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

---
 181/CH3/EX3.4/example3_4.sce | 36 ++++++++++++++++++++++++++++++++++++
 181/CH3/EX3.4/example3_4.txt |  6 ++++++
 2 files changed, 42 insertions(+)
 create mode 100755 181/CH3/EX3.4/example3_4.sce
 create mode 100755 181/CH3/EX3.4/example3_4.txt

(limited to '181/CH3/EX3.4')

diff --git a/181/CH3/EX3.4/example3_4.sce b/181/CH3/EX3.4/example3_4.sce
new file mode 100755
index 000000000..5e51bff29
--- /dev/null
+++ b/181/CH3/EX3.4/example3_4.sce
@@ -0,0 +1,36 @@
+// Calculate the peak load current
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 3-4 in page 153
+
+clear; clc; close;
+
+// Given data
+Rf=30; // Internal resistance in ohms
+Rl=990; // Load resistance in ohms
+Vm=110; // Rms supply voltage in in V
+
+// Calculation
+Im=(Vm/2)/(Rf+Rl);
+I_dc=Im/%pi;
+I_rms=Im/2;
+V_dc=(Im*Rl)/%pi;
+Pi=I_rms^2*(Rf+Rl);
+R=(((Vm/%pi)-(I_dc*Rl))/(I_dc*Rl))*100;
+
+printf("(a)Im = %0.2e A\n",Im);
+printf("(b)I_dc = %0.2e A\n",I_dc);
+printf("(c)I_rms = %0.2e A\n",I_rms);
+printf("(d)V_dc = %0.3e V\n",V_dc);
+printf("(e)Input power = %0.2f W\n",Pi);
+printf("(f)Percentage regulation = %0.3f percent",R);
+
+// Result
+// (a) Im=53.9mA
+// (b) Idc=17.2mA
+// (c) Irms=27mA
+// (d) Vdc=16.99V 
+// (e) Pi=0.74W 
+// (f) Percentage regulation=106%
\ No newline at end of file
diff --git a/181/CH3/EX3.4/example3_4.txt b/181/CH3/EX3.4/example3_4.txt
new file mode 100755
index 000000000..b2a2280e2
--- /dev/null
+++ b/181/CH3/EX3.4/example3_4.txt
@@ -0,0 +1,6 @@
+ (a)Im = 5.39e-002 A
+(b)I_dc = 1.72e-002 A
+(c)I_rms = 2.70e-002 A
+(d)V_dc = 1.699e+001 V
+(e)Input power = 0.74 W
+(f)Percentage regulation = 106.061 percent 
\ No newline at end of file
-- 
cgit