From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 608/CH13/EX13.15/13_15.sce | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)
 create mode 100755 608/CH13/EX13.15/13_15.sce

(limited to '608/CH13/EX13.15')

diff --git a/608/CH13/EX13.15/13_15.sce b/608/CH13/EX13.15/13_15.sce
new file mode 100755
index 000000000..4e0a951d6
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+++ b/608/CH13/EX13.15/13_15.sce
@@ -0,0 +1,21 @@
+//Problem 13.15: Determine the current flowing in the 2ohm resistance in the network shown in Figure 13.37(a).
+
+//initializing the variables:
+I = 15; // in amperes
+R1 = 6; // in ohms
+R2 = 4; // in ohms
+R3 = 8; // in ohms
+R4 = 2; // in ohms
+R5 = 7; // in ohms
+
+//calculation:
+//The 2ohm resistance branch is short-circuited as shown in Figure 13.37(b).
+//Figure 13.37(c) is equivalent to Figure 13.37(b).
+Isc = (R1/(R1 + R2))*I
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = ((R1 + R2)*(R3 + R5)/(R1 + R2 + R3 + R5))
+//From the Norton equivalent network shown in Figure 13.37(e) the current in the 2ohm resistance is given by:
+I = (r/(r + R4))*Isc
+
+printf("\n\n Result \n\n") 
+printf("\n the current in the 2ohm resistance is given by %.2f A",I)
\ No newline at end of file
-- 
cgit