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+// Exa 2.2
+
+clc;
+clear;
+
+// Given data
+
+// An amplifier as given in Fig. 2.5(b)
+R1 = 10*10^3;   //Input resistance of amplifier (Ω)
+Rf = 100*10^3;  // Feedback resistance of amplifier (Ω)
+vi = 1;         // Input voltage applied (Volts)
+RL = 25*10^3;    // Load resistance (Ω)
+
+// Solution
+
+i1 = vi/R1; //Input current(A)
+printf(' The value of input current = i1 = %.1f mA. \n ',i1*1000);
+vo = -1*(Rf/R1)*vi; // output voltage(V)
+printf(' The value of output voltage = vo = %d V. \n ',vo);
+iL = abs(vo)/RL;  // Load current(A)
+printf(' The value of load current = iL = %.1f mA.',iL*1000);
+disp(" The direction of iL is as shown in Fig. 2.5(b).");
+// iTot = i1 + iL;
+iTot = i1+iL;  // Total current(A)
+printf('  The value of total current = io = %.1f mA.',iTot*1000);
+disp(" In an inverting amplifier, for a +ive input, output will be -ive, therefore the direction of io is as shown in Fig. 2.5(b).");