1 files changed, 19 insertions, 0 deletions
diff --git a/Working_Examples/154/CH5/EX5.8/ch5_8.sce b/Working_Examples/154/CH5/EX5.8/ch5_8.sce
new file mode 100755
index 0000000..4c4e229
--- /dev/null
+++ b/Working_Examples/154/CH5/EX5.8/ch5_8.sce
@@ -0,0 +1,19 @@
+clc
+//From figure 5.13(a)
+//Applying KVL equation to the loop 
+I=(20+10)/(3+6)
+//As current will not flow in upper 3 ohm resistor so Thevenin voltage is equal to either of the two parallel branches
+V1=20-I*3
+printf("Thevenin voltage = %dV\n",V1)
+
+// Left 3 ohm and 6 ohm resistor are in parallel and their equivalent is in series with 3 ohm
+R1=3+(3*6)/(3+6)
+printf("Thevenin resistance =%dohm\n",R1)
+
+//Now to find Norton's equivalent
+I1=V1/R1
+printf("\n Norton current =%dA\n",I1)
+disp("The value of resistance in Norton equivalent will not change but will come in parallel with current source")
+
+
+