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+//chapter3
+//example3.4
+//page50
+
+Eb=250 // V
+Ec=-3 // V
+
+// given that Ib=0.003*(Eb+30*Ec)^1.5 mA
+// differentiating w.r.t Ec with Eb=constant, we get
+gm=0.003*1.5*(Eb+30*Ec)^0.5*30*10^-3
+mutual_inductance_micro=gm*10^6
+
+printf("mutual conductance = %f mho or %.3f micro mho \n",gm,mutual_inductance_micro)
+
+// differentiating given equation w.r.t Ec with Ib=constant, we get
+// 0=0.003*10^-3*1.5*(Eb+Ec)^1.5*(mu+30) where mu is equal to ratio of changes in Eb and Ec i.e. amplification factor
+// thus mu+30=0 hence we get
+mu=-30
+    printf("here negative sign of amplification factor indicates that Eb and Ec are in opposite direction. \n \n")
+// here we need not worry as to if mu may be positive because the equation given in problem statement will always give mu+30=0 i.e. mu=-30
+
+printf("amplification factor = %.3f \n",mu)  
+  
+rp=mu/gm
+if rp<0    // rp can not be negative
+    rp=-rp
+end
+
+printf("plate resistance = %.3f ohm \n",rp)
+
+//in book, the answers are less accurate. The accurate answers are
+// gm=1707.630 micro mho
+// plate resistance=17568.209 ohm