# Chapter 5 : Primary Sensing Elements and Transducers¶

## Example 5.1 Page No : 179¶

In :
# 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 :
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 :
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 :
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 :
%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:
[<matplotlib.lines.Line2D at 0x103fd9ad0>] ## Example 5.6 Page No : 189¶

In :
# 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 :
# 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 :
# Variables
Rwga = 1./400;

# Calculations
Re = Rwga/5;

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

Resolution (mm) =  0.0005


## Example 5.9 Page No : 195¶

In :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
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 :
# 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 :
# 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 :
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 :
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 :
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 :
# 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 :
# 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 :
# 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 :
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 :
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 :
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 :
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 :
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 :
# 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 :
# 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 :
# 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 :
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 :
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 :
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 :
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