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/EX13.01/13_01.sce | 30 ++++++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)
 create mode 100755 608/CH13/EX13.01/13_01.sce

(limited to '608/CH13/EX13.01')

diff --git a/608/CH13/EX13.01/13_01.sce b/608/CH13/EX13.01/13_01.sce
new file mode 100755
index 000000000..754cd0c34
--- /dev/null
+++ b/608/CH13/EX13.01/13_01.sce
@@ -0,0 +1,30 @@
+//Problem 13.01: (a) Find the unknown currents marked in Figure 13.3(a). (b) Determine the value of e.m.f. E in Figure 13.3(b).
+
+//initializing the variables:
+Iab = 50; // in ampere
+Ibc = 20; // in ampere
+Iec = 15; // in ampere
+Idf = 120; // in ampere
+Ifg = 40; // in ampere
+Iab = 50; // in ampere
+I = 2; // in ampere
+V1 = 4; // in volts
+V2 = 3; // in volts
+V3 = 6; // in volts
+R1 = 1; // in ohms
+R2 = 2; // in ohms
+R3 = 2.5; // in ohms
+R4 = 1.5; // in ohms
+
+//calculation:
+I1 = Iab - Ibc
+I2 = Ibc + Iec
+I3 = I1 - Idf
+I4 = Iec - I3
+I5 = Idf - Ifg
+// Applying Kirchhoff’s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:
+E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1
+
+printf("\n\n Result \n\n") 
+printf("\n (a) unknown currents I1, I2, I3, I4, I5 are %.0fA, %.0fA, %.0fA, %.0fA, %.0fA respetively",I1, I2, I3, I4, I5)
+printf("\n (b) value of e.m.f. E = %.0f Volts",E)
\ No newline at end of file
-- 
cgit