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+//example6.4
+clc
+disp("R_L=8ohm, I_CQ=140 mA, V_CC=10V")
+disp("P_ac= 0.48 W")
+disp("The turns ratio are specified as N1/N2:1 i.e 3:1")
+disp("Therefore N1/N2=3")
+n=1/3
+disp(n,"n=N2/N1=1/3=")
+r=8/(0.333)^2
+disp(r,"Therefore R''_L=R_L/n^2=")
+disp("1. As the transformer is ideal, whatever is the power delivered to the load,same is the power developed across primary.")
+disp("Therefore P_ac(across primary)=0.48W")
+disp("2. Using equation (9),")
+disp("we get, P_ac=(V_1rms^2)/(R''_L)")
+disp("Therefore 0.48=(V_1rms^2)/72")
+v=sqrt(34.56)
+disp(v,"Therefore V_1rms(in V)=")
+disp("But rms value of the load voltage is V_2rms")
+disp("So (V1_rms)/(V2_rms)=N1/N2=3/1")
+v=5.8787/3
+disp(v,"Therefore (V2_rms)(in V)=(V1_rms)/3=")
+disp("This is the rms value of the load voltage.")
+disp("3. The rms value of the primary voltage is (V1_rms) as calculated above.")
+disp("Therefore (V1_rms)=5.8787 V")
+disp("4. The power delivered to the load = (I_2rms^2)*R_L       ..Refer equation 13.")
+disp("0.48=(I_2rms^2)*8")
+i=sqrt(0.06)
+disp(i,"(I_2rms)[in A]=")
+disp("This is the rms value of the load current as the resistance value used is R_L and not R''_L")
+disp("5. The rms values of primary and secondary are related through the transformation ratio.")
+disp("Therefore (I_1rms)/(I_2rms)=N2/N1=n=0.333")
+i=0.2449*0.333
+disp(i,"Thererfore (I_1rms)[in A]=0.2449*0.333= ")
+disp("6. The dc power input is,")
+p=140*10^-2
+disp(p,"P_DC(in W)=(V_CC)*(I_CQ)=")
+n=(0.48*100)/1.4
+disp(n,"7. %eta=(P_ac *100)/(P_dc)=")
+d=1.4-0.48
+disp(d,"P_d(in W)=")
+disp("This is the power dissipation.")