From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 1319/CH1/EX1.10/1_10.sce | 39 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)
 create mode 100644 1319/CH1/EX1.10/1_10.sce

(limited to '1319/CH1/EX1.10/1_10.sce')

diff --git a/1319/CH1/EX1.10/1_10.sce b/1319/CH1/EX1.10/1_10.sce
new file mode 100644
index 000000000..f57c46c9e
--- /dev/null
+++ b/1319/CH1/EX1.10/1_10.sce
@@ -0,0 +1,39 @@
+// RLC circuit problems on resonace
+
+clc;
+clear;
+
+R=6.28;
+L=20*(10^-3);
+f=5*(10^3);
+
+w=2*%pi*f;
+
+C=1/(L*(w^2));
+
+Xc=1/(w*C);
+Xl=L*w;
+
+Vc=5;
+
+Z=Xc+R+Xl;
+
+I=Vc/Xc // Total current
+
+V=I*R;
+
+// frequency is inversely proportional to square root of capacitance
+// So if C is halved; f will increase square root of 2 times more.
+
+fn=sqrt(2)*f;
+
+Xln=2*%pi*fn*L;
+
+Q=Xln/R;
+
+//Note under resonance conditions Vl and Vc are much greater than the supply voltage.
+
+mprintf('i) The value of capacitor = %f micro F \n',(10^6)*C)
+mprintf('ii) The supply voltage = %f V \n',V)
+mprintf('iii) The frequency of resonance when C is halved = %f Hz \n',fn)
+mprintf('     The Q of the new circuit = %f \n',Q) 
-- 
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