Chapter 5 : Primary Sensing Elements and Transducers

Example 5.1 Page No : 179

In [1]:
# Variables
D = 15.*10**-3;
P = 300.*10**3;
sm = 300.*10**6;

# Calculations and Results
t = (3*D**2*P/(16*sm))**0.5;
print " thickness(m) = ", t
P = 150*10**3;
v = 0.28;
E = 200*10**9;
dm = 3*(1-v**2)*D**4*P/(256*E*t**3);
print " deflection at center for Pressure of 150 kN/m2(m) = ", dm

 thickness(m) =  0.000205395959064
 deflection at center for Pressure of 150 kN/m2(m) =  4.73232289684e-05

Example 5.2 Page No : 181

In [4]:
import math

T = 100.;
G = 80.*10**9;
d = 2.*15*10**-3;

# Calculations
th = 16*T/(math.pi*G*d**3)

# Results
print " angle of twist(rad) = ", th

 angle of twist(rad) =  0.000235785100877

Example 5.3 Page No : 183

In [5]:
import math 

# Variables
E = 110*10**9;
t = 0.073*10**-3;
b = 0.51*10**-3;
l = 370*10**-3;

# Calculations
th = math.pi/2;
T = (E*b*t**3)*th/(12*l);

# Results
print " Controlling torque(Nm) = ", T

 Controlling torque(Nm) =  7.72090748389e-06

Example 5.4 Page No : 185

In [6]:
import math 

# Variables
Rnormal = 10000./2;
Rpl = 10000./50;
Rc1 = Rnormal-3850;

# Calculations and Results
Dnormal = Rc1/Rpl;
print "Displacement(mm) = ", Dnormal
Rc2 = Rnormal-7560;
Dnormal = Rc2/Rpl;
print "Displacement(mm) = ", Dnormal
print "since one lacement is positive and other is negative so two lacements are in the opposite direction"
Re = 10*1./200;
print "Resolution (mm) = ", Re

Displacement(mm) =  5.75
Displacement(mm) =  -12.8
since one lacement is positive and other is negative so two lacements are in the opposite direction
Resolution (mm) =  0.05

Example 5.5 Page No : 187

In [8]:
%matplotlib inline
import math 
from matplotlib.pyplot import plot

# Variables
# plot the graph of error versus K
K = [0,    0.25,    0.5,    0.75,    1];
V = [0,    -0.174,    -0.454,    -0.524,    0];

# Results
plot(K,V)
Out[8]:
[<matplotlib.lines.Line2D at 0x103fd9ad0>]

Example 5.6 Page No : 189

In [9]:
# Variables
RAB = 125.;
Rtotal = 5000.;
R2 = 75/125.*Rtotal;
R4 = 2500.;
ei = 5.;

# Calculations
eo = ((R2/Rtotal)-(R4/Rtotal))*ei;

# Results
print "output voltage (V) = ", eo

output voltage (V) =  0.5

Example 5.7 Page No : 191

In [10]:
# Variables
Rm = 10000.;
Rp = Rm/15.;
R = 600.;
P = 5.;

# Calculations and Results
ei =  (P*R)**0.5;
print "Maximum excitation voltage (V) = ", ei
S = ei/360;
print "Sensitivity (V/degree) = ", S

Maximum excitation voltage (V) =  54.7722557505
Sensitivity (V/degree) =  0.152145154863

Example 5.8 Page No : 193

In [11]:
# Variables
Rwga = 1./400;

# Calculations
Re = Rwga/5;

# Results
print "Resolution (mm) = ", Re

Resolution (mm) =  0.0005

Example 5.9 Page No : 195

In [12]:
# Variables
mo = 0.8;
sr = 250;
sm = sr/mo;

# Calculations and Results
R = sm*1;
print "resolution of 1mm movement", R
Rq = 300/1000;
print "resolution required = ", Rq
print "since the resolution of potentiometer is higher than the resolution required so it is suitable for the application",

resolution of 1mm movement 312.5
resolution required =  0
since the resolution of potentiometer is higher than the resolution required so it is suitable for the application

Example 5.10 Page No : 197

In [13]:
# Calculations and Results
Pd = (10.**2)/150;
print "Power dissipation (W) = ", Pd
th_pot = 80.+Pd*30*10**-3;
PDa = 1-(10*10**-3)*(th_pot-35);
print "Power dissipation allowed(W) = ", PDa
print "Since power dissipation is higher than the dissipation allowed so potentiometer is not suitable"

Power dissipation (W) =  0.666666666667
Power dissipation allowed(W) =  0.5498
Since power dissipation is higher than the dissipation allowed so potentiometer is not suitable

Example 5.11 Page No : 199

In [14]:
# Variables
Gf = 4.2;

# Calculations
v = (Gf-1)/2;

# Results
print "Possion s ratio = ", v

Possion s ratio =  1.6

Example 5.12 Page No : 201

In [15]:
# Variables
strain = -5.*10**-6;
Gf = -12.1;
R = 120.;

# Calculations and Results
dR_nickel = Gf*R*strain;
print "change in resistance of nickel(ohm) = ", dR_nickel
Gf = 2;
R = 120;
dR_nicrome = Gf*R*strain;
print "change in resistance of nicrome(ohm) = ", dR_nicrome

change in resistance of nickel(ohm) =  0.00726
change in resistance of nicrome(ohm) =  -0.0012

Example 5.13 Page No : 203

In [16]:
# Variables
s = 100.*10**6;
E = 200.*10**9;

# Calculations
strain = s/E;
Gf = 2;
r_perunit = Gf*strain*100;

# Results
print "Percentage change in resistance = ", r_perunit

Percentage change in resistance =  0.1

Example 5.14 Page No : 205

In [17]:
# Variables
b = 0.02;
d = 0.003;
I = (b*d**3)/12;
E = 200*10**9;
x = 12.7*10**-3;
l = 0.25;
F = 3*E*I*x/l**3;
x = 0.15;
M = F*x;
t = 0.003;

# Calculations
s = (M*t)/(I*2);
strain = s/E;
dR = 0.152;
R = 120;
Gf = (dR/R)/strain;

# Results
print "Gauge factor = ", Gf

Gauge factor =  2.30873918538

Example 5.15 Page No : 207

In [18]:
# Variables
dR = 0.013;
R = 240;
l = 0.1;
Gf = 2.2;

# Calculations and Results
dl = (dR/R)*l/Gf;
print "change in length(m) = ", dl
strain = dl/l;
E = 207*10**9;
s = E*strain;
A = 4*10**-4;
F = s*A;
print "Force(N)", F

change in length(m) =  2.46212121212e-06
Force(N) 2038.63636364

Example 5.16 Page No : 209

In [19]:
# Variables
th1 = 30.;
th2 = 60.;
th0 = th1+th2/2;
Rth1 = 4.8;
Rth2 = 6.2;
Rth0 = 5.5;

# Calculations
ath0 = (1/Rth0)*(Rth2-Rth1)/(th2-th1);

# Results
print "alpha at o degree(/degree C) = ", ath0
print "5.5[1+0.0085(th-45)]",

alpha at o degree(/degree C) =  0.00848484848485
5.5[1+0.0085(th-45)]

Example 5.17 Page No : 211

In [20]:
# Variables
th1 = 100.;
th2 = 130.;

# Calculations
th0 = th1+th2/2;
Rth1 = 573.40;
Rth2 = 605.52;
Rth0 = 589.48;
ath0 = (1/Rth0)*(Rth2-Rth1)/(th2-th1);

# Results
print "alpha at o degree(/degree C) = ", ath0
print "Linear approximation is: Rth =  589.48[1+0.00182(th-115)]",

alpha at o degree(/degree C) =  0.00181629006356
Linear approximation is: Rth =  589.48[1+0.00182(th-115)]

Example 5.18 Page No : 213

In [22]:
# Variables
Rth0 = 100.;
ath0 = 0.00392;
dth = 65.-25;

# Calculations and Results
R65 = Rth0*(1+ath0*dth);
print "resistance at 65 degree C(ohm) = ", R65

th = (((150./100)-1)/ath0)+25;
print "Temperature (degree C)", th

resistance at 65 degree C(ohm) =  115.68
Temperature (degree C) 152.551020408

Example 5.19 Page No : 215

In [23]:
# Variables
Rth0 = 10;
ath0 = 0.00393;
dth = 150-20;

# Calculations
R150 = Rth0*(1+ath0*dth);

# Results
print "resistance at 150 degree C(ohm) = ", R150

resistance at 150 degree C(ohm) =  15.109

Example 5.20 Page No : 217

In [24]:
# Variables
th = 30.;
th0 = 50.;
tc = 120.;

# Calculations
t = -120*(math.log(1-(th/th0)));

# Results
print "time(s) = ", t

time(s) =  109.954887825

Example 5.21 Page No : 219

In [25]:
# Variables
R25 = 100;
ath = -0.05;
dth = 35-25;

# Calculations
R35 = R25*(1+ath*dth);

# Results
print "resistance at 35 degree C(ohm) = ", R35

resistance at 35 degree C(ohm) =  50.0

Example 5.22 Page No : 221

In [26]:
import math
# Variables
Ro = 3980.;
Ta = 273.;

# Calculations and Results
print "3980 =  a*3980*math.exp(b/273)",
Rt50 = 794;
Ta50 = 273+50;
print "794 =  a*3980*math.exp(b/323)",
print "on solving",
print " b = 2843", "a = 30*10**-6"
Ta40 = 273+40;
Rt40 = (30*10**-6)*3980*math.exp(2843./313);
print "resistance at 40 degree C (ohm)", Rt40
Rt100 = (30*10**-6)*3980*math.exp(2843./373);
print "resistance at 100 degree C (ohm)", Rt100

3980 =  a*3980*math.exp(b/273) 794 =  a*3980*math.exp(b/323) on solving  b = 2843 a = 30*10**-6
resistance at 40 degree C (ohm) 1051.3086649
resistance at 100 degree C (ohm) 243.887690913

Example 5.23 Page No : 223

In [27]:
# Calculations
th = ((1-1800./2000)/0.05)+70;
dth = th-70;

# Results
print "change in temperature (degree C)", dth

change in temperature (degree C) 2.0

Example 5.24 Page No : 225

In [28]:
# Variables
C = 500.*10**-12;
R20 = 10000.*(1-0.05*(20-25));

# Calculations and Results
f20 = 1/(2*math.pi*R20*C);
print "Frequency of oscillation at 20 degree C (Hz)", f20
R25 = 10000*(1-0.05*(25-25));
f25 = 1./(2*math.pi*R25*C);
print "Frequency of oscillation at 25 degree C (Hz)", f25
R30 = 10000*(1-0.05*(30-25));
f30 = 1./(2*math.pi*R30*C);
print "Frequency of oscillation at 30 degree C (Hz)", f30

Frequency of oscillation at 20 degree C (Hz) 25464.7908947
Frequency of oscillation at 25 degree C (Hz) 31830.9886184
Frequency of oscillation at 30 degree C (Hz) 42441.3181578

Example 5.25 Page No : 227

In [29]:
# Calculations and Results
Se_thermocouple = 500-(-72);
print "Sensitivity of thermocouple (micro V/degree C) = ", Se_thermocouple
Vo = Se_thermocouple*100*10**-6;
print "maximum output voltage(V) = ", Vo

Sensitivity of thermocouple (micro V/degree C) =  572
maximum output voltage(V) =  0.0572

Example 5.26 Page No : 229

In [30]:
# calculations and results
ET = 27.07+0.8;
print 'Required e.m.f.(mV)',ET
print "temperature corresponding to 27.87 mV is 620 degree C",

Required e.m.f.(mV) 27.87
temperature corresponding to 27.87 mV is 620 degree C

Example 5.27 Page No : 231

In [31]:
# Variables
Rm = 50;
Re = 12;

# Calculations and Results
E = 33.3*10**-3;
i = 0.1*10**-3;
Rs = (E/i)-Rm-Re;
print "series resistance (ohm) = ", Rs
Re = 13;
i1 = E/(Rs+Re+Rm);
AE = ((i1-i)/i)*800;
print "approximate error due to rise in resistance of 1 ohm in Re (degree C) = ", AE
R_change = 50*0.00426*10;
i1 = E/(Rs+Re+Rm+R_change);
AE = ((i1-i)/i)*800;
print "approximate error due to rise in Temp. of 10 (degree C) = ", AE

series resistance (ohm) =  271.0
approximate error due to rise in resistance of 1 ohm in Re (degree C) =  -2.39520958084
approximate error due to rise in Temp. of 10 (degree C) =  -7.44949870586

Example 5.28 Page No : 233

In [32]:
# Variables
E_20 = 0.112*10**-3;# emf at 20degree C
E_900 = 8.446*10**-3;
E_1200 = 11.946*10**-3;

# Calculations and Results
E1 = E_900-E_20;
E2 = E_1200-E_20;
print "E1 = 1.08*R1/(R1+2.5+R2)      (i)",
print "E2 = 1.08*(R1+2.5)/(R1+2.5+R2)      (ii)",
print "on solving (i) and (ii)",
R1 = 5.95;
R2 = 762.6;
print "value of resistance R1 (ohm) = ", R1
print "value of resistance R2 (ohm) = ", R2

E1 = 1.08*R1/(R1+2.5+R2)      (i) E2 = 1.08*(R1+2.5)/(R1+2.5+R2)      (ii) on solving (i) and (ii) value of resistance R1 (ohm) =  5.95
value of resistance R2 (ohm) =  762.6

Example 5.29 Page No : 235

In [33]:
# Calculations
linearity_percentage = (0.003/1.5)*100;

# Results
print "percentage linearity = ", linearity_percentage

percentage linearity =  0.2

Example 5.30 Page No : 237

In [34]:
# Variables
print "lacement = 0.5"
Vo = 2.*10**-3;
Se_LVDT = Vo
print "lacement "
print "senstivity of the LVDT (V/mm)", Se_LVDT

# Calculations and Results
Af = 250.;
Se_instrument = Se_LVDT*Af;
print "senstivity of instrument (V/mm)", Se_instrument
sd = 5./100;
Vo_min = 50./5;
Re_instrument = 1*1/1000;
print "resolution of instrument in mm", Re_instrument

lacement = 0.5
lacement 
senstivity of the LVDT (V/mm) 0.002
senstivity of instrument (V/mm) 0.5
resolution of instrument in mm 0

Example 5.31 Page No : 239

In [35]:
# Variables
b = 0.02;
t = 0.004;
I = (1./12)*b*t**3;
F = 25.;
l = 0.25;
E = 200.*10**9;

# Calculations and Results
x = (F*l**3)/(3*E*I);
print "deflection (m)", x
DpF = x/F;
Se = DpF*0.5*1000;
Re = (10./1000)*(2./10);
F_min = Re/Se;
F_max = 10/Se;
print "minimum force (N)", F_min
print "maximum force (N)", F_max
print "", Se

deflection (m) 0.006103515625
minimum force (N) 0.016384
maximum force (N) 81.92
 0.1220703125

Example 5.32 Page No : 241

In [37]:
print "permittivity of the air e0 = 8.85*10**-12"
# Variables
e0 = 8.85*10**-12;
w = 25*10**-3;

# Calculations
d = 0.25*10**-3;
Se = -4*e0*w/d;

# Results
print "sensitivity of the transducer (F/m) = ", Se

permittivity of the air e0 = 8.85*10**-12
sensitivity of the transducer (F/m) =  -3.54e-09

Example 5.33 Page No : 243

In [38]:
# Variables
C1 = 370*10**-12;
d1 = 3.5*10**-3;

# Calculations
d2 = 2.9*10**-3;
C2 = C1*d1/d2;

# Results
print "the value of the capacitance afte the application of pressure (F) = ", C2

the value of the capacitance afte the application of pressure (F) =  4.46551724138e-10

Example 5.34 Page No : 245

In [39]:
# Variables
fo1 = 100.*10**3;
d1 = 4.;
d2 = 3.7;

# Calculations
fo2 = ((d2/d1)**0.5)*fo1;
dfo = fo1-fo2;

# Results
print "change in frequency of the oscillator (Hz)", dfo

change in frequency of the oscillator (Hz) 3823.07969164

Example 5.35 Page No : 247

In [40]:
import math 

# Variables
L_air = (3.1-3)/2;
D_stress = 100/L_air;
e0 = 8.85*10**-12;
l = 20.*10**-3;
D2 = 3.1;
D1 = 3;

# Calculations and Results
C = (2*math.pi)*e0*l/(math.log(D2/D1));
print "capacitance(F) = ", C
l = (20*10**-3)-(2*10**-3);
C_new = (2*math.pi)*e0*l/(math.log(D2/D1));
C_change = C-C_new;
print "change in capacitance(F) = ", C_change

capacitance(F) =  3.39167370734e-11
change in capacitance(F) =  3.39167370734e-12

Example 5.36 Page No : 249

In [41]:
import math 

# Variables
M = 0.95;
w = 2.*math.pi*20;

# Calculations and Results
tc = (1/w)*((M**2)/(1-M**2))**0.5;
print "time constant (s)", tc
ph = ((math.pi/2)-(math.atan(w*tc)))*(180/math.pi);
print "phase shift(deg)", ph
C = (8.85*10**-12*300*10**-6)/(0.125*10**-3);
R = tc/C;
print "series resistance (ohm)", R
M = 1./(1+(1./(2*math.pi*5*tc)**2))**0.5;
print "amplitude ratio = ", M
Eb = 100;
x = 0.125*10**-3;
Vs = Eb/x;
print "voltage sensitivity (V/m)", Vs

time constant (s) 0.0242109278459
phase shift(deg) 18.1948723388
series resistance (ohm) 1139874192.37
amplitude ratio =  0.605390338483
voltage sensitivity (V/m) 800000.0

Example 5.37 Page No : 251

In [42]:
import math 

# Variables
e0 = 8.85*10**-12;
A = 500*10**-6;
d = 0.2*10**-3;
C = e0*A/d;
d1 = 0.18*10**-3;
C_new = e0*A/d1;

# Calculations and Results
C_change = C_new-C;
Ratio = (C_change/C)/(0.02/0.2);
print "ratio of per unit change of capacitance to per unit change of diaplacement", Ratio
d1 = 0.19*10**-3;
e1 = 1;
d2 = 0.01*10**-3;
e2 = 8;
C = (e0*A)/((d1/e1)+(d2/e2));
d1_new = 0.17*10**-3;
C_new = (e0*A)/((d1_new/e1)+(d2/e2));
C_change = C_new-C;
Ratio = (C_change/C)/(0.02/0.2);
print "ratio of per unit change of capacitance to per unit change of diaplacement", Ratio

ratio of per unit change of capacitance to per unit change of diaplacement 1.11111111111
ratio of per unit change of capacitance to per unit change of diaplacement 1.16788321168

Example 5.40 Page No : 253

In [43]:
# Variables
g = 0.055;
t = 2*10**-3;
P = 1.5*10**6;

# Calculations and Results
Eo = g*t*P;
print "output voltage(V) = ", Eo
e = 40.6*10**-12;
d = e*g;
print "charge sensitivity (C/N) = ", d

output voltage(V) =  165.0
charge sensitivity (C/N) =  2.233e-12

Example 5.41 Page No : 255

In [44]:
# Variables
g = 0.055;
t = 1.5*10**-3;
Eo = 100;

# Calculations
P =  Eo/(g*t);
A = 25.*10**-6;
F = P*A;

# Results
print "Force(N) = ", F

Force(N) =  30.303030303

Example 5.42 Page No : 257

In [45]:
# Variables
A = 25.*10**-6;
F = 5.;
P = F/A;
d = 150*10**-12;
e = 12.5*10**-9;
g = d/(e);
t = 1.25*10**-3;

# Calculations
Eo = (g*t*P);
strain = P/(12*10**6);
Q = d*F;
C = Q/Eo;

# Results
print "strain = ", strain
print "charge(C) = ", Q
print "capacitance(F) = ", C

strain =  0.0166666666667
charge(C) =  7.5e-10
capacitance(F) =  2.5e-10

Example 5.43 Page No : 259

In [47]:
import math 

# Variables
d = 2.*10**-12;
t = 1.*10**-3;
Fmax = 0.01;
e0 = 8.85*10**-12;
er = 5.;

# Calculations and Results
A = 100*10**-6;
Eo_peak_to_peak = 2*d*t*Fmax/(e0*er*A);
print "peak voltage swing under open conditions", Eo_peak_to_peak
Rl = 100*10**6;
Cl = 20*10**-12;
d1 = 1*10**-3;
Cp = e0*er*A/d1;
C = Cp+Cl;
w = 1000;
m = (w*Cp*Rl/(1+(w*C*Rl)**2)**0.5);
El_peak_to_peak = (2*d*t*Fmax/(e0*er*A))*m;

print "peak voltage swing under loaded conditions", El_peak_to_peak
E = 90*10**9;
dt = 2*Fmax*t/(A*E);
print "maximum change in crystal thickness (m)", dt

peak voltage swing under open conditions 0.0090395480226
peak voltage swing under loaded conditions 0.00151556405344
maximum change in crystal thickness (m) 2.22222222222e-12

Example 5.44 Page No : 261

In [49]:
import math 

# Variables
M = 0.95;
tc = 1.5*10**-3;

# Calculations and Results
w = (1/tc)*((M**2)/(1-M**2))**0.5;
print "minimum frequency (rad/s)", w
ph = ((math.pi/2)-(math.atan(w*tc)))*(180/math.pi);
print "phase shift(deg)", ph

minimum frequency (rad/s) 2028.2899482
phase shift(deg) 18.1948723388

Example 5.45 Page No : 263

In [50]:
import math 

# Variables
Kq = 40.*10**-3;
Cp = 1000.*10**-12;

# Calculations and Results
K = Kq/Cp;
print "sensitivity of the transducer(V/m)", K
Cc = 300*10**-12;
Ca = 50*10**-12;
C = Cp+Cc+Ca;
Hf = Kq/C;
print "high frequency sensitivity (V/m)", Hf
R = 1*10**6;
tc = R*C;
M = 0.95;
w = (1/tc)*((M**2)/(1-M**2))**0.5;
f = w/(2*math.pi);
print "minimum frequency (s)", w
print "now f = 10Hz",
f = 10;
w = 2*math.pi*f;
tc = (1/w)*((M**2)/(1-M**2))**0.5;
C_new = tc/R;
Ce = C_new-C;
print "external shunt capacitance(F)", Ce
Hf_new = Kq/C_new;
print "new value of high frequency sensitivity (V/m)", Hf_new

sensitivity of the transducer(V/m) 40000000.0
high frequency sensitivity (V/m) 29629629.6296
minimum frequency (s) 2253.655498
now f = 10Hz external shunt capacitance(F) 4.70718556919e-08
new value of high frequency sensitivity (V/m) 826073.256145

Example 5.46 Page No : 265

In [55]:
import math 
#

# Variables
R = 10.**6;
C = 2500.*10**-12;
tc = R*C;
t = 2.*10**-3;
d = 100.*10**-12;

# Calculations and Results
F = 0.1;
el = 10.**3*(d*F*(math.exp(-t/tc))/C);
print "voltage just before t = 2ms (mV)", el
el_after = 10.**3*(d*F*(math.exp(-t/tc)-1)/C);
print "voltage just after t = 2ms (mV)", el_after
print "when t = 10ms",
t = 10.*10**-3;
T = 2.*10
e_10 = 10.**3*(d*F*(math.exp((-T/tc)-1))*(math.exp(-(t-T))/tc)/C)
print "output voltage 10 ms after the application of impulse(mV) ", e_10

voltage just before t = 2ms (mV) 1.79731585647
voltage just after t = 2ms (mV) -2.20268414353
when t = 10ms output voltage 10 ms after the application of impulse(mV)  0.0

Example 5.47 Page No : 267

In [56]:
import math 
# to prove time constant should be approximately 20T to keep undershoot within 5%
# Variables
print "Let T = 1",
T = 1.;
el = 0.95;

# Calculations
tc = -T/math.log(el);

# Results
print "time constant", tc
print "as T = 1 so time constant should be approximately equal to 20T",

Let T = 1 time constant 19.4957257462
as T = 1 so time constant should be approximately equal to 20T

Example 5.48 Page No : 269

In [57]:
# Variables
Kh = -1*10**-6;
I = 3;
B = 0.5;

# Calculations
t = 2.*10**-3;
Eh = Kh*I*B/t;

# Results
print "output voltage (V)", Eh

output voltage (V) -0.00075

Example 5.49 Page No : 271

In [58]:
# Calculations
Th_wavelength = 1.24*10**-6/1.8

# Results
print "Threshold wavelength (m)", Th_wavelength

Threshold wavelength (m) 6.88888888889e-07

Example 5.50 Page No : 273

In [59]:
# Variables
E_imparted = (1.24*10**-6)/(0.2537*10**-6);
B_energy = E_imparted-4.30;
em_ratio = 0.176*10**12;

# Calculations
v = (2*B_energy*em_ratio)**0.5;

# Results
print "maximum velocity of emitted photo electrons (m/s)", v

maximum velocity of emitted photo electrons (m/s) 454815.603242

Example 5.51 Page No : 275

In [60]:
import math

# Variables
Ri = 30;
Rf = 100;
t = 10.;
tc = 72.;

# Calculations
Rt = Ri+(Rf-Ri)*(1-math.exp(-t/tc));

# Results
print "resistance of the cell (K ohm)", Rt

resistance of the cell (K ohm) 39.0772691917

Example 5.52 Page No : 277

In [61]:
import math 

# Calculations and Results
I_power = 250*0.2*10**-6;
print "incident power (W)", I_power

Rl = 10.*10**3;
C = 2.*10**-12;
fc = 1./(2*math.pi*Rl*C);
print "cut off frequency (Hz)", fc

incident power (W) 5e-05
cut off frequency (Hz) 7957747.15459

Example 5.53 Page No : 279

In [62]:
import math 

# Variables
I = 2.2*10**-3;
Eo = 0.33;
Rl = 100.;

# Calculations and Results
Ri = (Eo/I)-100;
print "internal resistance of cell (ohm)", Ri
Vo = 0.33*(math.log(25)/math.log(10));
print "open circuit voltage for a radiant incidence of 25 W/m2 (V) = ", Vo

internal resistance of cell (ohm) 50.0
open circuit voltage for a radiant incidence of 25 W/m2 (V) =  0.461320202862

Example 5.54 Page No : 281

In [63]:
import math 

# Variables
A = 1935*10**-6;
r = 0.914;
S_angle = A/r**2;
I = 180.;

# Calculations
L_flux = I*S_angle;

# Results
print "lumnious flux = ", L_flux
print "Corresponding to lumnious flux o.417 lm and a load resistance of 800 ohm the current is 120 micro Ampere",

lumnious flux =  0.416928019765
Corresponding to lumnious flux o.417 lm and a load resistance of 800 ohm the current is 120 micro Ampere