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+// Example 1.9
+//Calculating Branch Variables
+// From fig 1.33
+v_4=24;
+i_4=v_4/8; // current through 8 Ohm resistor
+v_3=7*i_4;
+// Applying KVL around the Loop on the right
+v_2=v_3+v_4;// Voltage across Current source
+v_1=v_2-25;// voltage across 10 Ohm resistor
+i_2=v_2/9;//Current across 9 Ohm resistor
+i_1=-v_1/10;// Current across 10 Ohm resistor
+p_v=25*i_1;// Power supplied by the Voltage Source
+i_s=i_2+i_4-i_1;// Current supplied by current source
+// Power supplied by Current source is given by,
+p_i=v_2*i_s;
+// Power Dissipated aross various Resistors
+p_r=10*(i_1)^2+9*(i_2)^2+7*(i_4)^2+8*(i_4)^2;
+disp(i_s,"Current supplied by Current source(in Amps)=")
+disp(p_v,"Power supplied by Voltage source(in Watt)=")
+disp(p_i,"Power supplied by Current source(in Watt)=")
+disp(p_r,"Power Dissipated across various Resistors(in Watt)=")