# Chapter 13: D.c. circuit theory

### Example 1, page no. 168

In [1]:
from __future__ import division
import math
#initializing  the  variables:
Iab  =  50;#  in  ampere
Ibc  =  20;#  in  ampere
Iec  =  15;#  in  ampere
Idf  =  120;#  in  ampere
Ifg  =  40;#  in  ampere
Iab  =  50;#  in  ampere
I  =  2;#  in  ampere
V1  =  4;#  in  volts
V2  =  3;#  in  volts
V3  =  6;#  in  volts
R1  =  1;#  in  ohms
R2  =  2;#  in  ohms
R3  =  2.5;#  in  ohms
R4  =  1.5;#  in  ohms

#calculation:
I1  =  Iab  -  Ibc
I2  =  Ibc  +  Iec
I3  =  I1  -  Idf
I4  =  Iec  -  I3
I5  =  Idf  -  Ifg
#  Applying  Kirchhoff’s  voltage  law  and  moving  clockwise  around  the  loop  of  Figure  13.3(b)  starting  at  point  A:
E  =  I*R2  +  I*R3  +  I*R4  +  I*R1  -  V2  -  V3  +  V1

#Results
print  "\n\n  Result  \n\n"
print  "\n (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are  ",I1,"A,  ",  I2,"A,  ",  I3,"A,  ",  I4,"A,  ",  I5,"A  respetively\n"
print  "\n (b)  value  of  e.m.f.  E  =  ",E,"  Volts\n"


Result

(a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are   30 A,   35 A,   -90 A,   105 A,   80 A  respetively

(b)  value  of  e.m.f.  E  =   9.0   Volts


### Example 2, page no. 168

In [2]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  4;#  in  volts
E2  =  2;#  in  volts
r1  =  2;#  in  ohms
r2  =  1;#  in  ohms
R  =  4;#  in  ohms

#calculation:
#E1 = I1*(r1 + R) + I2*R,
#E2 = I1*R + (R + r2)*I2,
I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))
I1 = (E1 - I2*R)/(r1 + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n  currents flowing are, I1 = ",round(I1,3),"A,  and I2 = ",round(-1*I2,3),"A, "
print   "and current flowing in middle branch is I1 - I2 = ", round(I1 + I2, 3),"A"


Result

currents flowing are, I1 =  0.857 A,  and I2 =  0.286 A, and current flowing in middle branch is I1 - I2 =  0.571 A

### Example 3, page no. 169

In [3]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  4;#  in  volts
E2  =  12;#  in  volts
R1  =  0.5;#  in  ohms
R2  =  2;#  in  ohms
R3  =  5;#  in  ohms

#calculation:
#E1 + E2 = I1*R1 + I2*R2
#E2 = - I1*R3 + I2*(R2 + R3)
I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)
I1 = (E1 + E2 - I2*R2)/R1

#Results
print  "\n\n  Result  \n\n"
print  "\n  currents flowing are, I1 = ",round(I1,2),"A,  and I2 = ",round(I2,2),"A"
print   " and in R3 branch is I1 - I2", round(I1 - I2,2)," A"


Result

currents flowing are, I1 =  6.52 A,  and I2 =  6.37 A and in R3 branch is I1 - I2 0.15  A

### Example 4, page no. 170

In [4]:
from __future__ import division
import math
#initializing  the  variables:
V  =  54;#  in  volts
I  =  8;#  in  Amps
R1  =  2;#  in  ohms
R2  =  11;#  in  ohms
R3  =  14;#  in  ohms
R4  =  3;#  in  ohms
R5  =  32;#  in  ohms

#calculation:
#V = (R1 + R2)*I1 - R2*I2
#0 = (R1 + R3)*I1 - R5*I2 - R3*I
I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)
I2 = -1*(V - I1*(R2 + R1))/R2

#Results
print  "\n\n  Result  \n\n"
print  "\n  the current flowing in the 2 ohm resistor = ",round(I1,0),"A,"
print  "\n  the current flowing in the 14 ohm resistor = ",round(I - I1,0),"A,"
print  "\n  the current flowing in the 32 ohm resistor = ",round(I2,0),"A,"
print  "\n  the current flowing in the 11 ohm resistor = ",round(I1 - I2,0),"A,"
print  "\n  the current flowing in the 3 ohm resistor = ",round(I - I1 + I2,0),"A"


Result

the current flowing in the 2 ohm resistor =  5.0 A,

the current flowing in the 14 ohm resistor =  3.0 A,

the current flowing in the 32 ohm resistor =  1.0 A,

the current flowing in the 11 ohm resistor =  4.0 A,

the current flowing in the 3 ohm resistor =  4.0 A

### Example 5, page no. 171

In [5]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  4;#  in  volts
E2  =  2;#  in  Volts
R1  =  2;#  in  ohms
R2  =  1;#  in  ohms
R =  4;#  in  ohms

#calculation:
I1 = E1/(R1 + (R*R2/(R + R2)))
I2 = (R2/(R + R2))*I1
I3 = (R/(R + R2))*I1
I4 = E2/(R2 + (R*R1/(R + R1)))
I5 = (R1/(R + R1))*I4
I6 = (R/(R + R1))*I4

#Results
print  "\n\n  Result  \n\n"
print  "\n  Resultant current flowing through source 1 = ",round(I1 - I6,3),"A,"
print  "\n  Resultant current flowing through source 2 = ",round(I4 - I3,3),"A,"
print  "\n  Resultant current flowing through resistor R, = ",round(I2 + I5,3),"A,"


Result

Resultant current flowing through source 1 =  0.857 A,

Resultant current flowing through source 2 =  -0.286 A,

Resultant current flowing through resistor R, =  0.571 A,

### Example 6, page no. 173

In [6]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  8;#  in  volts
E2  =  3;#  in  Volts
R1  =  3;#  in  ohms
R2  =  2;#  in  ohms
R =  18;#  in  ohms

#calculation:
I1 = E1/(R1 + (R*R2/(R + R2)))
I2 = (R/(R + R2))*I1
I3 = (R2/(R + R2))*I1
I4 = E2/(R2 + (R*R1/(R + R1)))
I5 = (R/(R + R1))*I4
I6 = (R1/(R + R1))*I4

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resultant current in the 18 ohm resistor= ",round(I3 - I6,3),"A "
print   "and P.d. across the 18 ohm resistor",round((I3-I6)*R,3),"V"
print  "\n  (b)the current in the 8 V battery= ",round(I1 + I5,3),"A"
print  "\n  (c)current in the 3 V battery = ",round(I2 + I4,3),"A"


Result

(a)Resultant current in the 18 ohm resistor=  0.073 A and P.d. across the 18 ohm resistor 1.313 V

(b)the current in the 8 V battery=  2.229 A

(c)current in the 3 V battery =  2.156 A

### Example 7, page no. 176

In [7]:
from __future__ import division
import math
#initializing  the  variables:
V  =  10;#  in  volts
R1  =  2;#  in  ohms
R2  =  8;#  in  ohms
R3 =  5;#  in  ohms
R =  10;#  in  ohms

#calculation:
I1 = V/(R1 + R2)
E = I1*R2
r = R3 + R1*R2/(R1 + R2)
I = E/(R + r)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 10 ohm resistor = ",round(I,3),"A"


Result

the current flowing in the 10 ohm resistor =  0.482 A

### Example 8, page no. 177

In [8]:
from __future__ import division
import math
#initializing  the  variables:
V  =  12;#  in  volts
R1  =  1;#  in  ohms
R2  =  4;#  in  ohms
R3 =  5;#  in  ohms
R =  0.8;#  in  ohms

#calculation:
I1 = V/(R1 + R2 + R3)
E = I1*R2
r = R2*(R1 + R3)/(R1 + R2 + R3)
I = E/(R + r)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 0.8 ohm resistor = ",round(I,3),"A"


Result

the current flowing in the 0.8 ohm resistor =  1.5 A

### Example 9, page no. 178

In [9]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  4;#  in  volts
E2  =  2;#  in  volts
R1  =  2;#  in  ohms
R2  =  1;#  in  ohms
R =  4;#  in  ohms

#calculation:
I1 = (E1 - E2)/(R1 + R2)
E = E1 - I1*R1
r = R2*R1/(R1 + R2)
I = E/(R + r)
P = R*I**2

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 4 ohm resistor = ",round(I,3),"A and power dissipated in the 4 ohm resistor = ",round(P,3),"W"


Result

the current flowing in the 4 ohm resistor =  0.571 A and power dissipated in the 4 ohm resistor =  1.306 W

### Example 10, page no. 178

In [10]:
from __future__ import division
import math
#initializing  the  variables:
V  =  24;#  in  volts
R  =  3;#  in  ohms
R1  =  20;#  in  ohms
R2 =  5;#  in  ohms
R3  = 10;#  in  ohms
R4 =  5/3;#  in  ohms

#calculation:
E = R3*V/(R3 + R2)
r = R4 + R3*R2/(R3 + R2)
I = E/(R + r)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 3 ohm resistor = ",round(I,0),"A"


Result

the current flowing in the 3 ohm resistor =  2.0 A

### Example 11, page no. 179

In [11]:
#determine the current in the 32 ohm resistor using Th´evenin’s theorem
from __future__ import division
import math
#initializing  the  variables:
E  =  54;#  in  volts
R1  =  2;#  in  ohms
R2  =  14;#  in  ohms
R3  =  3;#  in  ohms
R4  =  11;#  in  ohms
R5  =  32;#  in  ohms

#calculation:
Vac = R1*E/(R1 + R4)
Vbc = R2*E/(R2 + R3)
V = Vbc - Vac
r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)
I = V/(R5 + r)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 32 ohm resistor = ",round(I,0),"A flowing from A to B"


Result

the current flowing in the 32 ohm resistor =  1.0 A flowing from A to B

### Example 12, page no. 181

In [12]:
from __future__ import division
import math
#initializing  the  variables:
E  =  10;#  in  volts
R1  =  2;#  in  ohms
R2  = 8;#  in  ohms
R3  =  5;#  in  ohms
R  =  10;#  in  ohms

#calculation:
Isc = E/R1
r = R1*R2/(R1 + R2)
I = r*Isc/(r + R3 + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 10 ohm resistor = ",round(I,3),"A"


Result

the current flowing in the 10 ohm resistor =  0.482 A

### Example 13, page no. 182

In [13]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  4;#  in  volts
E2  =  2;#  in  volts
R1  =  2;#  in  ohms
R2  =  1;#  in  ohms
R  =  4;#  in  ohms

#calculation:
Isc = E1/R1 + E2/R2
r = R1*R2/(R1 + R2)
I = r*Isc/(r + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 4 ohm resistor = ",round(I,3),"A"


Result

the current flowing in the 4 ohm resistor =  0.571 A

### Example 14, page no. 182

In [14]:
from __future__ import division
import math
#initializing  the  variables:
V  =  24;#  in  volts
R  =  3;#  in  ohms
R1  =  20;#  in  ohms
R2 =  5;#  in  ohms
R3  = 10;#  in  ohms
R4 =  5/3;#  in  ohms

#calculation:
Isc = V/R2
r = R3*R2/(R3 + R2)
I = r*Isc/(r + R4 + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 3 ohm resistor = ",round(I,0),"A"


Result

the current flowing in the 3 ohm resistor =  2.0 A

### Example 15, page no. 183

In [15]:
from __future__ import division
import math
#initializing  the  variables:
I  =  15;#  in  Amps
R  =  2;#  in  ohms
R1  = 6;#  in  ohms
R2 =  4;#  in  ohms
R3  = 8;#  in  ohms
R4 =  7;#  in  ohms

#calculation:
Isc = R1*I/(R1 + R2)
r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)
I = r*Isc/(r + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n the current flowing in the 2 ohm resistor = ",round(I,2),"A"


Result

the current flowing in the 2 ohm resistor =  6.75 A

### Example 16, page no. 185

In [16]:
from __future__ import division
import math
#initializing  the  variables:
V  =  10;#  in  Volts
R  =  2;#  in  ohms

#calculation:
Isc = V/R

#Results
print  "\n\n  Result  \n\n"
print  "\n equivalent Norton network contains Current Source of amp = ",round(Isc,0),"A and a resistor of ",R," ohm"


Result

equivalent Norton network contains Current Source of amp =  5.0 A and a resistor of  2  ohm

### Example 17, page no. 185

In [17]:
from __future__ import division
import math
#initializing  the  variables:
Isc  =  4;#  in  Amps
R  =  3;#  in  ohms

#calculation:
E = Isc*R

#Results
print  "\n\n  Result  \n\n"
print  "\n equivalent Thevenin network contains Voltage Source of ",round(E,0),"V and a resistor of ",R," ohm"


Result

equivalent Thevenin network contains Voltage Source of  12.0 V and a resistor of  3  ohm

### Example 18, page no. 185

In [18]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  12;#  in  Volts
E2  =  24;#  in  Volts
R1  =  3;#  in  ohms
R2  =  2;#  in  ohms
R  =  1.8;#  in  ohms

#calculation:
Isc1 = E1/R1
Isc2 = E2/R2
I1 = Isc1 + Isc2
r = R1*R2/(R1 + R2)
E = I1*r
I = E/(r + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (A)equivalent Norton network contains Current Source of ",round(I1,1),"A and a resistor of ",r," ohm"
print  "\n     equivalent Thevenin network contains Voltage Source of ",round(E,1),"V and a resistor of ",r," ohm"
print  "\n  (B)the current flowing in the 1.8 ohm resistor is ",round(I,1),"A"


Result

(A)equivalent Norton network contains Current Source of  16.0 A and a resistor of  1.2  ohm

equivalent Thevenin network contains Voltage Source of  19.2 V and a resistor of  1.2  ohm

(B)the current flowing in the 1.8 ohm resistor is  6.4 A

### Example 19, page no. 186

In [19]:
from __future__ import division
import math
#initializing  the  variables:
E1  =  10;#  in  Volts
E2  =  6;#  in  Volts
I1 = 0.001;#in Amp
R1  =  2000;#  in  ohms
R2  =  3000;#  in  ohms
R3  =  600;#  in  ohms
R  =  200;#  in  ohms

#calculation:
Isc1 = E1/R1
Isc2 = E2/R2
I2 = Isc1 + Isc2
r1 = R1*R2/(R1 + R2)
E = I2*r1 - I1*R3
r = r1 + R3
I = E/(r + R)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (A)equivalent Norton network contains Current Source of ",round(I2*1000,0),"mA and a resistor of ",round(r1/1000,1)," Kohm"
print  "\n     equivalent Thevenin network contains Voltage Source of ",round(E,1),"V and a resistor of ",round(r1/1000,1)," Kohm"
print  "\n  (B)the current flowing in the 200 ohm resistor is ",round(I*1000,1),"mA"


Result

(A)equivalent Norton network contains Current Source of  7.0 mA and a resistor of  1.2  Kohm

equivalent Thevenin network contains Voltage Source of  7.8 V and a resistor of  1.2  Kohm

(B)the current flowing in the 200 ohm resistor is  3.9 mA

### Example 20, page no. 188

In [1]:
from __future__ import division
import math
#from pylab import *
%matplotlib inline
#initializing  the  variables:
E  =  6;#  in  Volts
R  =  2.5;#  in  ohms

#calculation:
RL = []
P = []
k = []
for h in range(6):
RL.append(h - 0.5)
k = h - 0.5
P.append(k*((E/(R + k))**2))
fig  = plt.figure()
ax.plot(RL,P,'-')
#plot(RL,P,'-')
xlabel('RL(ohm)')
ylabel('Power(W)')
show()
Pmax= R*(E/(2*R))**2

#Results
print  "\n\n  Result  \n\n"
print  "\n  A graph of RL against P is shown in Figure"
print  "  The maximum value of power is", Pmax,"W which occurs when RL =",R,"ohm"

Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.

Result

A graph of RL against P is shown in Figure
The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm

### Example 21, page no. 188

In [21]:
from __future__ import division
import math
from pylab import *
#initializing  the  variables:
E  =  30;#  in  Volts
R  =  1.5;#  in  ohms

#calculation:
RL = R
I = E/(R + RL)
P = I**2*RL

#Results
print  "\n\n  Result  \n\n"
print  "  The maximum value of power is", P,"W which occurs when RL =",RL,"ohm"


Result

The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm

### Example 22, page no. 189

In [22]:
from __future__ import division
import math
from pylab import *
#initializing  the  variables:
V  =  15;#  in  Volts
R1  =  3;#  in  ohms
R2  =  12;#  in  ohms

#calculation:
E = (R2/(R2+ R1))*V
r = R1*R2/(R1 + R2)
RL = r
I = E/(r + RL)
P = I**2*RL

#Results
print  "\n\n  Result  \n\n"
print  "  The maximum value of power is", P,"W which occurs when Total Load RL =",RL,"ohm"


Result

The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm