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/CH43/EX43.07/43_07.sce | 22 ++++++++++++++++++++++
 1 file changed, 22 insertions(+)
 create mode 100755 608/CH43/EX43.07/43_07.sce

(limited to '608/CH43/EX43.07/43_07.sce')

diff --git a/608/CH43/EX43.07/43_07.sce b/608/CH43/EX43.07/43_07.sce
new file mode 100755
index 000000000..9919eaa3c
--- /dev/null
+++ b/608/CH43/EX43.07/43_07.sce
@@ -0,0 +1,22 @@
+//Problem 43.07: Two coils are connected in series and their effective inductance is found to be 15 mH. When the connection to one coil is reversed, the effective inductance is found to be 10 mH. If the coefficient of coupling is 0.7, determine (a) the self inductance of each coil, and (b) the mutual inductance.
+
+//initializing the variables:
+k = 0.7; // coefficient of coupling
+L1 = 15E-3; // in Henry
+L2 = 10E-3; // in Henry
+
+//calculation:
+//L1 = La + Lb + 2*k*(La*Lb)^0.5
+//L2 = La + Lb - 2*k*(La*Lb)^0.5
+//self inductance of coils
+a = ((L1 - (L1  + L2)/2)/(2*k))^2
+La1 =((L1  + L2)/2 + (((L1  + L2)/2)^2 - 4*a)^0.5)/2
+La2 =((L1  + L2)/2 - (((L1  + L2)/2)^2 - 4*a)^0.5)/2
+Lb1 = (L1 + L2)/2 - La1
+Lb2 = (L1 + L2)/2 - La2
+//mutual inductance, M
+M = (L1 - L2)/4
+
+printf("\n\n Result \n\n")
+printf("\nself inductance of coils are %.2E H and %.2E H",La1, Lb1)
+printf("\n mutual inductance, M is %.2E H",M)
-- 
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