Chapter 13:Alternating Current

Problem no:13.2, Page no:150

In [4]:
from __future__ import division

#Cal of Frequency,Velocity,Amplitude

#Initialization

T=4                              #Time period in s

Lambda=30                        #wavelength in m

f=0.25                           #Frequency in Hz

tot_range=2                      #in m

#Calculation

f=1/T

v=f*Lambda

print "(a) f=",f,"Hz"

print "(b) v=",v,"m/s"

print "(c) The amplitude is half the total range,hence A=",int(tot_range/2),"m"
(a) f= 0.25 Hz
(b) v= 7.5 m/s
(c) The amplitude is half the total range,hence A= 1 m

Problem no:13.3, Page no:150

In [1]:
from __future__ import division

#Cal of Frequency

#Initialization

v=25                         #Wave velocity in cm/s

Lambda=0.1                   #Wavelength in mm

#Calculation

Lambda=Lambda/1000           #Convert Wavelength in m

v=v/100                      #Convert wave velocity in m/s

f=v/Lambda

print "f=",int(f),"Hz"
f= 2500 Hz

Problem no:13.4, Page no:151

In [2]:
from __future__ import division

#Cal of Wavelength

#Initialization

v=5020                       #Velocity in ft/s

f=256                        #Frequency in Hz

#Calculation

Lambda=v/f

print "Lambda=",round(Lambda,1),"ft"
Lambda= 19.6 ft

Problem no:13.5, Page no:151

In [6]:
from __future__ import division

#Cal of Frequency

#Initialization

Lambda=3.2                     #Wavelength in cm

c=3E+8                         #Velocity of light in m/s

#Calculation

Lambda=Lambda/100              #Convert Wavelength in m

f=c/Lambda

print "f=",'%0.1E' % f,"Hz"
f= 9.4E+09 Hz

Problem no:13.7, Page no:152

In [5]:
from __future__ import division

#Cal of Effective ac potential difference

#Initialization

Vmax=300              #Potential difference in V

#Calculation

Veff=0.707*(Vmax)

print "Veff=",int(Veff),"V"
Veff= 212 V

Problem no:13.8, Page no:152

In [4]:
from __future__ import division

#Cal of Effective current

#Initialization

Imax=10                 #Current in A

R=20                    #Resistance in Ohm     

#Calculation

Ieff=0.707*Imax

P=(Ieff**2)*R

print "Ieff=",Ieff,"A"

print "P=",int(round(P)),"W"
Ieff= 7.07 A
P= 1000 W

Problem no:13.9, Page no:153

In [10]:
from __future__ import division

#Cal of Secondary voltage and current

#Initialization

N1=100                     #No.of turns in primary winding

N2=500                     #No. of turns in secondary winding

V1=120                     #Primary voltage in V

I1=3                       #Primary current in  A

#Calculation

V2=(N2/N1)*V1

I2=(N1/N2)*I1

print "V2=",int(V2),"V"

print "I2=",I2,"A"
V2= 600 V
I2= 0.6 A

Problem no:13.10, Page no:154

In [9]:
from __future__ import division

#Initialization

V1=5000                   #primary voltage in V

V2=240                    #Secondary voltage in V

P=10                      #Power in kW

#Cal of Ratio of turns

N1_N2=V1/V2

P=P*1000                  #Convert power in W

#Cal of Maximum current

I2=P/V2

print "(a) N1_N2=",round(N1_N2,1)

print "(b) I2=",round(I2,1),"A"
(a) N1_N2= 20.8
(b) I2= 41.7 A