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
```

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
```

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);
```

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);
```

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)
```

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)
```

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);
```

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)
```

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
#hence pf is leading
#CALCULATIONS
pf = math.cos(phi*math.pi/180); #in degrees--->leading
#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);
```

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
pf = 0.707; #powerfactor------>leading------>math.cos(phi)
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);
```

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
pf = R/Z; #power factor------>leading
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 leads voltage by phi
#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);
```

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);
```

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);
```

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
r = (V/I)-(R); #additional resistance to be added in series
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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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
print "admittance is %1.5f mho"%(y);
```

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);
```

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)
```

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);
```

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);
```

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);
```

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);
```

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);
```

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
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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)
```

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);
```

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);
#note:given answers are wrong in textbook.Please check the answers
```

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);
```

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);
```

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
```

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);
```

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);
```

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
```

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);
```

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);
```

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
```

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);
```

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
```