# Chapter 1 : FUNDAMENTALS OF ELECTRICITY AND DC CIRCUITS¶

## Example 1.1 Page No : 18¶

In [1]:
import math

#INPUT DATA
# details for the first wire
l1 = 1;						#length in m
R1 = 2;						#resistance in ohms
x = R1;						#say
d = 1;						#say
p = 1;						#say
d1 = d;						#say diameter in m
p1 = p;						#say specific resistance of  wire
#details for the second wire
l2 = 3;						#length in m
d = 1;						#say
p = 1;						#say
d2 = 2*d;						#say diameter in m
p2 = 2*p;						#say specific resistance of  wire

#CALCULATIONS
R1 = p1*l1/(math.pi*d*d/4);						#(R1 = p1*l1/a1), where a1 is cross sectional area of first wire with diameter d as (math.pi*d*d/4)---------------equation 1
R2 = p2*l2/(math.pi*(4*d*d)/4);						#(R2 = p2*l2/a2), where a2 is cross sectional area of  second wire with diameter 2d as (math.pi*((2*d)*(2*d))/4)-------------equation 2
#dividing equation 1 by equation 2
z = R1/R2;
R2 = x/z;
#OUTPUT
print "Thus the resistance of second wire is %1.0f ohm "%(R2);

Thus the resistance of second wire is 3 ohm


## Example 1.2 Page No : 18¶

In [2]:
import math

#INPUT DATA
l1 = 20;			#length in cm for first case
l2 = 0.4;			#length in cm for second case
w = 0.1;			#width in cm
t = 0.4;			#thickness in cm
p = 1.7*10**-6			#resistivity of copper in ohm cm
a1 = 0.1*0.4			#area(w*t) in cm**2 for first case
a2 = 0.1*20			#area(l*t) in cm**2 for second case

#CALCULATIONS
R1 = p*l1/a1;			#resistance in ohms for first case
R2 = p*l2/a2;			#resistance in ohms for second case

#OUTPUT
print "Thus the resistance in first and second cases are %g ohms and %g ohms"%(R1,R2);

Thus the resistance in first and second cases are 0.00085 ohms and 3.4e-07 ohms


## Example 1.3 Page No : 19¶

In [3]:
import math

#INPUT DATA
la = 1000;			#length  of aluminium wire in cm
da = 0.2;			#diameter in cm
pa = 2.6*10**-6;			#specific resistance of aluminium in ohm cm
pc = 1.6*10**-6;			#specific resistance of copper in ohm cm
lc = 600;			#length  of copper wire in cm
i = 2;			#current in A pasmath.sing through combination
ia = 1.25;			#current in A pasmath.sing through aluminium wire

#CALCULATIONS
ic = i-ia;			#current in A pasmath.sing through copper wire
#resistance of aluminium wire in ohms
Ra = pa*la/(math.pi*(da*da)/4);			#(Ra = pa*la/a), where a is cross sectional area of aluminum wire with diameter da
Rc = ia/ic*Ra;			#resistance of copper wire
dc = math.sqrt(4*pc*lc/Rc);			#diameter of copper wire

#OUTPUT
print "Thus the diameter of copper wire  is %1.3f cm "%(dc);
#note:The answer given for diameter in text book is wrong.please check the calculations

Thus the diameter of copper wire  is 0.167 cm


## Example 1.4 Page No : 20¶

In [5]:
#INPUT DATA
l = 10000.;			#length drawn from 10cc of copper in cm
p = 1.7*10**-6;			#Resistivity of copper in ohm cm
v = 10.;			#volume of copper in cc
s1 = 10.;			#square sheet side in second case in cm

#CALCULATIONS
a = v/l;			#area of cross-section in cm**2 in first case
R1 = p*l/a;			#resistance of wire in first case in ohm
a1 = s1*s1;			#area of cross-section in cm**2 in second case
l1 = v/a1;			#thickness in case 2 in cm
R2 = p*l1/a1;			#resistance of wire in second case in ohm

#OUTPUT
print "Thus the resistance in first and second cases are %g ohms and %g ohms"%(R1,R2);

Thus the resistance in first and second cases are 17 ohms and 1.7e-09 ohms


## Example 1.5 Page No : 21¶

In [6]:
#INPUT DATA
t1 = 40.;			#temperature in degree centigrade
t2 = 100.;			#temperature in degree centigrade
R1 = 3.146;			#resistance of platinum coli at t1
R2 = 3.767;			#resistance of platinum coli at t2

#CALCULATIONS
x = R1/R2;
a0 = ((R1-R2)/(R2*t1-R1*t2));			#temperature coefficient at 0 degree centigrade
R0 = R1/(1+(a0*t1));			#resistance at zero degree centigrade
a40 = a0/(1+(a0*t1));			#temperature coefficient at 40 degree centigrade

#OUTPUT
print "Thus the temperature coefficient at 0 degree centigrade, resistance at zero degree centigrade, \
\ntemperature coefficient at 40 degree centigrade are %f /degree centigrade , %f ohms, %f /degree centigrade\
\n respectively "%(a0,R0,a40);

Thus the temperature coefficient at 0 degree centigrade, resistance at zero degree centigrade,
temperature coefficient at 40 degree centigrade are 0.003788 /degree centigrade , 2.732000 ohms, 0.003290 /degree centigrade
respectively


## Example 1.6 Page No : 21¶

In [8]:
#INPUT DATA
t1 = 12;			#temperature in degree centigrade
t2 = 50;			#temperature in degree centigrade
R1 = 0.4;			#copper coil resistance in ohms
a0 = 0.004;			#temperature coefficient of copper at zero degree centigrade

#CALCULATIONS
a12 = 1/((1/a0)+t1);			#temperature coefficient at 12 degree centigrade
R2 = R1*(1+(a12*(t2-t1)));			#resistance of copper wire  in ohm at 52 degree centigrade

#OUTPUT
print "Thus the resistance copper wire at 52 degree centigrade is %1.2f  ohm "%(R2);

Thus the resistance copper wire at 52 degree centigrade is 0.46  ohm


## Example 1.7 Page No : 22¶

In [9]:
#INPUT DATA
t1 = 20;			#temperature in degree centigrade
R1 = 45;			#shunt motor resistance at t1
R2 = 48.5;			#new shunt resistance at t2
a0 = 0.004;			#temperature coefficient of resistance at 0 degree centigrade

#CALCULATIONS
x = R1/R2;
t2 = ((1+(a0*t1)-x)/(a0*x));

#OUTPUT
print "Thus the temperature for new resistance is %d degree centigrade "%(t2);

Thus the temperature for new resistance is 41 degree centigrade


## Example 1.8 Page No : 22¶

In [10]:
#INPUT DATA
V = 180.;			#supply voltage in volts
I1 = 4.;			#initial current of coil in A
t1 = 20.;			#initial temperature
I2 = 3.4;			#new decreased current of coil in A at temperature t2
a0 = 0.0043;			#temperature coefficient in per degree centigrade

#CALCULATIONS
R1 = V/I1;			#initial resistance of coil in ohms
R2 = V/I2;			#resistance of coil after some time in ohms
x = R1/R2;
t2 = (1+(a0*t1)-x)/(a0*x);
t = t2-t1;			#temperature rise

#OUTPUT
print "Thus the temperature rise is %2.2f degree centigrade "%(t);

Thus the temperature rise is 44.57 degree centigrade


## Example 1.9 Page No : 23¶

In [12]:
#INPUT DATA
t2 = 2750.;			#temperature in degree centigrade for tungsten lamp
P = 150.;			#power in watts
V = 230.;			#voltage in volts
t1 = 16.;			#temperature in degree centigrade
a0 = 0.0047;			#temperature coefficient of tungsten in per degree centigrade

#CALCULATIONS
R2 = (V*V)/P;
a1 = 1/((1/a0)+t1);			#temperature coefficient of resismath.tant at 16 degree centigrade
R2 = (V*V)/P;			#resistance of the filament of the lamp under normal working condition
R1 = R2/(1+(a1*(t2-t1)));			#resistance of copper wire  in ohm at 52 degree centigrade
I2 = V/R2;			#normal current taken by lamb
I1 = V/R1;			#current taken at the moment of switching on

#OUTPUT
print "Thus the normal current taken by lamb and current taken at the moment of switching\
\n on are %1.4f A and %1.4f A respectively "%(I2,I1);

Thus the normal current taken by lamb and current taken at the moment of switching
on are 0.6522 A and 8.4464 A respectively


## Example 1.10 Page No : 23¶

In [13]:
#INPUT DATA
m1 = 2.;			#mass of water in kg
theta1 = 20.;			#temperature 20 degree centigrade
theta2 = 100.;			#temperature 100 degree centigrade(boiling point of water)
t = 1./10;			#time taken to boil water in hr
x = 40.;			#math.cost of energy of 1kwh in paise for one unit
y = 12.;			#math.cost of energy consumed
S = 1.;			#specific heat of water

#CALCULATIONS
H = m1*S*(theta2-theta1);			#heat energy required to raise temperature from theta1 to theta2 in kcals
H = H/860;			#heat energy in Kwh
E = (12./40);			#electrical energy or input energy to kettle in Kwh
n = H/E*100;			#efficiency of kettle in percentage;
P = E/t;			#power rating of kettle

#OUTPUT
print "Thus the efficiency of kettle in percentage  and power rating of kettle is %d and %1.0f Kw"%(n,P);

Thus the efficiency of kettle in percentage  and power rating of kettle is 62 and 3 Kw


## Example 1.11 Page No : 24¶

In [14]:
#INPUT DATA
m = 2.;			#mass of water in kg
theta1 = 20.;			#temperature 20 degree centigrade
theta2 = 100.;			#temperature 100 degree centigrade(boiling point of water)
t = 0.25;			#time taken to boil water in hr
V = 240.;			#power supply in volts
n = 80.;			#efficiency of kettle in percentage
S = 1.;			#specific heat of water

#CALCULATIONS
H = m*S*(theta2-theta1);			#output energy from the kettle in kcal
H = H/860;			#output energy from the kettle in kwh
n = n/100;
E = H/n;			#electrical energy or input energy to kettle in Kwh
P = E/t;			#power rating of kettle in Kw
P = P*1000;			#power rating of kettle in w
R = (V*V)/P;			#resistance of heating element in ohms

#OUTPUT
print "Thus the resistance of heating element is %2.2f ohms"%(R);

Thus the resistance of heating element is 61.92 ohms


## Example 1.12 Page No : 24¶

In [15]:
#INPUT DATA
m = 20.;			#mass of aluminium in kg
S = 0.896;			#specific heat of aluminium in KJ/Kg degree centigrade
L = 402.;			#latent heat of fusion of aluminium in KJ/Kg
theta2 = 657.;			#final temperature
theta1 = 20.;			#initial temperature(assumed)
P = 25.;			#power of furnace in Kw
n = 80.;			#efficiency of kettle in percentage

#CALCULATIONS
H = m*S*(theta2-theta1)+(m*L);			#heat energy required to melt aluminium or energy output from the furnace in Kj
H = H/4.186;			#heat energy required to melt aluminium or energy output from the furnace in Kcal
H = H/860;			#heat energy required to melt aluminium or energy output from the furnace in KWh
n = n/100;
E = H/n;			#electrical energy or input energy to kettle in Kwh
t = E/P;			#time taken to melt the aluminium in hr
t = t*60;			# time taken to melt the aluminium in min

#OUTPUT
print "Thus the time taken to melt the aluminium is %2.2f min"%(t);

Thus the time taken to melt the aluminium is 16.21 min


## Example 1.13 Page No : 25¶

In [16]:
#INPUT DATA
m = 80000.;			#mass of water lifted by pump in Kg/min
g = 9.81;			#gravity consmath.tant in m/sec**2
h = 2.;			#pump is in operation for two hours a day
d = 30.;			#pump is in operation for 30 days
T = h*d;			#total time for which pump is in operation in hrs
n = 70.;			#efficeincy in percentage
h = 12.;			#the height in m to which pump lifts water
C = 50.;			#math.cost of energy in paise/Kwh

#CALCULATIONS
P = m*g*h;			#potential energy possessed by water per minute or workdone by motor pump/minute measured in joules
P = P/60;			#potential energy possessed by water per minute or workdone by motor pump/minute  measured in joules/sec or watts.
O = P/1000;			#output power of motor in Kw
n = n/100;
E = O/n;			#input power of motor in Kw
Et = E*T;			#total energy supplied or energy consumption in Kwh
C = C/100;			#math.cost of energy in Rs/Kwh
Ct = C*Et;			#Total math.cost of energy

#OUTPUT
print "Thus the total math.cost of energy is  Rs %4.0f"%(Ct);

Thus the total math.cost of energy is  Rs 6727


## Example 1.14 Page No : 26¶

In [19]:
#INPUT DATA
P = 100.;			#power of power station in MW
g = 9.81;			#gravity consmath.tant in m/sec**2
h = 200.;			#effective head of power station in m
n = 80.;			#efficiency of station in percentage
t = 10.;			#operation time of power station

#CALCULATIONS
E1 = P*t;			# energy output from the station in 10 hours measured in MWh
n = n/100;
E2 = P*t/n;			#energy input to the station in 10 hours measured in MWh
E2 = E2*10**6*60*60;			#energy input to the station in 10 hours measured in Wsec or joules
#energy input to the station is equal to potential energy supplied by water to station
m = E2/(g*h);			#mass in kg of water used
d = 1000;			#density of water in kg/m**3
V = m/d;			#volume of water used in 10 hours

#OUTPUT
print "Thus the volume of water used in 10 hours is %e cubic metre"%(V);

Thus the volume of water used in 10 hours is 2.293578e+06 cubic metre


## Example 1.15 Page No : 26¶

In [20]:
#INPUT DATA
I = 20.;			#current in A
V = 8.;			#supply voltage in V
t = 3600.;			#1hr = 3600sec
m = 1000;			#mass in kg(1 tonne =  1000 kg)
#kinetic energy  =  energy dissipated in the resistance----eqn(1)

#CALCULATIONS
E = V*I*t;			# energy dissipated in resistance in joules
v = math.sqrt(E/(0.5*m));			#kinetic energy possesed by body(K = 0.5*m*v*v) and umath.sing eqn(1),we found out velocity in m/sec

#OUTPUT
print "Thus the velocity is %2.2f m/sec"%(v);

Thus the velocity is 33.94 m/sec


## Example 1.16 Page No : 27¶

In [21]:
#INPUT DATA
I = 7.9;			#current in A
V = 240.;			#supply voltage in V
t = 55.;			#temperature in degree centigrade
a0 = 0.00029;			#temperature coefficient in ohm/ohm/degree centigrade
l = 15.6;			#length of wire in m
a = 12.;			#cross-sectional area in mm**2

#CALCULATIONS
R = V/I;			#resistance of wire in ohm
p = R*a/l;			#resistivity of wire in ohm metre
Rt = R*(1+(a0*t));			#resistance at 55 degree centigrade in ohm
I1 = V/Rt;			#current through wire at temperature 55 degree centigrade in A

#OUTPUT
print "Thus the resistivity and current through wire at temperature 55 degree centigrade are %2.2f micro\
\n ohm meter and %2.2f A respectively"%(p,I1);

Thus the resistivity and current through wire at temperature 55 degree centigrade are 23.37 micro
ohm meter and 7.78 A respectively


## Example 1.17 Page No : 27¶

In [22]:
#INPUT DATA
R1 = 0.031;			#resistance of wire in ohm
d1 = 11.7;			#diameter of wire in mm in case 1
r1 = d1/2;			#radius of wire in mm in case 1
d2 = 5;			#diameter of wire in mm in case 2
r2 = d2/2;			#radius of wire in mm in case 2
# we know that resistance is inversely proportional to square of area of cross-section

#CALCULATIONS
R2 = R1*(((math.pi*r1*r1)/(math.pi*r2*r2)))**2;			#resistance of wire in case 2

#OUTPUT
print "Thus the new resistance of wire is %1.4f ohms"%(R2);

Thus the new resistance of wire is 2.2692 ohms


## Example 1.18 Page No : 27¶

In [24]:
#INPUT DATA
p20 = 1.724*10**-8;			#specific resistance of copper in ohm m
a = 0.0043;			#temperature coefficient of copper at 0 degree centigrade measured in per degree centigrade
r1 = 8;			#inner radius of copper circular ring in cm
r2 = 6;			#axial thickness in cm
r3 = 4;			#radial thickness in cm
a1 = r2*r3*10**-4;			#area of cross-section of ring in m**2
r2 = r2*2;
l = math.pi*((r1+r2)/2)/100;			#length of semicircular ring between faces in m
t1 = 20;			#temperature 20 degree centigrade
t2 = 50;			#temperature 50 degree centigrade

#CALCULATIONS
R20 = p20*(l/a1);			#resistance of ring at 20 degree centigrade in ohm
R50 = R20*((1+(a*t2))/(1+(a*t1)));			#resistance of ring at 50 degree centigrade in ohm

#OUTPUT
print "Thus the resistance of wire at 50 degree centigrade is %g ohms"%(R50);

Thus the resistance of wire at 50 degree centigrade is 2.52477e-06 ohms


## Example 1.19 Page No : 28¶

In [25]:
#INPUT DATA
l1 = 0.5;			#length of copper rod in m
a = 0.00426;			#temperature coefficient of copper measured in per degree centigrade
R1 = 4.25*10**-4;			#resistance of wire at 15 degree centigrade in ohm
d1 = 5*10**-3;			#diameter of copper rod in m in case 1
r1 = 0.5*d1;			#radius of copper rod in m in case 1
a1 = math.pi*((r1)**2);			#area of cross-section in m**2 in case 1
t1 = 15;			#temperature in degree centigrade
t2 = 50;			#temperature in degree centigrade

#CALCULATIONS
p = R1*a1/l1;			#resistivity in ohm-m
d2 = 1*10**-3;			#diameter of copper rod in m in case 2
r2 = d2/2;			#radius of copper rod in m in case 2
a2 = math.pi*(r2)**2;			#area of cross-section in m**2 in case 2
R15 = (a1/a2)**2*R1;			#resistance at 15 degree centigrade
R50 = R15*((1+(a*t2))/(1+(a*t1)));

#OUTPUT
print "Thus the resistance of wire at 50 degree centigrade is %1.4f ohm"%(R50);

Thus the resistance of wire at 50 degree centigrade is 0.3029 ohm


## Example 1.20 Page No : 28¶

In [26]:
#INPUT DATA
l1 = 7.5;			#length of aluminium wire in m
d1 = 1*10**-3;			#diameter of aluminium wire in m
r1 = 0.5*d1;			#radius of aluminium wire in m
a1 = math.pi*((r1)**2);			#area of cross-section in m**2 for aluminium wire
p1 = 0.028;			#resistivity of aluminium in micro ohm-m
l2 = 6;			#length of copper wire in m
p2 = 0.017;			#resistivity of copper in micro ohm-m
I = 5;			#current through parallel combination in A
I1 = 3;			#current through aluminium wire in A
I2 = I-I1;			#current through copper wire in A

#CALCULATIONS
R1 = p1*l1/a1;			#resistance of aluminium wire in ohm
V1 = I1*R1;			#voltage drop across the end of Al wire in V
#math.since the wires are connected in parallel,so V1 = V2
a2 = I2*p2*l2/V1;			#area of cross-section in m**2 for copper wire
d2 = math.sqrt(4*a2/math.pi);			#diameter of copper wire in m

#OUTPUT
print "Thus the diameter of copper wire is %g m"%(d2);

Thus the diameter of copper wire is 0.000569043 m


## Example 1.22 Page No : 29¶

In [27]:
#INPUT DATA
R20 = 100.;			#resistance of coil at 20 degree centigrade in ohms
R45 = 110.;			#resistance of coil at 45 degree centigrade in ohms
Rt = 124.;			#resistance of coil at t degree centigrade in ohms
t1 = 20.;			#temperature in degree centigrade
t2 = 15.;			#temperature in degree centigrade
a = R45/R20;

#CALCULATIONS
a0 = (a-1)/(45-(20*a));			#temperature coefficient of coil at 0 degree centigrade
x = (Rt/R20);
t = (x)*(1+(a0*t1));
t = t-1;
t = (t)*(1/a0);			#			#temperature of coil when Rt = 124 ohms measured in degree centigrade
deltat = t-t2;			#mean temperature rise

#OUTPUT
print "Thus the mean temperature rise is %2.0f degree centigrade"%(deltat);

Thus the mean temperature rise is 65 degree centigrade


## Example 1.23 Page No : 30¶

In [28]:
#INPUT DATA
R20 = 18.;			#resistance of coil at 20 degree centigrade in ohms
R50 = 20.;			#resistance of coil at 50 degree centigrade in ohms
Rt = 21.;			#resistance of coil at t degree centigrade in ohms
t1 = 20.;			#temperature in degree centigrade
t2 = 50.;			#temperature in degree centigrade
t3 = 15.;			#temperature in degree centigrade
a = R50/R20;

#CALCULATIONS
a0 = (a-1)/(50-(20*a));			#temperature coefficient of coil at 0 degree centigrade
x = (Rt/R50);
t = (x)*(1+(a0*t2));
t = t-1;
t = (t)*(1/a0);			#			#temperature of coil when Rt = 21 ohms measured in degree centigrade
deltat = t-t3;			#mean temperature rise

#OUTPUT
print "Thus the mean temperature rise is %2.0f degree centigrade"%(deltat);

Thus the mean temperature rise is 50 degree centigrade


## Example 1.24 Page No : 40¶

In [29]:
#INPUT DATA
R1 = 4.;			#resistance in ohms
R2 = 6.;			#			#resistance in ohms
I = 30.;			#current through parallel combination in A

#CALCULATIONS
I1 = I*(R2/(R1+R2));			#current through resistor1 in A
I2 = I-I1;			#current through resistor2 in A

#OUTPUT
print "Thus the current through resistor1 and resistor2 are %d A and %d A  respectively"%(I1,I2);

Thus the current through resistor1 and resistor2 are 18 A and 12 A  respectively


## Example 1.25 Page No : 41¶

In [30]:
#INPUT DATA
R1 = 2.;			#resistance1 in ohms
R2 = 3.;			#resistance2 in ohms
R3 = 4.;			#resistance3 in ohms
R4 = 5.;			#resistance4 in ohms
P = 100.;			#total power absorbed in watts

#CALCULATIONS
RT = ((R2*R3*R4)+(R1*R3*R4)+(R1*R2*R4)+(R1*R2*R3))/(R1*R2*R3*R4);
RT = 1/RT;			#equivalent resistance of parallel combination of R1,R2,R3,R4 Resistors
V = math.sqrt(P*RT);			#voltage in volts that has to be applied to absorb 100w of power

#OUTPUT
print "Thus the voltage in volts that has to be applied to absorb 100w of power is %1.3f V "%(V);

Thus the voltage in volts that has to be applied to absorb 100w of power is 8.827 V


## Example 1.26 Page No : 41¶

In [31]:
#INPUT DATA
V = 230;			#supply voltage in volts
I1 = 12.;			#initial current in A
I2 = 16;			#final current in A

#CALCULATIONS
I = I2-I1;			#current through the resistance placed in parallel in A
R = V/I;			#resistance in ohms by ohm's law

#OUTPUT
print "Thus the resistance placed in parallel is %2.1f ohm "%(R);

Thus the resistance placed in parallel is 57.5 ohm


## Example 1.27 Page No : 41¶

In [32]:
#INPUT DATA
I = 12.1;			#current in A entering the parallel combination of resistors
I1 = 7.2;			#current in A in resistor 1
R1 = 50.;			#resistance1 in ohm
R2 = 100.;			#resistance2 in ohm

#CALCULATIONS
V = I1*R1;			#supply voltage in volts by ohms law(V = I*R)
I2 = V/R2;			#current through R2 in A  by ohms law
I3 = I-I1-I2;			#current through resistance3 R3 in A by ohms law
R3 = V/I3;			#resistance in ohm

#OUTPUT
print "Thus the value of third resistance placed is %3.2f ohm "%(R3);

Thus the value of third resistance placed is 276.92 ohm


## Example 1.28 Page No : 42¶

In [33]:
#INPUT DATA
R1 = 3.6;			#resistance in ohm
R2 = 4.56;			#resistance in ohm
RT = 6.;			#resistance in ohm

#CALCULATIONS
X = RT-(R2);
R3 = (X*R1)/(R1-X);

#OUTPUT
print "Thus the value of third resistance placed is %1.1f ohm "%(R3);

Thus the value of third resistance placed is 2.4 ohm


## Example 1.29 Page No : 42¶

In [34]:
#INPUT DATA
P = 70.;			#total power dissipated in circuit in watts
V = 22.;			#applied voltage in volts
I = P/V;			#total current through the circuit in Amps
R1 = 12.;			#resistance 1 of parallel combination in ohms
R2 = 8.;			#resistance 2 of parallel combination in ohms

#CALCULATIONS
RP = (R1*R2)/(R1+R2);			#equivalent resistance of parallel combination in ohms
VP = I*RP;			#voltage across parallel combination in volts
VR = V-VP;			#voltage across the resistance R			# in volts
R3 = VR/I;			#by ohm's law

#OUTPUT
print "Thus the value of third resistance placed is %1.2f ohms "%(R3);

Thus the value of third resistance placed is 2.11 ohms


## Example 1.30 Page No : 43¶

In [35]:
#INPUT DATA
P = 70.;			#total power dissipated in circuit in watts
V1 = 6.;			#math.since applied voltage  E is 6V,as per the characteristics of parallel circuit P.D across R1 is
V2 = 6.;			#V1 = V2,in volts
R1 = 12.;			#resistance1 in parallel combination in ohms
R2 = 6.;			#resistance2 in parallel combination in ohms
R3 = 6.25			#resistance3 in series with parallel combination in ohms
I1 = V1/R1;			# current through the resistance R1 in Amps
I2 = V2/R2;			#current through the resistance R2 in Amps
r = 0.25;			#internal resistance in ohm

#CALCULATIONS
I = I1+I2;			#total current through parallel combination
E = (I*r)+(I*R3)+V2;			#emf of battery in Volts

#OUTPUT
print "Thus the value of emf of battery in Volts is %2.2f volts "%(E);

Thus the value of emf of battery in Volts is 15.75 volts


## Example 1.31 Page No : 44¶

In [36]:
#INPUT DATA
E = 12.;			#emf of battery in volts
R1 = 3.;			#resistance1 in parallel combination in ohms
R2 = 4.;			#resistance2 in parallel combination in ohms
R3 = 6.;			#resistance3 in parallel combination in ohms
R4 = 4.;			#resistance4 in series with parallel combination in ohms
r = 6.;			#internal resistance in ohm

#CALCULATIONS
RP = ((R2*R3)+(R3*R1)+(R1*R2))/(R1*R2*R3);
RP = 1/RP;			#equivalent resistance of parallel combination in ohms
RT = RP+R4+r;			#total circuit resistance in ohms
I = E/RT;			#total circuit current in A
V = E-(I*r);			#terminal voltage of battery in volts

#OUTPUT
print "Thus the terminal voltage of battery is %1.3f volts "%(V);

Thus the terminal voltage of battery is 5.647 volts


## Example 1.32 Page No : 45¶

In [37]:
#INPUT DATA
V = 24;			#supply voltage of battery in volts
Rab = 13.;			#resistance between A and B points in ohms
Rbc = 11.;			#resistance between B and C points in ohms
Rbe = 18.;			#resistance between B and E points in ohms
Rce = 14.;			#resistance between C and E points in ohms
Red = 9.;			#resistance between E and D points in ohms
Rcd = 5.;			#resistance between C and D points in ohms
Rae = 22;			#resistance between A and E points in ohm
Rx = Rae;
Ry = Rbe;
Raf = 1;			#resistance between A and F points in ohms

#CALCULATIONS
Rce = ((Rcd+Red)*(Rce))/(Rcd+Red+Rce);			#equivalent resistance of Rce in ohms
Rbe = ((Rbc+Rce)*(Rbe))/(Rbc+Rce+Rbe);			#equivalent resistance of Rbe in ohms
Rae = ((Rab+Rbe)*(Rae))/(Rab+Rbe+Rae);			#equivalent resistance of Rae in ohms
RT = Rae+Raf;			#total equivalent circuit resistance in ohms
Iaf = V/RT;			#current through resistance Raf in A
Vae = V-(Iaf*Raf);			#P.D across AE in volts
Iae = Vae/Rx;			#current in AE in A
Iab = Iaf-Iae;			#current in AB in A
Vab = Rab*Iab;			#P.D across AB in volts
Vbe = Vae-Vab;			#Voltage across branch BE in volts
Pbe = ((Vbe)**2)/(Ry);			#power absorbed in branch Be in watts
Ibe = Vbe/Ry;			#current in BE in A
Ibc = Iab-Ibe;			#current in BC in A
Icde = (Ibc)/(2);			#current in CDE in A
Vcd = Icde*(Rcd);			#P.D across CD

#OUTPUT
print "Current in branch AF is %d A  \
\nPower absorbed in BE is %1.1f watts  \
\nP.D across CD is %1.2f volts "%(Iaf,Pbe,Vcd);

Current in branch AF is 2 A
Power absorbed in BE is 4.5 watts
P.D across CD is 1.25 volts


## Example 1.33 Page No : 46¶

In [38]:
from numpy.linalg import solve

#INPUT DATA
V1 = 25.;			#supply voltage1 of battery in volts
V2 = 45.;			#supply voltage2 of battery in volts
R1 = 6.;			#resistance1 in ohms
R3 = 4.;			#resistance2 in ohms
R2 = 3.;			#resistance3 in ohms
#let I1 be the current in loop1 and I2 current be in loop2

#CALCULATIONS
#V1 = ((R1+R3)*(I1)-(R3*I2));			#applying KVL in loop1 -------------eqn(1)
#V2 = ((R3)*(I1)-(R2+R3)*(I2));			#applying KVL in loop2 -------------eqn(2)
#solving both eqn(1) and eqn(2)
a = [[(R1+R3),-R3],[(R3),-(R2+R3)]]
b = [[V1],[-V2]]
c = solve(a,b)#solve(a,b)			#ax = b
c1 = c[0];			#c1 is current in branch FABC measured in A
c2 = c[1];			#c2 is current in branch CDEF measured in A
c3 = c1-c2;			#current in branch CF in A

#OUTPUT
print "Current in R1 is %1.4f A  \
\ncurrent in R2 is %2.3f A   \
\ncurrent in R3 is %1.3f A "%(c1,c2,c3);

Current in R1 is 6.5741 A
current in R2 is 10.185 A
current in R3 is -3.611 A


## Example 1.34 Page No : 46¶

In [39]:
import math
from numpy.linalg import solve

#INPUT DATA
V = 4.5;			#supply voltage of battery in volts
RAB = 1000.;			#resistance between A and B points in ohms
RBC = 100.;			#resistance between B and C points in ohms
RAD = 5000.;			#resistance between A and D points in ohms
RCD = 450.;			#resistance between C and D points in ohms
Rg = 500.;			#resistance of galvanometer in ohms

#let I1 be the current across RAB and I1-Ig across RBC and I2 across RAD and I2+Ig across RCD and I be the total current
#where I = I1+I2
#CALCULATIONS
#(-(RAB*I1)-(Rg*Ig)+(RAD*I2)) = 0;			#applying KVL to loop ABDA -------------eqn(1)
#(-(RBC*I1)+((Rg+RCD+RBC)*(Ig))+(RCD*I2)) = 0;			#applying KVL to loop BCDB -------------eqn(2)
#((RAD+RCD)*I2)+(RCD*Ig)) = V;			#applying KVL to loop EADCFE-------------eqn(3)
#solving  eqn(1),eqn(2) and eqn(3)
b = [[0],[0],[V]];
c = solve(a,b)   #solve(a,b)			#ax = b
I1 = c[0];			#c1 is current in branch FABC measured in A
Ig = c[1];			#c2 is current in branch CDEF measured in A
I2 = c[2];			#current in branch CF in A

#OUTPUT
print "Current through galvanometer is %g A  since the answer is positive our assumed direction is correct "%(Ig);

Current through galvanometer is 3.73909e-05 A  since the answer is positive our assumed direction is correct


## Example 1.35 Page No : 47¶

In [40]:
import math
from numpy.linalg import solve

#INPUT DATA
V1 = 8.;			#supply voltage of battery in loop1 in volts
V2 = 4.;			#supply voltage of battery in loop2 in volts
RED = 200.;			#resistance between E and D points in ohms
RAD = 20.;			#resistance between A and D points in ohms
RCD = 50.;			#resistance between C and D points in ohms

#let I1 be the current across path AFED and I2 across AD and I1-I2 across  path DCBA
#CALCULATIONS
#((RCD*I1)-((RAD+RCD)*I2)) = 4;			#applying KVL to loop ADCBA -------------eqn(1)
#((RED*I1)+(RAD*I2)) = 8;			#applying KVL to loop AFEDA -------------eqn(2)
#solving  eqn(1)and eqn(2)
b = [[4],[8]];
c = solve(a,b)			#ax = b
I1 = c[0];			#c1 is current across path AFED in A
I2 = c[1];			#c2 is current across AD in A

#OUTPUT
print "Current in 20 ohm resistor  is %f A  math.since the answer is negative,the current actually flows from A to D "%(I2);

Current in 20 ohm resistor  is -0.026667 A  math.since the answer is negative,the current actually flows from A to D


## Example 1.36 Page No : 48¶

In [41]:
import math

#INPUT DATA
Rs = 25.;			#total resistance when two resistances are connected in series in ohms
Rp = 6.;			#total resistance when two resistances are connected in parallel in ohms
#let individual resistances be R1 and R2 ohms

#CALCULATIONS
#Rs = (R1+R2)---eqn(1)
#Rp = ((R1*R2)/(R1+R2))---eqn(2)
#let (R1*R2) = x
#let(R1-R2) = y
#solving  eqn(1)and eqn(2)
x = Rs*Rp;			#in ohms
y = math.sqrt((Rs)**2-(4*x));			#eqn---(3)
#solving eqn(1) and eqn(3)
z = Rs+y;
R1 = z/2;			#resistance1 in ohms
R2 = Rs-R1;			#resistance2 in ohms

#OUTPUT
print "Thus the individual resistances are R1 = %d ohms and R2 = %d ohms "%(R1,R2);

Thus the individual resistances are R1 = 15 ohms and R2 = 10 ohms


## Example 1.37 Page No : 48¶

In [42]:
import math

#INPUT DATA
P = 16.;			#total power dissipated in circuit in Watts
R1 = 4.;			#resistance R1 in Ohms
R2 = 2.;			#resistance R2 in Ohms
R3 = 8.;			#resistance R3 in Ohms
V = 8.;			#supply voltage in volts
#let resistance parallel to R1 is R ohms

#CALCULATIONS
Reff = (((V)**2)/P);			#total effective resistance of circuit in ohms
x = ((R2*R3)/(R2+R3));			#effective resistance of 2nd parallel circuit in ohms
z = (Reff-x);			#effective resistance of 1st parallel circuit where z = ((R1*R)/(R1+R)) in ohms------eqn(1)
#solving for R in eqn(1)
R = (R1*z)/(R1-z);
Reff = ((R1*R)/(R1+R))+(x);			#in ohms
I = V/Reff;			#total current in A

print "Thus the total current is I = %d A "%(I);

Thus the total current is I = 2 A


## Example 1.38 Page No : 49¶

In [43]:
import math

#INPUT DATA
R1 = 2.;			#resistance R1 in Ohms
R2 = 12.;			#resistance R2 in parallel circuit measured in ohms
R3 = 20.;			#resistance R3 in parallel circuit measured in ohms
R4 = 30.;			#resistance R4 in parallel circuit measured  in ohms
R5 = 2.;			#resistance R5 in ohms
V = 100.;			#supply voltage in volts

#CALCULATIONS
Reff = (R1)+((1)/((1/R2)+(1/R3)+(1/R4)))+(R5);			#total effective resistance of circuit in ohms
I = V/Reff;			#total current in A
# let individual currents in 3 parallel resistance network be I1,I2,I3 respectively
#Then I1+I2+I3 = I---eqn(1)
#where I2 = (I1*R1/R2) and I3 = (I1*R1/R3)
#solving for I1 in eqn(1)
I1 = I/2;			#in A
I2 = (I1*R2/R3);			#in A
I3 = (I1*R2/R4);			#in A

# Results
print "I1 = %d A  I2 = %1.0f A  I3 = %1.0f A"%(I1,I2,I3);

I1 = 5 A  I2 = 3 A  I3 = 2 A


## Example 1.39 Page No : 49¶

In [45]:
#INPUT DATA
V = 100.;			#supply voltage in volts
I = 10.;			#total current in A
P1 = 600.;			#power dissipated in coil in Watts

#CALCULATIONS
#Reff = ((R1*R2)/(R1+R2)) is total effective resistance of circuit in ohms----eqn(1)
Reff = V/I;			#total effective resistance of circuit in ohms
R1 = ((V)**2)/(P1);			#in ohms----eqn(2)
#solving for R2 in eqn(1)
R2 = ((Reff*R1)/(R1-Reff));			#in ohms

# Results
print "R1 = %2.2f Ohms  R2 = %1.0f Ohms "%(R1,R2);

R1 = 16.67 Ohms  R2 = 25 Ohms


## Example 1.40 Page No : 50¶

In [46]:
import math
#INPUT DATA
R1 = 5.;			#resistance in ohms
P = 20.;			#power dissipated in R1 in Watts
R2 = 10.;			#resistance parallel to R1

#CALCULATIONS
I1 = math.sqrt(P/R1);			#current through R1 in A
I = ((R1+R2)/(R2))*(I1);			#total supply current in A

# Results
print "I = %d A"%(I);

I = 3 A


## Example 1.41 Page No : 50¶

In [47]:
import math

#INPUT DATA
I1 = 2.;			#current through R1 in A
I3 = 1.5;			#current through R3 in A
I5 = 0.5;			#current through R5 in A
P2 = 75.;			#power dissipated in R2 in W
P4 = 30.;			#power dissipated in R4 in W
V = 200;			#supply voltage in volts
#let the current through R2 and R4 be I2 and I4 respectively

#CALCULATIONS
I2 = I1-I3;			#current through R2 in A
I4 = I3-I5;			#current through R4 in A
R2 = P2/(I2)**2;			#resistance R2 in Ohms
R4 = P4/(I4)**2;			#resistance R4 in Ohms
R5 = (R4*I4)/(I5);			#resistance R5 in Ohms
#(R1*I1)+(R2*I2) = 200
#(R3*I3)+(R4*I4) = (R2*I2)
R1 = ((V-(R2*I2))/I1);			#resistance R1 in Ohms
R3 = ((R2*I2)-(R4*I4))/(I3);			#resistance R3 in Ohms

#OUTPUT
print "R1 = %d ohms  R2 = %d ohms  R3 = %d ohms  R4 = %d ohms  R5 = %d ohms "%(R1,R2,R3,R4,R5);

R1 = 25 ohms  R2 = 300 ohms  R3 = 80 ohms  R4 = 30 ohms  R5 = 60 ohms


## Example 1.42 Page No : 50¶

In [48]:
import math

#INPUT DATA
VA = 0.2;			#voltage across ammeter A in Volts
VB = 0.3;			#voltage across ammeter B in volts
I = 20.;			#total current in A

#CALCULATIONS
RA = VA/I;			#resistance through ammeter A in ohms
RB = VB/I;			#resistance through ammeter B in ohms
IA = ((RB*I)/(RA+RB));			#current through ammeter A in amps
IB = I-IA;			#current through ammeter B in amps

#OUTPUT
print "IA = %1.0f Amps  IB = %d Amps  "%(IA,IB);

IA = 12 Amps  IB = 8 Amps


## Example 1.43 Page No : 51¶

In [49]:
import math
#Chapter-1, Example 1.43, Page 51

#INPUT DATA
R1 = 10.;			#resistance R1 in ohms
R2 = 20.;			#resistance R2 in ohms
R3 = 40.;			#resistance R3 in ohms
#after certain manipulations the Resultant network can be evaluated as parallel combinaton of R1,R2,R3

#CALCULATIONS
RAD = 1/((1/R1)+(1/R2)+(1/R3));			#Resultant resistance in Ohms

#OUTPUT
print "Resultant resistance RAD is %1.3f ohms"%(RAD);

Resultant resistance RAD is 5.714 ohms


## Example 1.44 Page No : 51¶

In [50]:
import math
#Chapter-1, Example 1.44, Page 51

#INPUT DATA
V = 200.;			#supply voltage in volts
I = 25.;			#total current in A
P1 = 1500.;			#power dissipated in watts

#CALCULATIONS
R1 = (V)**2/(P1);			#resistance R1 in Ohms
Reff = (V)/(I);			#total effective resistance of R1 and R2 in parallel in Ohms
R2 = (Reff*R1)/(R1-Reff);			#resistance R2 in Ohms

#OUTPUT
print "R1 = %2.3f ohms  R2 =  %2.3f ohms"%(R1,R2);

R1 = 26.667 ohms  R2 =  11.429 ohms


## Example 1.45 Page No : 52¶

In [51]:
import math
#Chapter-1, Example 1.45, Page 52

#INPUT DATA
V = 15.;			#supply voltage in volts
VAB = 5.;			#voltage across AB in volts
R1 = 5.;			#resistance in ohms
R2 = 10.;			#resistance in ohms
R3 = 10.;			#resistance in ohms

#CALCULATIONS
VAC = ((R1)/(R1+R3))*V;			#Volatge across AC terminals in Volts
#VBC = (((R)/(R+2))*V)--------------eqn(1) by ohm's law
#VAB = (VAC-((VBC)*(R2-(((R1+R2)*R)/(R+2)))---------------eqn(2) by ohm's law
#solving equation 2 with Vab = 5V
R = 10./10;			#resistance R in ohms

#OUTPUT
print "R = %d ohms"%(R);

R = 1 ohms


## Example 1.46 Page No : 52¶

In [52]:
import math
#Chapter-1, Example 1.46, Page 52

#INPUT DATA
Ra = 4.;			#resistance in ohms
Rb = 9.;			#resistance in ohms
Rc = 18.;			#resistance in ohms
Rd = 2.;			#resistance in ohms
Re = 7.;			#resistance in ohms
Rf = 15.;			#resistance in ohms
V = 125.;			#voltage in volts

#CALCULATIONS
R1 = ((Ra)+((Rb*Rc)/(Rc+Rb)));			#resistance in branch1 in ohms
R2 = ((Rd)+(Re));			#resistance in branch2 in ohms
Reff = ((R1*R2)/(R1+R2))+Rf;			#effective resistance in ohms
I = V/Reff;			#current in Rf resistor in Amps
I1 = (I)*(Rb)/(Rb+R1);			#current in resistor Ra in Amps
Ix = (I1)*(Rb/(Rb+Rc)) ;			#current in resistor Rc in Amps
V2 = (Ix)*(Rc);			#voltage across Rc in volts
I2 = I-I1;			#current across Re in Amps
P4 = (I2)**2*Re;			#power dissipated across Re in W

#OUTPUT
print "current across 15 ohm resistor is %1.2f amps  \
\nvoltage across 18 ohm resistor is %dV  \
\npower dissipated in 7 ohm resistor is %2.1f Watts "%(I,V2,P4);

current across 15 ohm resistor is 6.33 amps
voltage across 18 ohm resistor is 18V
power dissipated in 7 ohm resistor is 77.8 Watts


## Example 1.47 Page No : 53¶

In [53]:
import math
from numpy.linalg import solve

#INPUT DATA
R1 = 20.;			#resistance in ohms
R2 = 50.;			#resistance in ohms
R3 = 30.;			#resistance in ohms
R4 = 15.;			#resistance in ohms
V = 100.;			# supply voltage in volts
#applying KVL to meshes I and II
#R1*(I1)+(R3)*(I1-I2) = V;-------eqn(1)
#(R2+R4)*(I2)+(R3)*(I2-I1) = 0;---------eqn(2)
#solving eqn(1) and eqn(2)

#CALCULATIONS
a = [[R2,-R3],[-R3,(R3+R4+R2)]];
b = [[V],[0]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A

#OUTPUT
print "current across 15 ohm resistor is %1.4f amps"%(I2);

current across 15 ohm resistor is 0.7792 amps


## Example 1.48 Page No : 53¶

In [54]:
import math
from numpy.linalg import solve

#INPUT DATA
RAB = 1.;			#resistance across AB in ohms
ROB = 4.;			#resistance across OB in ohms
RBC = 2.;			#resistance across BC in ohms
RAC = 1.5;			#resistance across AC in ohms
V = 10.;			# supply voltage in volts
#let ROC is R ohms
#applying KVL to meshes I,II and III
#RAB*(I1)+0+ROB*(I1-I3) = 0-------eqn(1)
#RAC*(I2)+ROC*(I2-I3) = 0---------eqn(2)
#ROB*(I3-I1)+R*(I3-I2)+RBC*I3 = 10------eqn(3)
#solving eqn(1) we get it as (I2 = I1) and applying it in eqn(2) we get R as 6 ohms
R = 6;			#resistance ROC
#solving eqn(1),(2)and (3)

#CALCULATIONS
a = [[RAB+ROB,0,-ROB],[0,(RAC+R),-R],[-ROB,-R,(RBC+R+ROB)]];
b = [[0],[0],[10]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A
I3 = c[2];			#current in mesh3 in A
I = (I3-I2);			#current flowing through R

#OUTPUT
print "current across resistor R is %1.1f amps"%(I);

current across resistor R is 0.5 amps


## Example 1.49 Page No : 54¶

In [55]:
import math
from numpy.linalg import solve

#INPUT DATA
R1 = 5.;			#resistance in ohms
R2 = 15.;			#resistance in ohms
R3 = 10.;			#resistance in ohms
R4 = 8.;			#resistance in ohms
R5 = 12.;			#resistance in ohms
V1 = 4.;			# supply voltage in volts
V2 = 6.;			#supply voltage in volts
#let currents in mesh I,II and III is I1,I2,I3 respectively
#applying KVL to meshes I,II and III
#(R1+R2)*(I1)-R2*(I2) = V1-------eqn(1)
#R2*(I1)-(R2+R3+R4)*(I2)+(R4)*(I3) = 0---------eqn(2)
#R4*(I2)-(R4+R5) = V2------eqn(3)
#solving eqn(1) we get it as (I2 = I1) and applying it in eqn(2) we get R as 6 ohms
R = 6;			#resistance ROC
#solving eqn(1),(2)and (3)

#CALCULATIONS
a = [[R1+R2,-R2,0],[R2,-(R2+R3+R4),R4],[0,R4,-(R4+R5)]];
b = [[V1],[0],[V2]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A
I3 = c[2];			#current in mesh3 in A
I = (I2-I3);			#current flowing through R4 in Amps

#OUTPUT
print "current across 8 ohm resistor is %1.3f amps"%(I);

current across 8 ohm resistor is 0.319 amps


## Example 1.50 Page No : 55¶

In [56]:
import math
from numpy.linalg import solve

#INPUT DATA
RAB = 25.;			#resistance in ohms
RBC = 10.;			#resistance in ohms
RAD = 20.;			#resistance in ohms
RCD = 15.;			#resistance in ohms
RG = 50.;			#resistance of galvanometer in ohms
REF = 2.;			#internal resistance in ohms
V = 25.;			# supply voltage in volts
#let currents in mesh I,II and III is I1,I2,I3 respectively
#applying KVL to meshes I,II and III
#-(RG)*(I1)-(RG+RCD+RBC)*(I2)-(RCD)*(I3) = 0---------eqn(2)
#solving eqn(1),(2)and (3)

#CALCULATIONS
b = [[0],[0],[-V]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A
I3 = c[2];			#current in mesh3 in A
I = (I1-I2);			#currentthrough galvanometer in Amps

#OUTPUT
print "current across galavanometer is %1.5f amps"%(I);

current across galavanometer is 0.04875 amps


## Example 1.51 Page No : 56¶

In [57]:
import math
from numpy.linalg import solve

#INPUT DATA
V1 = 100;			#source1 voltage in volts
V2 = 50.;			#source2 voltage in volts
R1 = 10.;			#resistance in ohms
R2 = 20.;			#resistance in ohms
R3 = 30.;			#resistance in ohms
R4 = 40.;			#resistance in ohms
#let currents in mesh I,II is I1,I2 respectively
#applying KVL to meshes I,II
#(R1+R3+R4)*(I1)-(R3)*(I2) = V1-------eqn(1)
#(R3)*(I1)-(R2+R3)*(I2) = -V2---------eqn(2)
#solving eqn(1),(2)

#CALCULATIONS
a = [[(R1+R3+R4),-R3],[R3,-(R2+R3)]];
b = [[V1],[-V2]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A
I = (I2-I1);			#current through R3 in Amps

#OUTPUT
print "current across 30 ohm resistor is %1.3f amps"%(I);

current across 30 ohm resistor is 0.161 amps


## Example 1.52 Page No : 57¶

In [58]:
import math
from numpy.linalg import solve

#INPUT DATA
V1 = 10;			#source1 voltage in volts
V2 = 5;			#source2 voltage in volts
V3 = 5;			#source3 voltage in volts
RAH = 2;			#resistance in ohms
RAB = 3;			#resistance in ohms
RBE = 5;			#resistance in ohms
REG = 5;			#resistance in ohms
RED = 5;			#resistance in ohms
RBC = 7;			#resistance in ohms
RCD = 3;			#resistance in ohms
RDF = 5;			#resistance in ohms
RHG = 5;			#resistance in ohms
#let currents in mesh I,II,III is I1,I2,I3 respectively
#applying KVL to meshes I,II
#(RAH+RHG+RAB+RBE+REG)*(I1)-(RBE)*(I2)-(REG)*(I3) = V1-------eqn(1)
#-(RBE)*(I1)+(RBC+RCD+RBE+RED)*(I2)-(RDF)*(I3) = -V2---------eqn(2)
#-(REG)*(I1)-(RED)*(I2)+(REG+RED+RDF)*(I3) = -V3--------eqn(3)
#solving eqn(1),(2) and (3)

#CALCULATIONS
a = [[(RAH+RHG+RAB+RBE+REG),-RBE,-REG],[-REG,(RBC+RCD+RBE+RED),-(RDF)],[-REG,-RED,(REG+RED+RDF)]];
b = [[V1],[-V2],[-V3]];
c = solve(a,b);
I1 = c[0];			#current in mesh1 in A
I2 = c[1];			#current in mesh2 in A
I3 = c[2];			#current in mesh3 in A
P1 = V1*I1;			#power output from V1 in W
P2 = V2*I2;			#power output from V2 in W
P3 = V3*I3;			#power output from V3 in W

#OUTPUT
print "power output from 10V is %1.1f W from 5V is %1.2fW from 5V is %1.2fW "%(P1,-P2,-P3);

power output from 10V is 3.7 W from 5V is 1.14W from 5V is 1.43W


## Example 1.54 Page No : 61¶

In [59]:
import math

#INPUT DATA
RAC = 10.;			#resistance in ohms
RCD = 10.;			#resistance in ohms
RCF = 50.;			#resistance in ohms
RDH = 50.;			#resistance in ohms
RDF = 30.;			#resistance in ohms
RHF = 10.;			#resistance in ohms
#umath.sing star to delta conversion,the star point D is eliminated

#CALCULATIONS
RCF1 = ((RCD*RDF)+(RDF*RDH)+(RDH*RCD))/(RDH);			#by umath.sing star to delta conversion technique
RFH = ((RCD*RDF)+(RDF*RDH)+(RDH*RCD))/(RCD);			#by umath.sing star to delta conversion technique
RHC = ((RCD*RDF)+(RDF*RDH)+(RDH*RCD))/(RDF);			#by umath.sing star to delta conversion technique
RCF2 = (RCF*RCF1)/(RCF+RCF1);
RCF = RCF2;			#equivalent resistance of RCF in ohms
RHF1 = (RHF*RFH)/(RHF+RFH);
RHF = RHF1;			#equivalent resistance of RHF in ohms
RAB = (RAC)+(RHC*(RCF+RHF))/(RHC+RCF+RHF);			#equivalent resistance of AB in ohms

#OUTPUT
print "Thus equivalent resistance of AB is %f ohms"%(RAB);
#note:given final answer is wrong in textbook.Please check the calculations

Thus equivalent resistance of AB is 33.333333 ohms


## Example 1.55 Page No : 62¶

In [60]:
import math

#INPUT DATA
RAB = 1.;			#resistance in ohms
RBE = 2.;			#resistance in ohms
RED = 3.;			#resistance in ohms
REF = 3.;			#resistance in ohms
RDF = 3.;			#resistance in ohms
RAD = 2.;			#resistance in ohms
RAC = 1.;			#resistance in ohms
RBC = 1.;			#resistance in ohms
RFC = 2.;			#resistance in ohms

#CALCULATIONS
#Delta DEF is converted into equivalent star where RDN = REN = RFN = 1 ohm
#series branches of inner star network are added
#Star ABCN is converted to equivalent delta
RDN = 1;
REN = RDN;
RFN = REN;
RAN = RAD+RDN;RBN = RBE+REN;RCN = RFC+RFN
RAB1 = ((RAN*RBN)+(RBN*RCN)+(RCN*RAN))/(RCN);			#by umath.sing star to delta conversion technique
RBC1 = ((RAN*RBN)+(RBN*RCN)+(RCN*RAN))/(RAN);			#by umath.sing star to delta conversion technique
RCA1 = ((RAN*RBN)+(RBN*RCN)+(RCN*RAN))/(RBN);			#by umath.sing star to delta conversion technique
#parallel resistances in each branch are converted to math.single resistance
RAB2 = (RAB*RAB1)/(RAB+RAB1);
RAB = RAB2;			#equivalent resistance of RAB in ohms
RBC2 = (RBC*RBC1)/(RBC+RBC1);
RBC = RBC2;			#equivalent resistance of RBC in ohms
RCA2 = (RAC*RCA1)/(RAC+RCA1);
RCA3 = ((RCA2)*(RAB+RBC))/(RBC+RAB+RCA2);
RCA = RCA3;

#OUTPUT
print "Thus equivalent resistance of CA is %1.1f ohms"%(RCA);
#TO FIND EQUIVALENT RESISTANCE BETWEEN DF

#node A is eliminated umath.sing star to delta conversion
RBC = (RAB*RAD)+(RAD*RAC)+(RAC*RAB)/(RAD);			#by umath.sing star to delta conversion technique
RCD = (RAB*RAD)+(RAD*RAC)+(RAC*RAB)/(RAB);			#by umath.sing star to delta conversion technique
#node C is eliminated umath.sing star to delta conversion
RDB = (0.72*5)+(5*2)+(2*0.72)/2;
RBF1 = (0.72*5)+(5*2)+(2*0.72)/5;
RFD = (0.72*5)+(5*2)+(2*0.72)/0.72;
#parallel branches between nodes B and D and nodes D and F are reduced as
RBD = (RDB*5)/(RDB+5);
RDF = (RFD*3)/(RFD+3);
#node E is eliminated umath.sing star to delta conversion technique
RBF = ((2*3)+(3*3)+(3*2))/3.;
RFD = ((2*3)+(3*3)+(3*2))/2.;
RDB = ((2*3)+(3*3)+(3*2))/3.;
RDF1 = 4.2;			#(R' = RDB+RDF)
RDF = ((RDF1)*(RDF1/2))/(RDF1+(RDF1/2));

#OUTPUT
print " Thus equivalent resistance of DF is %1.1f ohms"%(RDF);

Thus equivalent resistance of CA is 0.6 ohms
Thus equivalent resistance of DF is 1.4 ohms


## Example 1.56 Page No : 66¶

In [62]:
import math

#INPUT DATA
RAB = 6.;			#resistance in ohms
RBC = 3.;			#resistance in ohms
RBD = 4.;			#resistance in ohms
V1 = 25.;			#source voltage in volts
V2 = 45.;			#source voltage in volts

#CALCULATIONS
#applying kirchoff's current law at node B
#-I1-I2+I3 = 0
#I1 = (V1-VB)/RAB
#I2 = (V3-VB)/RBC
#I3 = VB/RBD
VB = ((V1/RAB)+(V2/RBC))/((1/RAB)+(1/RBC)+(1/RBD));
I1 = (V1-VB)/(RAB);			#current across AB
I2 = (V2-VB)/(RBC);			#current across BC
I3 = (VB)/(RBD);			#current across BD

#OUTPUT
print "Thus currents I1, I2 ,I3 are %1.1f A %1.2f A %1.1f A"%(I1,I2,I3);

Thus currents I1, I2 ,I3 are -0.1 A 6.48 A 6.4 A


## Example 1.57 Page No : 67¶

In [63]:
import math
from numpy.linalg import solve

#INPUT DATA
I1 = 25.;			#current source in A
I2 = 6.;			#current source in A
I3 = 5.;			#current source in A
RAB = 5.;			#resistance in ohms
RAC = 10.;			#resistance in ohms
RBC = 2.;			#resistance in ohms
#let currents across AC and BC and AB are Ix,Iy and Iz respectively
#applying kirchoff's current law at node A
#-I1+Ix+I3+Iz = 0------eqn(1)
#applying kirchoff's current law at node B
#-Iz-I3+Iy+I2 = 0------eqn(2)

#CALCULATIONS
a = [[((1/RAC)+(1/RAB)),(-1/RAB)],[(-1/RAB),((1/RAB)+(1/RBC))]];
b = [[20],[-1]];
c = solve(a,b)
VA = c[0];			#voltage at node A
VB = c[1];			#voltage at node B

#OUTPUT
print "Thus voltages at node A and B are %2.1f V and %2.1f V"%(-VA,VB);

Thus voltages at node A and B are -81.2 V and 21.8 V


## Example 1.58 Page No : 68¶

In [64]:
#INPUT DATA
RAB = 1.;			#resistance in ohms
RAD = 1;			#resistance in ohms
RDC = 2;			#resistance in ohms
RCB = 1;			#resistance in ohms
RAC = 1;			#resistance in ohms
#delta DAC has been converted to star DAC where RDN = 0.5 ohms,RNA = 0.25 ohms,RNC = 0.5 ohms

#CALCULATIONS
RDN = 0.5;
RNA = 0.25;
RNC = 0.5
RBD = ((RDN)+(((RNA+RAB)*(RNC+RCB))/(RNA+RAB+RNC+RCB)));			#equivalent resistance across BD

#OUTPUT
print "Thus equivalent resistance across BD is %1.2f ohms"%(RBD);

Thus equivalent resistance across BD is 1.18 ohms


## Example 1.59 Page No : 69¶

In [66]:
#INPUT DATA
RAB = 9.;			#resistance in ohms
RBC = 1.;			#resistance in ohms
RCA = 1.5;			#resistance in ohms
RAD = 6.;			#resistance in ohms
RBD = 4.;			#resistance in ohms
RCD = 3.;			#resistance in ohms
#star ABC has been converted to delta AnBnCn where RABn = 18 ohms,RBCn = 9 ohms,RCAn = 13.5 ohms

#CALCULATIONS
RABn = 18.;
RBCn = 9.;
RCAn = 13.5;
RAB1 = ((RAB*RABn)/(RAB+RABn));			#equivalent resistance across AB
RBC1 = ((RBC*RBCn)/(RBC+RBCn));			#equivalent resistance across BC
RAC1 = ((RCA*RCAn)/(RCA+RCAn));			#equivalent resistance across AC
#there are two parallel paths across points A and B.
#(a)one directly from A to B having a resistance of 6 ohms and
#(b)The other via C having a total resistance
RBA = ((RBC1+RAC1)*(RAB1))/(RBC1+RAC1+RAB1);			#final equivalent resistance across AB
RCB = ((RAC1+RAB1)*(RBC1))/(RAC1+RAB1+RBC1);			#final equivalent resistance across BC
RCA = ((RAB1+RBC1)*(RAC1))/(RAB1+RBC1+RAC1);			#final equivalent resistance across AC

#OUTPUT
print "Thus final equivalent resistance across AB, BC and CA are %1.2f ohms, %1.2f ohms, %1.2f ohms"%(RBA,RCB,RCA);
#note:answer given for RCA is wrong.Please check the calculations

Thus final equivalent resistance across AB, BC and CA are 1.64 ohms, 0.80 ohms, 1.13 ohms