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//Basic Circuit Concepts
//page no-1.14
//example1.8
I1=10/8;
printf("I1=2/5= %2f Ampere", I1)
I2=5;
printf("\nI2=4/8= %2f Ampere", I2)
disp("Applying KVL to the path from A to B") //Applying KVL to the path from A to B
disp("VA - 3*I1 - 8 + 3*I2 - VB = 0");
disp("VA - VB = 3*1.25 + 8 - 3*5")
VAB= (3*1.25)+8-(3*5);
printf("VAB = %.2f Volt", VAB);
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