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

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 3802/CH1/EX1.11/Ex1_11.sce |  40 ++++++++++++++++++++++++++++++++++++++++
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+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex1_11.sce.
+clc;
+clear;
+R=10;         //resistance in ohms
+L=5;          //inductance in henry
+V=100;        //supply voltage in volts
+t1=2;        //time at which k1 switch opened in seconds
+//CASE1
+printf("\n (a)")
+i=(V*(1-exp(-((R*t1)/L))))/R;
+printf("\n  The inductive current at the time k1 is opened=%1.2f A",i)
+
+//CASE2
+printf("\n (b)")
+v1=V*exp(-((R*t1))/L);
+printf("\n  The voltage across the inductor at t=2second=%1.2f V",v1)
+
+//CASE3
+printf("\n (c)")
+t2=3;     //time in seconds
+Imax=(V/R); 
+v2=Imax*R*(exp(-((R*t2))/L));
+printf("\n  The voltage across the inductor at t=3 second=%1.4f V",v2)
+//For v2 calculation ,the answer in the book is wrong
+
+//CASE4
+printf("\n (d)")
+t3=0;   //initial time in seconds
+it=(-R*(-Imax)*exp(-(R*t3)/L))/L;    //rate of decay of inductor current in amphere per seconds
+printf("\n  The initial value of rate of decay of inductor current=%d A/s",it)
+
+//CASE5
+printf("\n (e)")
+Energy=(1/2)*L*Imax^2;
+printf("\n  The energy dissipated in the resistor=%d J",Energy)
-- 
cgit