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+//To find the the resistance Rl
+// example 3-12 in page 58
+clc;
+// Given data
+Iav=1e-3;// Average current through the PMCC = 1 mA
+Ip=250e-3;// primary current= 250 mA
+Rm=1700;// coil resistance in ohm
+Ns=500;// number of secondary turns
+Np=4;//number of primary turns
+Vf=0.7;//diode forward drop in volts
+Rs=20e+3;// Rs=20 k ohm
+//calculation
+Im=Iav/0.637;//peak current
+Em=(Im*(Rs+Rm))+(2*Vf);//secondary peak voltage
+Es=Em*0.707;//secondary rms voltage
+Irms=1.11*Iav;// RMS meter current
+Is=Ip*(Np/Ns);//transformer rms secondary current
+Il=Is-Irms;//current through Rl
+Rl=Es/Il;
+printf("Rl=%.1f K-ohm\n",Rl/1000);
+//result
+//Rl=28.2 K-ohm
+\ No newline at end of file