Added new codeHEADmaster

10 files changed, 210 insertions, 0 deletions

diff --git a/3830/CH5/EX1.8/Ex5_1.sce b/3830/CH5/EX1.8/Ex5_1.sce
new file mode 100644
index 000000000..ae7ba3075
--- /dev/null
+++ b/3830/CH5/EX1.8/Ex5_1.sce
@@ -0,0 +1,23 @@
+// Exa 5.1

+

+clc;

+clear;

+

+// Given

+

+E1 = 1/100; // exposure set for grid line impression(sec)

+E2 = 10; // second exposure duration(sec)

+R = 10^-4; // persistence of CRO screen(sec)

+I1 = 1; // Trace intensity for exposure 1(candle power)

+I2_normal = 4 ; // trace intensity for normal settings(candle power)

+

+// Solution

+

+printf(' The emission of light that would be received by photographic paper in both exposures must be the same \n  Also, the product of time and light is to be the same. \n');

+I_req = I1*E1/R;

+printf('  Hence, the image intensity required = %d \n' ,I_req );

+I_boost = I_req/I2_normal;

+printf('  Therefore, the intensity boost required = %d times \n' , I_boost);

+

+printf('  The light emitted is proportional to the kinetic energy of the electron while it strikes the screen, which is equal to sqrt(V) , where V is the velocity while striking \n');

+ printf('  Therefore, the accelerating voltage of the accelerating anode should br increased by %d times \n',sqrt(I_boost));

diff --git a/3830/CH5/EX5.10/Ex5_10.sce b/3830/CH5/EX5.10/Ex5_10.sce
new file mode 100644
index 000000000..fb6531326
--- /dev/null
+++ b/3830/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,19 @@
+// Exa 5.10

+

+clc;

+clear;

+ 

+// Given

+

+Va = 2000; // voltage applied to anodes(V)

+l = 50*10^-3; // length of horizontal plates(m)

+m = 9.1*10^-31; // mass of electron in kg

+e = 1.6*10^-19; // velocity of electron in m/s

+// Max transit time is T/4

+

+// Solution

+

+V = sqrt(2*Va*e/m);

+Fc = V/(4*l);

+printf(' The velocity of electron = %.3f * 10^8 m/s \n',V*10^-8);

+printf('  The cutoff frequency = %.3f MHz \n',Fc/10^6);

diff --git a/3830/CH5/EX5.2/Ex5_2.sce b/3830/CH5/EX5.2/Ex5_2.sce
new file mode 100644
index 000000000..423ef15ad
--- /dev/null
+++ b/3830/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,12 @@
+// Exa 5.2

+

+clc;

+clear;

+

+// Given 

+

+// Teo magnetic coils

+

+// Solution

+

+printf('When the bullet passes through the first coil, a pulse is generated \n This is applied to an AND gate. The other input to the AND gate is from a crystal oscillator\n When the bullet passes through the second coil, another pulse is generated, which is used to stop counting; this is the disable pulse \n Therefore, the number of counts accelerated during this interval is a measure of the time taken by the bullet to traversea distance d between the coils \n Therefore, velocity = d/t');

diff --git a/3830/CH5/EX5.3/Ex5_3.sce b/3830/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..57f809943
--- /dev/null
+++ b/3830/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,29 @@
+// Exa 5.3

+

+clc;

+clear;

+

+// Given

+

+fs = 10000; // frequency of modulated signal(Hz)

+fm = 200*10^3; // modulation frequency(Hz)

+Ri = 10; // Input resistance(ohms)

+e2_by_e1 = 1.3; // limit for lowest frequency(in %)

+

+// Solution

+

+F_lower = fm - fs ; 

+

+printf(' For a double-section filter, \n  e2/e1 =  1/sqrt(1+(w*Rf*Cf)^2) \n');

+// Therefore,

+

+function y=f(x) 

+    y =(1/(sqrt(2*%pi*F_lower*x)^2+1))-e2_by_e1/100;

+endfunction

+[x,v,info] = fsolve(0,f);

+printf('  The product of Rf*Cf = %.4f  sec \n ',x);

+printf(' Let Rf = 10^5 Ohms, so that attenuation is 10:1. Therefore, Cf = ');

+Cf = x*10^-7;

+printf(' %.3f pf \n ',Cf*10^12);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.4/Ex5_4.sce b/3830/CH5/EX5.4/Ex5_4.sce
new file mode 100644
index 000000000..24ab7b7b6
--- /dev/null
+++ b/3830/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,17 @@
+// Exa 5.4

+

+clc;

+clear;

+

+// Given

+

+// The Lissajous pattern

+Y2 = 2.5; // slope of the major axis(in div)

+Y1 = 1.2; // slope of the vertical axis(in div)

+

+// Solution

+

+printf(' The phase shift V2 and V1 can be given as sin(Theta) = Y1/Y2 \n  -where V1 and V2 are voltages applied to X and Y axis respectively \n ');

+

+Theta = asind(Y1/Y2) ;

+printf(' Since, the ellipse is lying in the I and the III quadrant, \n  The angle is theta or 360-theta , i.e, %.2f or %.2f \n',Theta,360-Theta);

diff --git a/3830/CH5/EX5.5/Ex5_5.sce b/3830/CH5/EX5.5/Ex5_5.sce
new file mode 100644
index 000000000..309ea2dfb
--- /dev/null
+++ b/3830/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,31 @@
+// Exa 5.5

+

+clc;

+clear;

+

+// Given

+

+S = 0.6*10; // sensitivity of oscillograph in V per cm

+A = 2; // Area of oscilloscope area in cm^2

+dx = 4; // x-axis deflection in cm

+dy = 3; // y-axis deflection in cm

+

+// solution

+

+printf(' Ref fig. 5.5(a and b) -If Ic leads Vc by 90 degree, C will be a lossless ideal capacitor, and it will have infinite resistance R. Therefore, Ic is leading Vc by <90 degree . Theta is loss of the capacitor \n ');

+printf(' Power factor = cos(theta) = 1  when theta = o degree) \n');

+

+pf = 1;

+

+Vcondenser = (1/sqrt(2)) * S * dx*200; // since one-two thousandth od C voltage is applied to the x-plates

+Icondenser = (1/sqrt(2)) * S * 1/100000 ; // since Y-plates are impressed with voltage 100000 times the magniture of  condenser I.

+

+Pcondenser = Vcondenser * Icondenser;

+

+printf(' If p.f =1, the ellipse could have a major axis of %d cm and a minimum axis of %d cm \n',2*dx,2*dy);

+

+printf(' Total area = %.2f cm^2 \n',%pi/4 * 2*dx*2*dy);

+

+printf(' power loss of the capacitor = %.4f W \n',Pcondenser*A/(12/%pi));

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.6/Ex5_6.sce b/3830/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..91ea8c739
--- /dev/null
+++ b/3830/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,19 @@
+// Exa 5.6

+

+clc;

+clear;

+

+// Given

+

+// A stationary Lissajous pattern

+Vy = 6 ; // max value on vertical axis

+Vx = 5; // max value on horizontal axis

+fx = 1500; // horizontal input frequency(Hz)

+

+// Solution

+

+// fy/fx = No of pts the target meets per bottom(x-axis)/No of pts the target meets per bottom(y-axis)

+

+fy = (Vy/Vx)*fx;

+

+printf('The frequency of vertical axis = %d Hz \n',fy);

diff --git a/3830/CH5/EX5.7/Ex5_7.sce b/3830/CH5/EX5.7/Ex5_7.sce
new file mode 100644
index 000000000..e96248365
--- /dev/null
+++ b/3830/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,17 @@
+// Exa 5.7

+

+clc;

+clear;

+

+// Given

+

+b2 =2.5 ; // Max no of divisions on y-axis

+b1 = 1.25; // point of intersection on y-axis(div)

+

+// Solution

+

+printf(' Let theta be the phase angle of V2 w.r.t V1 where V1 and V2 are the voltages applied to x and y axis respectively \n');

+// Sin theta = b1/b2;

+Theta = asind(b1/b2);

+

+printf('  Therefore, the phase angle of V2 w.r.t V1 = %d degrees \n. But another possible value is(360-theta) i.e. %d degrees \n',Theta,360-Theta);

diff --git a/3830/CH5/EX5.8/Ex5_8.sce b/3830/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..e591eb0a6
--- /dev/null
+++ b/3830/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,26 @@
+// Exa 5.8

+

+clc;

+clear;

+

+// Given

+

+Va = 2000; // Anode voltage(Volts)

+Vd = 100; // Deflecting plates volage(Volts)

+a =1.5*10^-2; // axial length in m

+Sd= 30*10^-2; // screen distance in m

+Ld = 5*10^-2; // deflecting plates length in m

+

+// Solution

+

+// Let,

+ x = 1.76*10^11 ; // e/m ratio in c/kg

+L = Sd + Ld/2 ;

+D = (Ld*L*Vd)/(2*a*Va) ; // Deflection produced(m)

+Vo = sqrt(2*x*Va); // velocity of electrons in m/kg

+

+

+printf(' The deflection produced on screen = %.3f cm \n',D*100);

+printf('  The velocity of the electrons when they enter the field of the deflecting plates = %.4f * 10^7 m/kg \n', Vo/10^7);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.9/Ex5_9.sce b/3830/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..e9fab8f1b
--- /dev/null
+++ b/3830/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,17 @@
+// Exa 5.9

+

+clc;

+clear;

+

+// Given

+

+// A saw-tooth waveform is applied to an average diode voltmeter(Refer Fig. 5.24)

+

+printf('For a saw-tooth waveform, the rms value = Vm/T  \n -where Vm js max voltage value and T being the time period \n');

+

+printf(' Average value Va, 0.433 Vrms \n ');

+

+printf('Similarly, Iav = 0.433*Vrms/R \n');

+

+printf(' Error = 100 * (0.433*Vrms - (0.433/R)*0.45 / (0.45*(Vrms/R))) = -3.8 percent \n');

+printf(' The meter reading is 3.8 percent less than the expected value \n');