# Chapter 12: Resistance and DC Circuits¶

## Example 12.1, Page 237¶

In :
#Initialization
i1=8                                 #current in Amp
i2=1                                #current in Amp
i3=4                                #current in Amp

#Calculation
i4=i2+i3-i1                                #current in Amp

#Results
print'Magnitude, I4 = %d A'%i4

Magnitude, I4 = -3 A


## Example 12.2, Page 239¶

In [ ]:
#Initialization
e=12                   #EMF source in volt
v1=3                   #node voltage
v3=3                   #node voltage

#Calculation
v2=v1+v3-e                   #node voltage

#Results
print'V2 = %d V'%v2


## Example 12.4, Page 242¶

In :
import numpy as np

#We have used method II for solving our problem by using simultaneous equations

a = np.array([[25,-2],[400,-8]])
b = np.array([,])
c=np.linalg.solve(a,b)

print'Voc = %d V'%c
print'R = %d ohm'%c


Voc = 10 V
R = 100 ohm


## Example 12.5, Page 244¶

In :
#Initialization
r1=100                  #Resistance in Ohm
r2=200                 #Resistance in Ohm
r3=50                 #Resistance in Ohm
v1=15                 #voltage source
v2=20                 #voltage source

#Calculation
#Considering 15 V as a source & replace the other voltage source by its internal resistance,
r11=(r2*r3)*(r2+r3)**-1              #resistance in parallel
v11=v1*(r11/(r1+r11))               #voltage
#Considering 20 V as a source & replace the other voltage source by its internal resistance,
r22=(r1*r3)*(r1+r3)**-1              #resistance in parallel
v22=v2*(r22/(r2+r22))               #voltage

#output of the original circuit
v33=v11+v22

#Results
print'Voltage, V = %.2f'%v33

Voltage, V = 7.14


## Example 12.6, Page 246¶

In :
#Initialization
r1=10                  #Resistance in Ohm
r2=5                  #Resistance in Ohm
v2=5                 #voltage source
i=2                 #current in Amp

#Calculation
#Considering 5 V as a source & replace the current source by its internal resistance,
i1=v2*(r1+r2)**-1                   #current using Ohms law
#Considering current source & replace the voltage source by its internal resistance,
r3=(r1*r2)*(r1+r2)**-1              #resistance in parallel
v3=i*r3                         #voltage using Ohms law
i2=v3*r2**-1                   #current using Ohms law
i3=i1+i2                       #total current

#Results
print'Output Current, I = %.2f A'%i3

Output Current, I = 1.67 A


## Example 12.8, Page 251¶

In :
import numpy as np
r=25                    #resistance in ohm

#We have used for solving our problem by using simultaneous equations

a = np.array([[(-13*60**-1),(1*20**-1)],[(1*60**-1),(-9*100**-1)]])
b = np.array([[-5],[-100*30**-1]])
c=np.linalg.solve(a,b)
i1=c/r                        #required current

print'V2 = %.2f V'%c          #wrong answer in textbook
print'V3 = %.2f V'%c          #wrong answer in textbook
print'Current, I1 = %.2f A'%i1


V2 = 33.04 V
V3 = 43.15 V
Current, I1 = 1.73 A


## Example 12.9, Page 253¶

In :
import numpy as np
re=10                 #resistance in ohm

#We have used for solving our problem by using simultaneous equations

a = np.array([[(-160),(20), (30)],[(20),(-210), (10)], [(30),(10), (-190)]])
b = np.array([[-50],,])
c=np.linalg.solve(a,b)
ve=re*(c-c)

print'I1 = %d mA'%(c*10**3)             #current I1
print'I2 = %d mA'%(c*10**3)              #current I2
print'I3 = %d mA'%(c*10**3)              #current I3
print'Voltage, Ve = %.3f V'%ve


I1 = 326 mA
I2 = 33 mA
I3 = 53 mA
Voltage, Ve = 0.197 V

In [ ]: