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/EX13.11/13_11.sce | 25 +++++++++++++++++++++++++
 1 file changed, 25 insertions(+)
 create mode 100755 608/CH13/EX13.11/13_11.sce

(limited to '608/CH13/EX13.11/13_11.sce')

diff --git a/608/CH13/EX13.11/13_11.sce b/608/CH13/EX13.11/13_11.sce
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+++ b/608/CH13/EX13.11/13_11.sce
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+//Problem 13.11: A Wheatstone Bridge network is shown in Figure 13.32(a). Calculate the current flowing in the 32 ohm resistor, and its direction, using Th´evenin’s theorem. Assume the source of e.m.f. to have negligible resistance.
+
+//initializing the variables:
+E = 54; // in volts
+R1 = 2; // in ohms
+R2 = 14; // in ohms
+R3 = 3; // in ohms
+R4 = 11; // in ohms
+R5 = 32; // in ohms
+
+//calculation:
+//The 32ohm resistance branch is short-circuited as shown in Figure 13.32(b).
+//The p.d. between A and C,
+Vac = (R1/(R1 + R4))*E
+//The p.d. between B and C,
+Vbc = (R2/(R2 + R3))*E
+//Hence the p.d. between A and B
+Vab = Vbc - Vac
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R1*R4/(R1 + R4) + R2*R3/(R2 + R3)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.32(f), the current in the 32 ohm resistance is given by:
+I = E/(r + R5)
+
+printf("\n\n Result \n\n") 
+printf("\n the current in the 32 ohm resistance is given by %.0f A",I)
\ No newline at end of file
-- 
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