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+//Example 5.4 // potential difference across the heater element
+clc;
+clear;
+close;
+w=20;// in kg
+t1=10;// in degree celsius
+t2=90;//in degree celsius
+t=2*3600+19*60+34;// time in seconds
+I=4;// in ampere
+n=80;//efficiency of conversion in percentage
+ho=w*(t2-t1);//output heat required in kcal
+V=((ho*4187*100)/(I*t*n));// POTENTIAL DROP IN VOLTS
+disp(V,"potential drop across heater element in volts is")