# Chapter 5 : SINGLE PHASE AC CIRCUITS¶

## Example 5.1 Page No : 157¶

In [1]:
import math
from numpy import ones

def  r2p(x,y):			#function to convert recmath.tangular to polar
polar  =  ones(2)
polar[0]  =  math.sqrt ((x **2) +(y**2))
polar[1]  =  math.atan (y/x)
polar[1]  = (polar [1]*180)/math.pi
return polar

def  p2r(r,theta):			#function to convert polar to recmath.tangular
rect  =  ones(2)
theta  = ( theta *math.pi) /180
rect [0] = r* math.cos(theta)
rect [1] = r* math.sin(theta)
return rect

#CALCULATIONS
I1 = r2p(7,-5);
print (I1);
I2 = r2p(-9,6);
I2[1] = I2[1]+(180);			#this belongs to quadrant 2 and hence 180 degrees should be added
print (I2);
I3 = r2p(-8,-8);
I3[1] = I3[1]+(180);			#this belongs to quadrant 3 and hence 180 degrees should be added
print (I3);
I4 = r2p(6,6);
print (I4);
#note:here direct functions for converson are not available and hence we defined user defined functions for polar to rect and rect to polar conversions

[  8.60232527 -45.        ]
[  10.81665383  135.        ]
[  11.3137085  225.       ]
[  8.48528137  45.        ]


## Example 5.2 Page No : 157¶

In [3]:
import math
from numpy import ones

def  r2p(x,y):			#function to convert recmath.tangular to polar
polar  =  ones(2)
polar[0]  =  math.sqrt ((x **2) +(y**2))
polar[1]  =  math.atan (y/x)
polar[1]  = (polar [1]*180)/math.pi
return polar

def  p2r(r,theta):			#function to convert polar to recmath.tangular
rect  =  ones(2)
theta  = ( theta *math.pi) /180
rect [0] = r* math.cos(theta)
rect [1] = r* math.sin(theta)
return rect

#CALCULATIONS
#for subdivision 1
I1 = p2r(10,60);
I2 = p2r(8,-45);
I3 = I1+I2;
print (I3);
I4 = r2p(I3[0],I3[1]);
print (I4)
#for subdivision 2
I5 = r2p(5,4);
I6 = r2p(-4,-6);
I7 = ones(2)
I7[0] = (I5[0])*(I6[0]);
I7[1] = (I5[1]+I6[1]);
I7[1] = I7[1]-180;
print (I7);
#for subdivision 3
I8 = r2p(-2,-5);
I9 = r2p(5,7);
I10 = ones(2)
I10[0] = I8[0]/I9[0];
I10[1] = I8[1]-I9[1];
I10[1] = I10[1]-180
print (I10);
#note:here direct functions for converson are not available and hence we defined user defined functions for polar to rect and rect to polar conversions

[ 10.65685425   3.00339979]
[ 11.07198956  15.73932193]
[  46.17358552 -135.        ]
[   0.62601269 -161.56505118]


## Example 5.3 Page No : 160¶

In [4]:
import math

#given i(t) = 5*math.sin(314*t+(2*math.pi/3))&& v(t) = 20*math.sin(314*t+(5*math.pi/6))
#CALCULATIONS
P1 = 2*(math.pi/3);			#phase angle of current in radians
P1 = P1*(180/math.pi);			#phase angle of current in degrees
P2 = 5*(math.pi/6);			#phase angle of voltage in radians
P2 = P2*(180/math.pi);			#phase angle of voltage in degrees
P3 = P2-P1;			#current lags voltage by P3 degrees
P4 = P3*math.pi/180;
pf = math.cos(P4);			#lagging pf
Vm = 20;			#peak voltage
Im = 5;			#peak current
Z = Vm/Im;			#impedance in ohms
R = (Z)*math.cos(P4);			#resistance in ohms
Xl = math.sqrt((Z)**2-(R)**2);			#reacmath.tance
W = 314;
L = Xl/W;			#inductance in henry
V = Vm/math.sqrt(2);			#average value of voltage
I = Im/math.sqrt(2);			#average value of current
av = (V*I)*math.cos(P4);			#average power in watts
print "thus impedance, resistance, inductance, powerfactor and average power are %d ohms, %1.2f ohms, %g H,%1.3f and %2.1f W respectively"%(Z,R,L,pf,av);

thus impedance, resistance, inductance, powerfactor and average power are 4 ohms, 3.46 ohms, 0.00636943 H,0.866 and 43.3 W respectively


## Example 5.4 Page No : 161¶

In [5]:
import math
#Chapter-5, Example 5.4, Page 161

#INPUT DATA
I = 10.;			#given current in A
P = 1000.;			#power in Watts
V = 250.;			#voltage in volts
f = 25.;			#frequency in Hz

#CALCULATIONS
R = P/((I)**2);			#resistance in ohms
Z = V/I;			#impedance in ohms
Xl = math.sqrt((Z)**2-(R)**2);			#reacmath.tance in ohms
L = Xl/(2*math.pi*f);			#inductance in Henry
Pf = R/Z;			#power factor,lagging,pf = math.cos(phi)

# Results
print "thus impedance, resistance, inductance, reactance and powerfactor are %d ohms, %d ohms, %1.3f H, \
%2.2f ohms and %1.1f respectively"%(Z,R,L,Xl,Pf);

thus impedance, resistance, inductance, reactance and powerfactor are 25 ohms, 10 ohms, 0.146 H, 22.91 ohms and 0.4 respectively


## Example 5.5 Page No : 162¶

In [6]:
import math
#Chapter-5, Example 5.5, Page 162

#INPUT DATA
V = 250.;			#supply voltage in volts
f = 50.;			#frequency in hz
Vr = 125.;			#voltage across resistance in volts
Vc = 200.;			#voltage across coil in volts
I = 5.;			#current in A
#CALCULATIONS
R = Vr/I;			#resistance in ohms
Z1 = Vc/I;			#impedance of coil in ohms
#Z1 = math.sqrt((R1)**2+(Xl)**2)------eqn(1)
Z = V/I;			#total impedance in ohms
#Z = math.sqrt((R+R1)**2+(Xl)**2)-----eqn(2)
#solving eqn(1)and eqn(2) we get R1 as follows
R1 = (((Z)**2-(Z1)**2)-(R)**2)/(2*R);			#in ohms
Xl = math.sqrt((Z1)**2-(R1)**2);			#reacmath.tance of coil in ohms
P = ((I)**2*R1);			#power absorbed by the coil in Watts
Pt = ((I)**2)*(R+R1);			#total power in Watts

# Results
print "thus impedance, resistance, reactance are %d ohms, %d ohms, %2.2f ohms respectively"%(Z1,R,Xl);
print "power absorbed and total power are %3.1f W and %3.1f W respectively"%(P,Pt)

thus impedance, resistance, reactance are 40 ohms, 25 ohms, 39.62 ohms respectively
power absorbed and total power are 137.5 W and 762.5 W respectively


## Example 5.6 Page No : 163¶

In [7]:
import math
#Chapter-5, Example 5.6, Page 163

#INPUT DATA
V = 240;			#supply voltage in volts
Vl = 171;			#voltage across inductor in volts
I = 3;			#current in A
phi = 37;			#power factor laggging in degrees
#CALCULATIONS
Zl = Vl/I;			#impedance of coil in ohms
#Zl = math.sqrt((R1)**2+(Xl)**2)------eqn(1)
Z = V/I;			#total impedance in ohms
#Z = math.sqrt((R+R1)**2+(Xl)**2)-----eqn(2)
pf = math.cos(phi*math.pi/180);			#powerfactor
Rt = pf*Z;			#total resistance in ohms			#Rt = (R+R1)
#substituting Rt value in eqn(2) we find Xl as follows
Xl = math.sqrt((Z)**2-(Rt)**2);			#reacmath.tance of inductor in ohms
#ubstituting Xl value in eqn(1) we find R1 as follows
R1 = math.sqrt((Zl)**2-(Xl)**2);			#resistance of inductor in ohms
R = Rt-R1;			#resistance of resistor in ohms
print "Thus resistance of resistor  is %2.2f ohms"%(R);
print "Thus resisimath.tance and reacmath.tance of inductor are %2.2f ohms and %2.2f ohms respectively"%(R1,Xl)

Thus resistance of resistor  is 33.38 ohms
Thus resisimath.tance and reacmath.tance of inductor are 30.51 ohms and 48.15 ohms respectively


## Example 5.7 Page No : 164¶

In [8]:
import math
#Chapter-5, Example 5.7, Page 164

#INPUT DATA
V = 100;			#supply voltage in volts
#for COIL A
f = 50;			#frequency in Hz
I1 = 8;			#current in A
P1 = 120;			#power in Watts
#for COIL B
I2 = 10;			#current in A
P2 = 500;			#power in Watts
#CALCULATIONS
#FOR COIL A
Z1 = V/I1;			#impedance of coil A in ohms
R1 = P1/(I1)**2;			#resistance of coil A in ohms
X1 = math.sqrt(((Z1)**2-(R1)**2));			#reacmath.tance of coil A in ohms
#FOR COIL B
Z2 = V/I2;			#impedance of coil B in ohms
R2 = P2/(I2)**2;			#resistance of coil B in ohms
X2 = math.sqrt(((Z2)**2-(R2)**2));			#reacmath.tance of coil B in ohms
#When both COILS A and B are in series
Rt = R1+R2;			#total resistance in ohms
Xt = X1+X2;			#total reacmath.tance in ohms
Zt = math.sqrt((Rt)**2+(Xt)**2);			#total impedance in ohms
It = V/Zt;			#current drawn in A
P = ((It)**2)*(Rt);			#power taken in watts
print "Thus current drawn and power taken in watts are %2.2f A and %3.2f W respectively"%(It,P);

Thus current drawn and power taken in watts are 4.66 A and 130.12 W respectively


## Example 5.8 Page No : 167¶

In [9]:
import math
#Chapter-5, Example 5.8, Page 167

#INPUT DATA
R = 100;			#resistance in ohms
C = 50*10**-6;			#capacitance in F
V = 200;			#voltage in Volts
f = 50;			#frequency in Hz
#Z = R-(1j)*(Xc)------>impedance
Xc = 1/(2*math.pi*f*C);			#capacitive reacmath.tance in ohms
Z = math.sqrt((R)**2+(Xc)**2);			#impedance in ohms
I = V/Z;			#current in A
pf = R/Z;			#power factor ------>math.cos(phi)---->leading
phi = math.acos(0.844);			#phase angle in radians
phi = phi*180/math.pi;			#phase angle in degrees
Vr = (I)*(R);			#voltage across resistor
Vc = (I)*(Xc);			#votage across capacitor
print "Thus impedance, current, powerfactor and phaseangle are %3.2f ohms, %1.2f A, %1.3f and %2.2f degrees respectively"%(Z,I,pf,phi);
print "voltage across resistor and capacitor are %d V and %3.2f V respectively"%(Vr,Vc)

Thus impedance, current, powerfactor and phaseangle are 118.54 ohms, 1.69 A, 0.844 and 32.44 degrees respectively
voltage across resistor and capacitor are 168 V and 107.41 V respectively


## Example 5.9 Page No : 169¶

In [10]:
import math
#Chapter-5, Example 5.9, Page 169

#INPUT DATA
phi = 40;			#phase in degrees
V = 150;			#voltage in Volts
I = 8;			#current in A
#the applied voltage lags behind the current .That means the current leads the voltage
#CALCULATIONS
#hence it is a capacitive circuit
pa = V*I*pf;			#active power in W
pr = V*I*math.sin(phi*math.pi/180);			#reactive power in VAR
print "Thus active and reactive power are %3.1f W and %3.1f VAR respectively"%(pa,pr);

Thus active and reactive power are 919.3 W and 771.3 VAR respectively


## Example 5.10 Page No : 169¶

In [13]:
import math
#Chapter-5, Example 5.10, Page 169

#INPUT DATA
#given v = 141.4*math.sin(314*t)
P = 700.;			#power in Watts
Vm = 141.4;			#maximum value of supply voltage
#CALCULATIONS
Vr = Vm/(math.sqrt(2));			#rms value of supply voltage
I = P/(Vr*pf);			#current in A
Z = Vr/I;			#impedance in ohms
R = (Z)*(pf);			#resistance in ohms
phi = math.acos(pf)*180/math.pi;			#angle in degrees
Xc = (Z)*(math.sin(phi));			#reacmath.tance in ohms
C = 1/(3.14*7.13);			#capacitance in F
print "Thus resistance and capacitance are %1.2f ohms and %g F respectively"%(R,C);

Thus resistance and capacitance are 7.14 ohms and 0.0446664 F respectively


## Example 5.11 Page No : 169¶

In [14]:
import math
#Chapter-5, Example 5.11, Page 169

#INPUT DATA
V = 200.;			#supply voltage in volts
f = 50.;			#freq in hz
P = 7000.;			#power in Watts
Vr = 130.;			#volatge across resistor in volts
P = 7000.;			#power in Watts

#CALCULATIONS
R = ((Vr)**2)/P;			#resistance in ohms
I = Vr/R;			#current in A
Z = V/I;			#total impedance in ohms
Xc = math.sqrt((Z)**2-(R)**2);
C = 1/(2*math.pi*f*Xc);			#capacitance in F
phi = math.acos(pf);			#angle in radians
phi = phi*180/math.pi;			#angle in degrees
Vm = V*math.sqrt(2);			#maximum value of voltage
#voltage equation v = Vm*math.sin(2*math.pi*f*t)------>282.84*math.sin(314.16*t)
#current equation ------>i = 76.155*math.sin(314.16*t+phi)
print "Thus current, resistance, p.f, capacitance, impedance are %2.2f A , %1.2f ohms, %2.1f , \
%g F and %1.2f ohms respectively"%(I,R,pf,C,Z);

Thus current, resistance, p.f, capacitance, impedance are 53.85 A , 2.41 ohms, 0.6 , 0.00112771 F and 3.71 ohms respectively


## Example 5.12 Page No : 170¶

In [15]:
import math

#INPUT DATA
C = 50.;			#capacitance in uf
R = 100.;			#resistance in ohms
V = 200.;			#supply voltage in volts
f = 50.;			#freq in hz
#CALCULATIONS
Xc = 1/(2*math.pi*f*C*10**-6);			#capacitive reacmath.tance in ohms
Z = R-((1j)*Xc);			#impedance in ohms
print (Z);
z1 = math.sqrt((R)**2+(Xc)**2);
theta = math.atan(Xc/R);
pf = math.cos(theta);			#powerfactor
I = V/z1;			#current in A
P = V*I*pf;			#power in Watts
print "Thus current, power factor, power are % 1.2f A ,%1.3f ,%d W respectively"%(I,pf,P);

(100-63.6619772368j)
Thus current, power factor, power are  1.69 A ,0.844 ,284 W respectively


## Example 5.13 Page No : 170¶

In [16]:
import math

#INPUT DATA
C = 0.05;			#capacitance in uf
F = 500;			#freq in hz
#CALCULATIONS
Xl = 1/(2*math.pi*F*C*10**-6);			#capacitive reacmath.tance in ohms
#at resonance Xl = Xc
L = (Xl/(2*math.pi*F));			#inductance in H
print "Thus value of L is %1.2f H"%(L);

Thus value of L is 2.03 H


## Example 5.14 Page No : 171¶

In [18]:
import math

#INPUT DATA
V = 200;			#voltage in V
R = 50;			#resistance in ohms
L = 0.5;			#inductance in Henry
F = 50;			#freq in hz
#CALCULATIONS
Xl = 2*math.pi*F*L;			#inductive reacmath.tance
Z = (R)+((1j)*Xl)			#impedance
print (Z);
z1 = math.sqrt((R)**2+(Xl)**2);			#magnitude
theta = math.atan(Xl/R);			#angle in radians
I = V/z1;			#current in A
P = V*I*math.cos(theta);			#power supplied in W
#here capacitive reacmath.tance equals inductive reacmath.tance
#hence Xc = Xl
C = 1/(2*math.pi*F*Xl);			#capacitance in uf
print "Thus current and power required are % 1.2f A and %2.2f W respectively"%(I,P);
print "Thus additional resistance that neede to be connected in series with R and C to have\
same current at unity power factor is %1.1f ohms"%(r);

(50+157.079632679j)
Thus current and power required are  1.21 A and 73.60 W respectively
Thus additional resistance that neede to be connected in series with R and C to have same current at unity power factor is 114.8 ohms


## Example 5.15 Page No : 171¶

In [19]:
import math

#INPUT DATA
R = 50.;			#resistance in ohms
L = 9.;			#inductance in Henry
I0 = 1.;			#current in A
f = 75.;			#ferquency in Hz
#at resonance Xl = Xc
#CALCULATIONS
Xl = 2*math.pi*f*L;			#inductive reacmath.tance
Xc = Xl;			#capacitive reacmath.tance
C = 1/(2*math.pi*f*Xc);			#capacitance in uf
print "Thus capacitance is %g F"%(C);

Thus capacitance is 5.00352e-07 F


## Example 5.16 Page No : 175¶

In [20]:
import math

#INPUT DATA
R = 10.;			#resistance in ohms
L = 0.1;			#inductance in Henry
C = 150.;			#capacitor in uf
V = 200.;			#voltage in V
f = 50.;			#frequency in hz
#CALCULATIONS
Xc = 1/(2*math.pi*f*C*10**-6);			#Capacitive reacmath.tance in ohms
Xl = (2*math.pi*f*L);			#inductive reacmath.tance in ohms
Z = R+((1j)*(Xl-Xc));			#impedance in ohms
z1 = math.sqrt((R)**2+(Xl-Xc)**2);			#magnitude of Z
I = V/z1;			#current in A
pf = R/z1;			#power factor----->math.cos(phi)
#As Xl-Xc is inductive,pf is lagging
z2 = math.sqrt((R**2)+(Xl)**2);			#impedance of coil in ohms
Vl = I*(z2);			#voltage across coil in volts
Vc = I*(Xc);			#voltage across capacitor in volts
print "Thus inductive reacmath.tance, capacitive reacmath.tance, impedance, current, powerfactor are %2.2f ohms, \
%2.2f ohms, %2.2f ohms, %d A, %1.1f respectively"%(Xl,Xc,z1,I,pf);

Thus inductive reacmath.tance, capacitive reacmath.tance, impedance, current, powerfactor are 31.42 ohms, 21.22 ohms, 14.28 ohms, 14 A, 0.7 respectively


## Example 5.17 Page No : 176¶

In [21]:
import math

#INPUT DATA
L = 10;			#inductance in milliHenry
C = 5;			#capacitor in uf
phi = 50;			#phase in degrees-------->lagging
f = 500;			#frequency in hz
V = 200;			#supply voltage in volts

#CALCULATIONS
Xc = 1/(2*math.pi*f*C*10**-6);			#Capacitive reacmath.tance in ohms
Xl = (2*math.pi*f*L*10**-3);			#inductive reacmath.tance in ohms
R = (Xc-Xl)/(math.tan(phi*math.pi/180));			#resistance in ohms
Z = math.sqrt((R)**2+(Xc-Xl)**2);			#impedance in ohms
I = V/Z;			#current in A
Vr = (I)*(R);			#voltage across resistance
Vl = (I)*(Xl);			#voltage across inductance
Vc = (I)*(Xc);			#voltage across capacitance
print "Thus voltages across resistance, inductance, capacitance are %3.2f volts, %3.2f volts, %3.2f volts respectively"%(Vr,Vl,Vc);

Thus voltages across resistance, inductance, capacitance are 128.56 volts, 149.26 volts, 302.47 volts respectively


## Example 5.18 Page No : 176¶

In [22]:
import math
from sympy import Symbol,solve
#Chapter-5, Example 5.18, Page 176

#INPUT DATA
L = 5;			#inductance in Henry
f = 50;			#frequency in hz
V = 230;			#supply voltage in volts
R = 2;			#resistance in ohms
V1 = 250;			#voltage across coil in V

#CALCULATIONS
Xl = (2*math.pi*f*L);			#inductive reacmath.tance in ohms
Z1 = math.sqrt((R)**2+(Xl)**2);			#impedance of coil in ohms
I = V1/Z1;			#current in A
Z = V/I;			#total impedance in ohms
#Z = math.sqrt((R)**2+(Xl-Xc)**2) and solving for Xc
Xc = Symbol("Xc");
p = (Xc**2)-3141.58*(Xc)+378004
roots2  =  solve(p);
r2  =  roots2[1];
#Xc cannot be greater than Z
C = 1/(2*math.pi*f*r2);			#capacitance in F
print "Thus value of C that must be present suct that voltage across coil is 250 volts is %g F respectively"%(C);

Thus value of C that must be present suct that voltage across coil is 250 volts is 1.05531e-06 F respectively


## Example 5.19 Page No : 178¶

In [23]:
import math
#Chapter-5, Example 5.19, Page 178

#v = 350*math.cos(3000*t-20)
#i = 15*math.cos(3000*t-60)
#INPUT DATA
L = 0.5;			#inductance in Henry
phi = -40;			#phase difference between applied voltage and current
#Xl>Xc(P.f is lagging)
w = 3000;			#freq in hz
Vm = 350;			#peak voltage in volts
Im = 15;			#peak current in amps
#CALCULATIONS
Z = Vm/Im;			#total impedance in ohms
#Xl-Xc = 0.839*R = X
#Z = math.sqrt((R)**2+(X)**2)
#Z = 1.305*R
R = Z/1.305;			#resistance in ohms
X = 0.839*R;			#
#X = Xl-Xc
Xl = w*L;			#reactive inductance in ohms
Xc = Xl-X;			#capacitive reacmath.tance in ohms
C = 1/(w*Xc);			#capacitance in uf
print "Thus resistance and capacitance are %2.2f ohms and %g F respectively"%(R,C);

Thus resistance and capacitance are 17.62 ohms and 2.24435e-07 F respectively


## Example 5.20 Page No : 182¶

In [12]:
import math
from numpy.linalg import inv
from scipy.optimize import fsolve
from sympy.solvers import solve

#INPUT DATA
R = 10;			#resistance in ohms
L = 0.1;			#inductance in henry
f = 50;			#frequency in hz
#CALCULATIONS
Xl = (2*math.pi*f*L);			#inductive reacmath.tance in ohms
Z = R+((1j)*(Xl));			#impedance in ohms
Y = inv([[Z]])#[0];			#admittance in mho
y = abs(Y);			#admittance in mho

admittance is 0.03033 mho


## Example 5.21 Page No : 182¶

In [15]:
import math
from numpy.linalg import inv

#INPUT DATA
#CALCULATIONS
Z = 10+((1j)*(5));			#impedance in ohms
Y = inv([[Z]]);			#admittance in mho
print (Y);

[[ 0.08-0.04j]]


## Example 5.22 Page No : 182¶

In [14]:
import math

#INPUT DATA
Z1 = 7.+((1j)*5);			#impedance of branch1 in ohms
Z2 = 10.-((1j)*8);			#impedance  of branch2 in ohms
V = 230.;			#supply voltage in volts
f = 50.;			#frequency in hz
#CALCULATIONS
Y1 = 1/(Z1);			#admittance of branch1 in mho
Y2 = 1/(Z2);			#admittance of branch2 in mho
Y = Y1+Y2;			#admittance of combined circuit
print (Y);
g = abs(Y);			#conductance in mho;
B = math.atan(Y.imag/Y.real);			#susceptance in mho
I = V*(Y);			#current
print (I);			#total current taken from mains in A
z = math.atan(I.imag/I.real);
pf = math.cos(z);			#power factor
print "thus conductance and susceptance of the circuit is %1.3f mho and %1.3f mho respectively"%(g,B);
print "power factor is %1.3f lagging"%(pf)

(0.155570204351-0.0187870797627j)
(35.7811470007-4.32102834542j)
thus conductance and susceptance of the circuit is 0.157 mho and -0.120 mho respectively
power factor is 0.993 lagging


## Example 5.23 Page No : 183¶

In [29]:
import math
#Chapter-5, Example 5.23, Page 183

#INPUT DATA
V = 240.;			#voltage in volts
f = 50.;			#frequency in Hz
R = 15.;			#resisimath.tance in ohms
I = 22.1;			#current in A
#CALCULATIONS
G = 1/R;			#conductance in mho
#susceptance of the circuit,B = 1/(Xl) = 0.00318/L
#admittance of the circuit,(G-jB) = (0.067-j(0.00318/L))
Y = I/V;			#admittance in mho;
#Y = math.sqrt((0.067)**2+(0.00318/L)**2) = 0.092-----eqn(1)
#solving eqn(1) for L we have it as
L = math.sqrt((0.00318)**2/((Y)**2-(G)**2));			#inductance in henry
#when current is 34A
I1 = 34;			#current in A
Y1 = I1/V;			#admittance in mho
#for Y1 we need to find f
f1 = math.sqrt((3.183)**2/((Y1)**2-(G)**2));			#frequency in hz
print "Thus value of frequency when current is 34A is %2.1f Hz"%(f1);

Thus value of frequency when current is 34A is 25.5 Hz


## Example 5.24 Page No : 184¶

In [18]:
import math
from numpy.linalg import inv

#Chapter-5, Example 5.24, Page 184

#INPUT DATA
L = 0.05;			#inductance in henry
R2 = 20.;			#resistance in ohms
R1 = 15.;			#resistance in ohms
V = 200.;			#supply voltage in volts
f = 50.;			#frequency in hz
#CALCULATIONS
#for branch 1
Z1 = (R1)+((1j)*(2*math.pi*f*L));			#impedance in ohms
Y1 = inv([[Z1]]);			#admittance in branch
I1 = V*(Y1);			#current in branch
print (I1);
i1 = abs(I1);			#magnitude of current
#for branch 2
Y2 = 1/R2;			#admittance in branch
I2 = V*Y2;			#current in branch
i2 = abs(I2);			#magnitude of current
I = I1+I2;			#total current in A
i = abs(I);			#magnitude of total current
theta = math.atan(I.imag/I.real);			#angle in radians
theta = theta*(180)/(math.pi);			#angle in degrees
print "Thus current in branch1,branch2 abd total currents are %1.2f A, %d A, %2.2f A respectively"%(i1,i2,i);
print "phase angle of the combination is %2.1f degrees"%(theta);

[[ 6.3594338-6.6595835j]]
Thus current in branch1,branch2 abd total currents are 9.21 A, 10 A, 17.66 A respectively
phase angle of the combination is -22.2 degrees


## Example 5.25 Page No : 185¶

In [21]:
import math
from numpy.linalg import inv

#INPUT DATA
L = 6.;			#inductance in millihenry
R2 = 50.;			#resistance in ohms
R1 = 40.;			#resistance in ohms
C = 4.;			#capacitance in uf
V = 100.;			#voltage in volts
f = 800.;			#frequency in hz
#CALCULATIONS
Xl = (2*math.pi*f*L*10**-3);			#inductive reacmath.tance in ohms
Xc = 1/(2*math.pi*f*C*10**-6);			#capacitive reacmath.tance in ohms
Y1 = inv([[(R1)+(1j*Xl)]]);			#admittance of branch1 in mho
Y2 = inv([[(R2)-(1j*Xc)]]);			#admittance of branch2 in mho
I1 = V*(Y1);			#current in branch 1
I2 = V*(Y2);			#current in branch 2
I = I1+I2;			#total curremt in A
theta = (math.atan(I1.imag/I1.real))-math.atan(I2.imag/I2.real);
theta = theta*180/math.pi;			#angle in degrees
print "Thus total current taken from supply is %2.2f"%(abs(I));
print "phase angle between currents of coil and capacitor is %2.2f degrees"%(theta);

Thus total current taken from supply is 2.61
phase angle between currents of coil and capacitor is -81.86 degrees


## Example 5.26 Page No : 186¶

In [24]:
import math

#INPUT DATA
Z1 = 10+(1j*15);			#impedance in ohms
Z2 = 6-(1j*8);			#impedance in ohms
I = 15.;			#current in A
#CALCULATIONS
I1 = ((Z2)/(Z1+Z2))*(I);			#umath.sing current division rule
I2 = ((Z1)/(Z1+Z2))*(I);			#umath.sing current division rule
i1 = abs(I1);			#magnitude of current 1
i2 = abs(I2);			#magnitdude of current 2
P1 = ((i1)**2)*(Z1*(1));			#power consumed by branch 1
P2 = ((i2)**2)*(Z2*(1));			#power consumed by branch 2
print "Thus power consumed by branches 1 and 2 are %3.2f W and %4.1f W respectively"%(P1.real,P2.real);

Thus power consumed by branches 1 and 2 are 737.70 W and 1438.5 W respectively


## Example 5.27 Page No : 187¶

In [26]:
import math
from numpy.linalg import inv
#Chapter-5, Example 5.27, Page 187

#INPUT DATA
V = 200.;			#voltage in volts
f = 50.;			#frequency in hz
R1 = 10.;			#resistance in ohms
L1 = 0.0023;			#inductance in henry
R2 = 5.;			#resistance in ohms
L2 = 0.035;			#inductance in henry
#CALCULATIONS
Xl1 = (2*math.pi*f*L1);			#inductive reacmath.tance in branch 1 in ohm
Xl2 = (2*math.pi*f*L2);			#inductive reacmath.tance in branch 2 in ohm
Y1 = inv([[10+(1j*7.23)]]);			#admittance of branch 1 in mho
Y2 = inv([[5+(1j*10.99)]]);			#admittance of branch 2 in mho
Y = Y1+Y2;			#total admittance in mho
I1 = V*(Y1);			#current through branch1
I2 = V*(Y2);			#current through branch2
I = I1+I2;			#total current in A
theta = math.atan(I.imag/I.real);			#angle in radians
pf_of_combination = math.cos(theta);			#powerfactor---->lagging
print "Thus currents in branch1, branch2 and total current are %2.1f A, %2.1f A and %2.2f A respectively"%(abs(I1),abs(I2),abs(I));
print "pf of combination is %1.3f"%(pf_of_combination);

Thus currents in branch1, branch2 and total current are 16.2 A, 16.6 A and 31.68 A respectively
pf of combination is 0.631


## Example 5.28 Page No : 189¶

In [27]:
import math

#INPUT DATA
f = 50.;			#freq in hz
V = 100.;			#volatge in V
L1 = 0.015;			#inductance in branch 1 in henry
L2 = 0.08;			#inductance in branch 2 in henry
R1 = 2.;			#resistance of branch 1 in ohms
x1 = 4.71;			#reacmath.tance of branch 1 in ohms
R2 = 1.;			#resistance of branch 2 in ohms
x2 = 25.13;			#reacmath.tance of branch 2 in ohms
Z1 = (R1)+(1j*x1);			#impedance of branch1 in ohms
Z2 = (R2)+(1j*x2);			#impedance of branch1 in ohms
I1 = V/Z1;			#current in branch 1 in A
print "current in branch 1 in A"
print (I1);
I2 = V/Z2;			#current in branch 2 in A
print "current in branch 2 in A"
print (I2);
I3 = I1+I2;			#total current in A
print "total current in A"
print (I3);
#note:Answer for real part of total current given in textbook is wrong.Please check the calculations

current in branch 1 in A
(7.63822319652-17.9880156278j)
current in branch 2 in A
(0.158098542505-3.97301637316j)
total current in A
(7.79632173903-21.961032001j)


## Example 5.29 Page No : 189¶

In [28]:
import math
from numpy.linalg import inv

#CALCULATIONS
R = 8;			#resistance in ohms
Xc = -(1j)*12;			#capacitive reacmath.tance in ohms
Y = (inv([[R]])+inv([[Xc]]));			#admittance in mho
print (Y);

[[ 0.125+0.08333333j]]


## Example 5.30 Page No : 189¶

In [29]:
import math
from numpy.linalg import inv

#CALCULATIONS
R = 3;			#resistance in ohms
Xl = (1j)*4;			#inductive reacmath.tance in ohms
Y = (inv([[R]])+inv([[Xl]]));			#admittance in mho
print (Y);

[[ 0.33333333-0.25j]]


## Example 5.31 Page No : 196¶

In [31]:
import math

#INPUT DATA
R = 10.;			#resistance in ohms
L = 10.;			#inductance in milli henry
C = 1.;			#capacitance in uF
V = 200.;			#applied voltage in volts

#CALCULATIONS
fr = 1/(2*math.pi*(math.sqrt(L*C*10**-3*10**-6)));			#resonant frequency in hz
I0 = V/(R);			#current at resonance in A
Vr = I0*R;			#voltage across resistance in volts
Xl = 2*math.pi*fr*L*10**-3;			#inductance in ohms
Vl = I0*Xl;			#voltage across inductor in volts
Xc = inv([[2*math.pi*fr*C*10**-6]]);			#capacitance in ohms
Vc = I0*Xc;			#voltage across capacitor in volts
wr = 2*math.pi*fr			#angular resonant frewuency in rad/sec
Q = (wr*L*10**-3)/(R);			#quality factor
Bw = (fr/Q);			#bandwidth in hz
print "Thus resonant frequency and current are %4.2f hz and %d A respectively"%(fr,I0);
print "voltages across resistance, inductance and capacitance are %d V, %d V and %d V respectively"%(Vr,Vl,Vc);
print "bandwidth and quality factor are %3.2f hz and %d respectively"%(Bw,Q);

Thus resonant frequency and current are 1591.55 hz and 20 A respectively
voltages across resistance, inductance and capacitance are 200 V, 2000 V and 2000 V respectively
bandwidth and quality factor are 159.15 hz and 10 respectively


## Example 5.32 Page No : 196¶

In [1]:
import math
from numpy.linalg import inv
#Chapter-5, Example 5.32, Page 196

#INPUT DATA
V = 220.;			#applied voltage in volts
f = 50.;			#frequency in hz
Imax = 0.4;			#maximum current in A
Vc = 330.;			#voltage across capacitance in volts
#at resonance condition I0 = 0.4 A
I0 = 0.4			#current in A
#CALCULATIONS
Xc = (Vc)/(I0);			#capacitive reacmath.tance in ohms
C = inv([[2*math.pi*f*Xc]]);			#capacitance in F
#at resonance condition Xc = Xl, hence
L = Xc/(2*math.pi*f);			#inductance in henry
R = V/(Imax);			#resistance in ohms
print "Thus resistance, inductance and capacitance are %d ohms, %1.2f H and %g F respectively"%(R,L,C);

Thus resistance, inductance and capacitance are 550 ohms, 2.63 H and 3.8583e-06 F respectively


## Example 5.33 Page No : 197¶

In [2]:
import math

#INPUT DATA
R1 = 5;			#resistance of branch1 in ohms
R2 = 2;			#resistance of branch2 in ohms
L = 10;			#inductance in mH
C = 40;			#capacitance in uF
#CALCULATIONS
fr = (1./(2*math.pi*(math.sqrt(L*C*10**-9))))*(math.sqrt(((C*10**-6*(R1)**2)-L*10**-3)/((C*10**-6*(R2)**2)-L*10**-3)));			#resonant frequency in hz
print "Thus resonant frequency is %f hz"%(fr);

Thus resonant frequency is 240.665502 hz


## Example 5.34 Page No : 197¶

In [3]:
import math

#INPUT DATA
R = 20;			#resistance in ohms
L = 0.2;			#inductance in H
C = 100;			#capacitance in uF
#resistance will be non-inductive only at reosnant frequency
#CALCULATIONS
fr = (1./(2*math.pi*(math.sqrt(L*C*10**-6))))*(math.sqrt((L-(C*10**-6*(R)**2))/(L)));			#resonant frequency in hz
print "Thus resonant frequency is %2.2f hz"%(fr);
Rf = (L)/(C*R*10**-6);			#non-inductive resistance
print "Thus value of non-inductive resistance is %d ohms"%(Rf);

Thus resonant frequency is 31.83 hz
Thus value of non-inductive resistance is 100 ohms


## Example 5.35 Page No : 198¶

In [4]:
#INPUT DATA
Q = 250;			#quality factor
fr = 1.5*10**6;			#resonant freq in hertz

#CALCULATIONS
Bw = (fr)/(Q);			#bandwidth in Hz
hf1 = fr+Bw;			#half power freq 1
hf2 = fr-Bw;			#half power freq 2
print "Thus bandwidth is %d hz"%(Bw);
print "Thus value of half-power frequencies are %g hz and %g hz"%(hf1,hf2);

Thus bandwidth is 6000 hz
Thus value of half-power frequencies are 1.506e+06 hz and 1.494e+06 hz


## Example 5.36 Page No : 198¶

In [5]:
import math

#INPUT DATA
L = 40*10**-3;			#inductance in henry
C = 0.01*10**-6;			#capacitance in uf
#CALCULATIONS
fr = 1./(2*math.pi*math.sqrt(L*C));			#resonant frequency
print "Thus resonant frequency is %d hz"%(fr);

Thus resonant frequency is 7957 hz


## Example 5.37 Page No : 198¶

In [8]:
import math
from numpy.linalg import inv
#Chapter-5, Example 5.37, Page 198

#INPUT DATA
V = 120.;			#source voltage in volts
R = 50.;			#resistance in ohms
L = 0.5;			#inductance in Henry
C = 50.;			#capacitance in uF

#CALCULATIONS
#at Resonance
fr = (1./(2*math.pi*(math.sqrt(L*C*10**-6))));			#resonant frequency in hz
I0 = V/R;			#current at resonance in A
Vl = (1j)*(I0*L);			#voltage developed across inductor in volts
Vc = (-1j)*(I0*L);			#voltage developed across capacitor in volts
Q = (inv([[R]]))*(math.sqrt(L/(C*10**-6)));			#quality factor
Bw = (fr)/(Q);			#Bandwidth in Hz
#given resonance is to occur at 300 rad/sec,then
wr = 300;			#wr = (2*math.pi*f*r)------->measured in Hz
#wr = inv(math.sqrt(L*Cn))
Cr = inv([[L*(wr)**2]]);			#capacitance required in uF
print "Thus resonant frequency, current, quality factor and bandwidth are %2.1f Hz, \
%1.1f A, %d and %2.1f hz respectively"%(fr,I0,Q,Bw);
print "New value of capacitance at 300 rad/sec is %g F"%(Cr)

Thus resonant frequency, current, quality factor and bandwidth are 31.8 Hz, 2.4 A, 2 and 15.9 hz respectively
New value of capacitance at 300 rad/sec is 2.22222e-05 F


## Example 5.38 Page No : 199¶

In [9]:
import math
#INPUT DATA
Q = 45.;			#quality factor
f1 = 600.*10**3;			#freq in Hz
f2 = 1000.*10**3;			#freq in Hz
#given new resistance is 50% greater than former.let us consider two reismath.tances as R1 = 1 ohm and R2 = 1.5 ohm for ease of calculation.Then
R1 = 1;			#resistance in ohm
R2 = 1.5;			#resistance in ohm

#CALCULATIONS
W1 = 2*math.pi*f1;			#angular freq 1 in rad/sec
W2 = 2*math.pi*f2;			#angular freq 2 in rad/sec
Q = 45;			#quality factor
L = (Q*R1)/(W1);			#inductance in henry
Q1 = (W2*L)/(R2);			#new quality factor
print "Thus new quality factor is %d"%(Q1);

Thus new quality factor is 50


## Example 5.39 Page No : 199¶

In [10]:
import math
from numpy.linalg import inv

#INPUT DATA
R = 4.;			#resistance in ohm
L = 100.*10**-6;			#inductance in henry
C = 250.*10**-12;			#capacitance in Farads

#CALCULATIONS
fr = inv([[2*math.pi*math.sqrt(L*C)]]);			#resonant frequency in Hz
Q = (inv([[R]]))*(math.sqrt(L/C));			#Q-factor
Bw = fr/Q;			#bandwidth in Hz
hf1 = fr+Bw;			#halfpower freq1 in Hz
hf2 = fr-Bw;			#halfpower freq2 in Hz

print "Thus resonant freq, Q-factor and new halfpower frequencies are %d hz , %d, %g hz, %g hz respectively"%(fr,Q,hf1,hf2);

Thus resonant freq, Q-factor and new halfpower frequencies are 1006584 hz , 158, 1.01295e+06 hz, 1.00022e+06 hz respectively


## Example 5.40 Page No : 200¶

In [11]:
import math
from numpy.linalg import inv

#INPUT DATA
R = 10;			#resistance in ohm
L = 10**-3;			#inductance in henry
C = 1000*10**-12;			#capacitance in Farads
V = 20;			#voltage in volts
#CALCULATIONS
fr = inv([[2*math.pi*math.sqrt(L*C)]]);			#resonant frequency in Hz
Q = (inv([[R]]))*(math.sqrt(L/C));			#Q-factor
Bw = fr/Q;			#bandwidth in Hz
hf1 = fr+Bw;			#halfpower freq1 in Hz
hf2 = fr-Bw;			#halfpower freq2 in Hz
print "Thus resonant freq, Q-factor and new halfpower frequencies are %d hz , %d , %g hz, %g hz respectively"%(fr,Q,hf1,hf2);

Thus resonant freq, Q-factor and new halfpower frequencies are 159154 hz , 100 , 160746 hz, 157563 hz respectively


## Example 5.41 Page No : 208¶

In [14]:
import math

#INPUT DATA
P1 = 1000.;			#power1 in watts
P2 = 1000.;			#power2 in watts
#CALCULATIONS
#for case(1)
Pt = P1+P2;			#total power in watts
phi = math.atan(math.sqrt(3)*((P2-P1)/(P2+P1))*(180/math.pi));			#math.since math.tan(phi) = math.sqrt(3)*((P2-P1)/(P2+P1)))
pf = math.cos(phi);
print "Thus power and powerfactor are %d W ,%d respectively"%(Pt,pf);
#for case(2)
P3 = 1000;			#power3 in watts
P4 = -1000;			#power4 in watts
Pt1 = P3+P4;			#total power in watts
pf1 = 0;			#math.since we cannot perform division by zero in scilab,it doesn't consider it as infinite quantity to yield 90 degree angle and hence powerfactor 0
print "Thus power and powerfactor are %d W ,%d respectively"%(Pt1,pf1);

Thus power and powerfactor are 2000 W ,1 respectively
Thus power and powerfactor are 0 W ,0 respectively


## Example 5.42 Page No : 209¶

In [15]:
import math

#INPUT DATA
V1 = 400.;			#voltage in volts
Z1 = (3.+((1j)*4));			#impedance in ohms
#CALCULATIONS
#in star connected system,phase voltage = (line voltage)
Ep = V1/(math.sqrt(3));			#voltage in volts
Ip = Ep/Z1;			#current in A
ip1 = abs(Ip);			#line current in A
theta = math.atan(Ip.imag/Ip.real);
Pt = math.sqrt(3)*V1*ip1*math.cos(theta);			#total power consumed in load in W
print "Thus total power consumed in load is %f W"%(Pt);
#note:for line current the answer given is 46.02A instead of 46.2 A and hence total power consumed changes

Thus total power consumed in load is 19200.000000 W


## Example 5.43 Page No : 209¶

In [16]:
#INPUT DATA
V1 = 400;			#voltage in volts
Il = 10;			#current in A
#CALCULATIONS
#in star connected system,phase current = (line current) = I1
phase_voltage = (V1)/(math.sqrt(3));			#voltage in Volts
print "Thus phase voltage is %1.0f V"%(phase_voltage);

Thus phase voltage is 231 V


## Example 5.44 Page No : 209¶

In [19]:
import math
#Chapter-5, Example 5.44, Page 209

#INPUT DATA
Z1 = (6-((1j)*8));			#impedance1 in ohms
Z2 = (16+((1j)*12));			#impedance2 in ohms
I1 = (12+((1j)*16));			#current in A
#CALCULATIONS
V = I1*Z1;			#applied voltage in volts
I2 = V/(Z2);			#current in other branch in A
print "current in other branch in Amps"
print (I2);
I = I1+I2;			#total current in A
print "total current in Amps";
print (I);
i1 = abs(I);			#magnitude in A
i2 = math.atan(I.imag/I.real);
P = V*i1*math.cos(i2);			#power consumed in circuit
print "Thus voltage applied and power consumed are %d V and %d W respectively"%(V.real,P.real);

current in other branch in Amps
(8-6j)
total current in Amps
(20+10j)
Thus voltage applied and power consumed are 200 V and 4000 W respectively


## Example 5.45 Page No : 210¶

In [20]:
import math

#INPUT DATA
Vl = 415.;			#voltage in volts
Z = (4+((1j)*6));			#impedance in each phase in ohm
#CALCULATIONS
Ip = Vl/Z;			#current in each phase in A
ip1 = abs(Ip);			#magnitude of Ip
Il = (math.sqrt(3))*(ip1);			#line current in A
phi = math.atan(Ip.imag/Ip.real)
P = (math.sqrt(3))*Vl*Il*math.cos(phi);			#power supplied in W
print "Thus power supplied is %d W"%(P);
#note:the math.cosfunction of scilab and calculator will differ slightly

Thus power supplied is 39744 W


## Example 5.46 Page No : 210¶

In [21]:
#INPUT DATA
Vl = 400;			#voltage in volts
Il = 20;			#current in A
f = 50;			#freq in hz
pf = 0.3			#power factor
#CALCULATIONS
Ip = Il/math.sqrt(3);			#phase current in A
Z = Vl/Ip;			#impedance in each phase in ohms
phi = math.acos(0.3);			#angle in radians
Zb = Z*(math.cos(phi)+(1j)*math.sin(phi));			#impedance connected in each phase
print "Thus impedance connected in each phase in ohms";
print (Zb);

Thus impedance connected in each phase in ohms
(10.3923048454+33.0454232837j)


## Example 5.47 Page No : 210¶

In [22]:
import math

#INPUT DATA
P1 = 6*10**3;			#power in Kw
P2 = -1*10**3;			#power in Kw
#CALCULATIONS
P = P1+P2;			#total power in Kw
a = math.atan(math.sqrt(3)*((P2-P1)/(P2+P1)));
pf = math.cos(a);			#power factor
print "Thus power and power factor are %d W and %1.2f respectively"%(P,pf);

Thus power and power factor are 5000 W and 0.28 respectively


## Example 5.48 Page No : 211¶

In [23]:
import math

#INPUT DATA
Z = 3-((1j)*4);			#impedance in ohms
Vl = 400;			#line voltage in volts
#CALCULATIONS
Vp = Vl/(math.sqrt(3));			#phase voltage in volts
Ip = Vp/abs(Z);			#phase current in Amps
#line current(Il) = phase current(Ip)
Il = Ip;			#line current in A
power_factor = math.cos(math.atan(Z.imag/Z.real));
power_consumed = math.sqrt(3)*Vl*Il*power_factor;
print "Thus power consumed and power factor are %f W and %1.1f respectively"%(power_consumed,power_factor);
#note:answer computed for power consumed in textbook is wrong.Please check the calculations

Thus power consumed and power factor are 19200.000000 W and 0.6 respectively


## Example 5.49 Page No : 211¶

In [24]:
import math

#INPUT DATA
Il = 10.;			#current in Amps
Vl = 400.;			#line voltage in volts
#CALCULATIONS
Vp = Vl/(math.sqrt(3));			#line to neutral voltage
Ip = Il;			#phase current in Amps
print "Thus line to neutral voltage  and phase current are %1.0f V and %d A respectively"%(Vp,Ip);

Thus line to neutral voltage  and phase current are 231 V and 10 A respectively


## Example 5.50 Page No : 211¶

In [26]:
import math

#INPUT DATA
P1 = 2000;			#power in watts
P2 = 1000;			#power in watts
Vl = 400;			#line voltage in volts
#CALCULATIONS
P = P1+P2;			#power in Watts
a = math.sqrt(3*(P1-P2)/(P1+P2));
b = math.atan(math.sqrt(a));
power_factor = math.cos(b);
kVA = P/power_factor;
print "Thus power, power factor and kVA are %d W , %1.3f and %1.2f respectively"%(P,power_factor,kVA);
#note:computed value for powerfactor and kVA in textbook are wrong.Please check the calculations

Thus power, power factor and kVA are 3000 W , 0.707 and 4242.64 respectively