohm
+gain_beta=100
+Vbe=0.7 // V
+R1=10 // kilo ohm
+R2=5 // kilo ohm
+
+// when Ic=0, Vce=Vcc i.e. Vce=6 and when Vce=0, Ic=Vcc/(Rc+Re) i.e. Ic=15/(1+2)
+// so equation of load line becomes Ic=-(1/3)*Vce+5
+
+clf()
+x=linspace(0,15,5)
+y=-(1/3)*x+5
+plot2d(x,y,style=3,rect=[0,0,16,6])
+xtitle("dc load line","Vce(volts)","Ic(mA)")
+
+V2=Vcc*R2/(R1+R2) // voltage across R2 i.e. 5 kilo ohm
+Ie=(V2-Vbe)/Re
+Ic=Ie
+Vce=Vcc-Ic*(Rc+Re)
+
+printf("the operating point is %.3f V and %.3f mA \n",Vce,Ic)
diff --git a/2459/CH12/EX12.11/Figure12_11.JPG b/2459/CH12/EX12.11/Figure12_11.JPG
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