Chapter 21 : Inductive Circuits

Example No. 21_1 Page No. 650

In [4]:
from math import sqrt,atan,pi
# If a R=30 ohms and Xl=40 ohms are in series with 100V applied, find the following: Zt, I, Vr, Vl and Theta z. What is the phase angle between Vl and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt

# Given data

R = 30.#     # Resistance=30 Ohms
Xl = 40.#    # Inductive reactance=40 Ohms
Vt = 100.#   # Applied voltage=100 Volts


R1 = R*R#
Xl1 = Xl*Xl#

Zt = sqrt(R1+Xl1)#
print 'Zt = %0.2f ohms'%Zt

I = (Vt/Zt)#
print  'I = %0.2f Ampers'%I

Vr = I*R#
print  'Vr = %0.2f Volts'%Vr

Vl = I*Xl#
print  'Vl = %0.2f Volts'%Vl

Oz = atan(Xl/R)*180/pi
print  'Theta z = %0.2f degree'%Oz

#Prove that the sum of the series voltage drop equals the applied voltage Vt

Vt = sqrt((Vr*Vr)+(Vl*Vl))#
print  'Sum of Voltage Drop = %0.f which is Equal to Applied Voltage 100V '%Vt

Zt = 50.00 ohms
I = 2.00 Ampers
Vr = 60.00 Volts
Vl = 80.00 Volts
Theta z = 53.13 degree
Sum of Voltage Drop = 100 which is Equal to Applied Voltage 100V

Example No. 21_2 Page No. 654

In [5]:
from math import sqrt
# What is the total Z of a 600-Ohms R in parallel with a 300 Ohms Xl? Assume 600 V for the applied voltage.

# Given data

R = 600.#    # Resistance=600 Ohms
Xl = 300.#   # Inductive reactance=300 Ohms
V = 600.#    # Applied voltage=600 Volts

Ir = V/R#
Il = V/Xl#
A = Ir*Ir#
B = Il*Il#
It = sqrt(A+B)#

Zeq = V/It#
print 'The Total Impedence = %0.2f Ohms'%Zeq

The Total Impedence = 268.33 Ohms

Example No. 21_3 Page No. 656

In [6]:
# An air-core coil has an Xl of 700 Ohms and an Re of 2 Ohms. Calculate the value of Q for this coil.

# Given data

Xl = 700#   # Inductive reactance=700 Ohms
Re = 2#     # AC effective resistance=2 Ohms

Q = Xl/Re#
print 'The Q of Coil = %0.2f'%Q

The Q of Coil = 350.00

Example No. 21_4 Page No. 658

In [7]:
from math import pi
# A 200 uH coil has a Q of 40 at 0.5 MHz. Find Re.

# Given data

L = 200.*10**-6#  # L of coil=200 uHenry
Q = 400#         # Q=40
f = 0.5*10**6#   # Frequency=0.5 MHz
pi = 3.14#

Xl = 2*pi*L*f#

Re = Xl/Q#
print 'The AC Effective Resistance = %0.2f Ohms'%Re

The AC Effective Resistance = 1.57 Ohms