Chapter 14:Series AC Circuits

Problem no:14.2, Page no:160

In [1]:
from __future__ import division

import math

#Initialization

C=10                 #Capacitance in muF

V=15                 #Voltage in V

f=5                  #Frequency in kHz

#Cal of reactance of capacitor

f=f*1000

C=C*(10**(-6))

Xc=round(1/(2*math.pi*f*C),2)

print "(a) Xc=",Xc,"Ohm"

#Cal of Current

I=V/Xc

print "(b) I=",round(I,2),"A"

(a) Xc= 3.18 Ohm
(b) I= 4.72 A

Problem no:14.4, Page no:160

In [2]:
from __future__ import division

import math

#Initializtion

V=24                        #Voltage in V

f=60                        #Frequency in Hz

L=0.3                       #Inductance in H

#Cal of Reactance

XL=2*math.pi*f*L

print "(a) XL=",int(XL),"Ohm"

#Cal of Current

I=V/XL

print "(b) I=",round(I,2),"A"

(a) XL= 113 Ohm
(b) I= 0.21 A

Problem no:14.5, Page no:160

In [4]:
from __future__ import division

import math

#Cal of Inductance

#Initialization

f=500                       #Frequency in Hz

XL=80                       #Reactance in Ohm

#Calculation

L=XL/(2*math.pi*f)

L=L*1000 #Convert L in mH

print "L=",round(L,1),"mH"

L= 25.5 mH

Problem no:14.11, Page no:167

In [5]:
from __future__ import division

import math

#Initialization

L=5                                #inductance in mH

C=5                                #Capacitance in pF

R=5                                #Resistance in Ohm

V=5E-4

#Cal of frequency

L=L/1000                           #Convert L in H

C=C/(10**12)                       #Convert C in F


fo=1/(2*math.pi*(math.sqrt(L*C)))  #Frequency

fo=fo/1000                         #Convert fo in kHz

print "(a) fo=",int(math.ceil(fo)),"kHz"

#Cal of Current

#At Resonance XL=XC and Z=R

I=V/R

I=I*1000

print "(b) I=",I,"mA"

(a) fo= 1007 kHz
(b) I= 0.1 mA