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// Example 4.3
// From the diagram 4.5
// Using super position theorem
// 4-A current source is active
i=4/{1+(2+3)}; // Current
R=3; // Rsistance of 3 Ohms
V4=i*R; // Voltage across 3 Ohms resistance in Case-1
// 5-A current source is active
i5=5; // 5-A current source
V5=(-i5)*{1/[1+(2+3)]*3}; // Voltage across 3 Ohms resistance in Case-2
// 6-V voltage source is active
i6=6; // 6-A current source
V6=i6*{3/[1+(2+3)]}; // Voltage across 3 Ohms resistance in Case-3
V=V4+V5+V6; // Voltage across 3 Ohms resistance
disp(' Voltage across 3 Ohms resistance is = '+string(V)+' Volt');
// p 106 4.3
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