path: root/1430/CH3/EX3.2/exa3_2.sce

// Example 3.2
// A Paradox Resolved
// From figure 3.4(c)
i=((40+20)*(10^-3))/(5+10);// Applying KVL in figure 3.4(c)
disp(i,"Circuit Current(in Amps)=")