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// Examle 3.22
// From the diagram (3.40b) Apply KCL to node a
// will get { (va-0)/2+ (va-vb)/3 = 5 }............(1
// Similarly apply KCL at node b
// will get { (vb-va)/3+ vb-0)/4 = -6 }............(2
// After solving these 2 equation will have
Va=2.44; // Voltage at node a
Vb=-8.89; // Voltage at node b
Vab=Va-Vb; // Voltage across 3 ohm resistor
disp(' Voltage across 3 ohm resistor = '+string(Vab)+' Volt');
// p 80 3.22
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