Chapter 19 : Inductance

Example No. 19_1 Page No. 580

In [1]:
# The current in an inductor changes from 12 to 16 A in 1s. How much is the di/dt rate of current change in amperes per second?

# Given data

di = 4#     # Differential current=16-12=4 Amps
dt = 1#     # Differential time=1 sec

A = di/dt#
print 'The di/dt Rate of Current change = %0.f A/s'%A
The di/dt Rate of Current change = 4 A/s

Example No. 19_2 Page No. 580

In [2]:
# The current in an inductor changes by 50 mA in 2 us. How much is the di/dt rate of current change in amperes per second?

# Given data

di = 50.*10**-3#    # Differential current=50 mAmps
dt = 2.*10**-6#     # Differential time=2 usec

A = di/dt#
print 'The di/dt Rate of Current change = %0.2e A/s'%A
The di/dt Rate of Current change = 2.50e+04 A/s

Example No. 19_3 Page No. 581

In [3]:
# How much is the inductance of a coil that induces 40 V when its current changes at the rate of 4 A/s?

# Given data

Vl = 40#    # Induced voltage=40 Volts
R = 4       # Current changing rate=di/dt=4 A/s

L = Vl/R#
print 'The Value of Inductance = %0.f Henry'%L
The Value of Inductance = 10 Henry

Example No. 19_4 Page No. 582

In [6]:
# How much is the inductance of a coil that induces 1000 V when its current changes at the rate of 50 mA in 2us?

# Given data

Vl = 1000#      # Induced voltage=1000 Volts
di = 50*10**-3#  # differential current=50 mAmps
dt = 2*10**-6#   # differential time=2 usec

A = di/dt#

L = Vl/A#
print 'The Value of Inductance = %0.2e Henry'%L
print 'OR 40 mH'
The Value of Inductance = 4.00e-02 Henry
OR 40 mH

Example No. 19_5 Page No. 582

In [7]:
# How much is the self-induced voltage across a 4-H inductance produced by a current change of 12 A/s?

# Given data

L = 4#    # Inductor=4 H
R = 12#   # current change=di/dt=12 A/s

Vl = L*R#
print 'The Value of Self-Induced Voltage = %0.f Volts'%Vl
The Value of Self-Induced Voltage = 48 Volts

Example No. 19_6 Page No. 583

In [8]:
# The current through a 200-mH L changes from 0 to 100 mA in 2 us. How much is Vl ?

# Given data

L = 200*10**-3#      # Inductor=200 mH
di = 100*10**-3#     # differential current=100 mAmps
dt = 2*10**-6#       # differectial time=2 usec

A = di/dt#

Vl = L*A#
print 'The Value of Self-Induced Voltage = %0.2e Volts'%Vl
print 'OR 10 kVolts'
The Value of Self-Induced Voltage = 1.00e+04 Volts
OR 10 kVolts

Example No. 19_7 Page No. 583

In [10]:
# A coil L1 produces 80 uWb of magnetic flux. Of this total flux, 60 uWb arelinked with L2. How much is k between L1 and L2?

# Given data

lf1 = 80.*10**-6# # Magnetic flux of coil L1=80 uWb
lf2 = 60.*10**-6# # Magnetic flux of coil L2=60 uWb

k = lf2/lf1#
print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.2f'%k
The Coefficient of Coupling k between Coil L1 and Coil L2 = 0.75

Example No. 19_8 Page No. 584

In [11]:
# A 10-H inductance L1 on an iron core produces 4 Wb of magnetic flux. Another coil L2 is on the same core. How much is k between L1 and L2?

# Given data

lf1 = 4# # Magnetic flux of coil L1=4 Wb
lf2 = 4# # Magnetic flux of coil L2=4 Wb

k = lf2/lf1#
print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.f'%k
The Coefficient of Coupling k between Coil L1 and Coil L2 = 1

Example No. 19_9 Page No. 585

In [13]:
from math import sqrt
# Two 400-mH coils L1 and L2 have a coefficient of coupling k equal to 0.2. Calculate Lm.

# Given data

L1 = 400*10**-3# # L1=400 mH
L2 = 400*10**-3# # L2=400 mH
k = 0.2#        # Coupling coefficient=0.2

Lm = k*sqrt(L1*L2)#
print 'The mutual inductance = %0.2f Henry'%Lm
print 'i.e 80*10**-3 H OR 80 mH'
The mutual inductance = 0.08 Henry
i.e 80*10**-3 H OR 80 mH

Example No. 19_10 Page No. 586

In [15]:
from math import sqrt
# If the two coils had a mutual inductance LM of 40 mH, how much would k be?

# Given data

L1 = 400.*10**-3# # Coil Inductance 1=400 mH
L2 = 400.*10**-3# # Coil Inductance 2=400 mH
Lm = 40.*10**-3#  # Mutual inductance=40 mH

lt = sqrt(L1*L2)#

k = Lm/lt#
print'The Coupling Coefficient k = %0.2f'% k
The Coupling Coefficient k = 0.10

Example No. 19_11 Page No. 590

In [21]:
# A power transformer has 100 turns for Np and 600 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs if the primary voltage Vp is 120 V?

# Given data

np = 100.#       # Turns in primary coil=100
ns = 600.#       # Turns in secondary coil=600
vp = 120.#       # Primary voltage=120 Volts

Tr = np/ns#
print 'The Turns Ratio = %0.4f'%Tr
print 'OR 1:6'

vs = vp*(ns/np)#
print 'The Secondary Voltage = %0.2f Volts'%vs
The Turns Ratio = 0.1667
OR 1:6
The Secondary Voltage = 720.00 Volts

Example No. 19_12 Page No. 590

In [23]:
# A power transformer has 100 turns for Np and 5 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs with a primary voltage of 120 V?

# Given data

np = 100.#       # Turns in primary coil=100
ns = 5.#         # Turns in secondary coil=5
vp = 120.#       # Primary voltage=120 Volts

Tr = np/ns#
print 'The Turns Ratio 20:1 or %0.2f'%Tr

vs = vp*(ns/np)#
print 'The Secondary Voltage = %0.2f Volts'%vs
The Turns Ratio 20:1 or 20.00
The Secondary Voltage = 6.00 Volts

Example No. 19_13 Page No. 591

In [24]:
# A transformer with a 1:6 turns ratio has 720 V across 7200 Ohms in the secondary. (a) How much is Is? (b) Calculate the value of Ip.

# Given data

vs = 720.#       # Secondary voltage=720 Volts
Rl = 7200.#      # Secondary load=7200 Ohms
tr = 1./6#       # Turns ratio=1:6

Is = vs/Rl#
print 'The Secondary Current = %0.2f Amps'%Is

Ip = Is/tr#
print 'The Primary Current = %0.2f Amps'%Ip
The Secondary Current = 0.10 Amps
The Primary Current = 0.60 Amps

Example No. 19_14 Page No. 591

In [25]:
# A transformer with a 20:1 voltage step-down ratio has 6 V across 0.6  in the secondary. (a) How much is Is? (b) How much is Ip?

# Given data

vs = 6.#         # Secondary voltage=6 Volts
Rl = 0.6#       # Secondary load=0.6 Ohms
tr = 20./1#      # Turns ratio=20:1

Is = vs/Rl#
print 'The Secondary Current = %0.2f Amps'%Is

Ip = Is/tr#
print 'The Primary Current = %0.2f Amps'%Ip
The Secondary Current = 10.00 Amps
The Primary Current = 0.50 Amps

Example No. 19_15 Page No. 593

In [26]:
# Calculate the primary current I P if the secondary current Is equals its rated value of 2 A.

# Given data

vs = 25.2#      # Secondary voltage=25.2 Volts
vp = 120.#       # Primary voltage=120 Volts
Is = 2.#         # Secondary current=2 Amps

Ip = Is*(vs/vp)#
print 'The Primary current = %0.2f Amps'%Ip
print 'OR 420 mAmps'
The Primary current = 0.42 Amps
OR 420 mAmps

Example No. 19_16 Page No. 593

In [27]:
# Determine the Primary Impedence Zo

# Method 1
# Given data

Vp = 32.#    # Primary Voltage = 32 Volts
Rl = 8.#     # Load Resistance = 8 Ohms
TR = 4.#     # Turns Ratio Np/Ns = 4/1

Vs = Vp/TR#

Is = Vs/Rl#

Ip = ((Vs/Vp)*Is)#

Zp = Vp/Ip#
print 'Primary Impedence = %0.2f Ohms by Method 1'%Zp

# Method 2

Zp = TR*TR*Rl#
print 'Primary Impedence = %0.2f Ohms by Method 2'%Zp
Primary Impedence = 128.00 Ohms by Method 1
Primary Impedence = 128.00 Ohms by Method 2

Example No. 19_17 Page No. 594

In [29]:
from math import sqrt
# Calculate the turns ratio Np/Ns that will produce a reflected primary impedance Zp of (a) 75 Ohms# (b) 600 Ohms.

# Given data

Zs = 300.#   # Secondary impedence=300 Ohms
Zp1 = 75.#   # Primary impedence=75 Ohms
Zp2 = 600.#  # Primary impedence=600 Ohms

tra = sqrt (Zp1/Zs)#
print 'The Turns ratio Np/Ns = %0.2f'%tra
print 'OR 1/2'

trb = sqrt (Zp2/Zs)#
print 'The Turns ratio Np/Ns is 1.414/1 or %0.2f'%trb
The Turns ratio Np/Ns = 0.50
OR 1/2
The Turns ratio Np/Ns is 1.414/1 or 1.41

Example No. 19_18 Page No. 595

In [32]:
# Inductance L1 is 5 mH and L2 is 10 mH. How much is Lt?

# Given data

l1 = 5*10**-3#   # Inductor 1=5 mH
l2 = 10*10**-3#  # Inductor 2=10 mH

Lt = l1+l2#
print 'The Total Inductance = %0.2e Henry'%Lt
print 'i.e 15 mH'
The Total Inductance = 1.50e-02 Henry
i.e 15 mH

Example No. 19_19 Page No. 597

In [33]:
# Inductances L1 and L2 are each 8 mH. How much is Leq?

# Given data

l1 = 8*10**-3#   # Inductor 1=8 mH
l2 = 8*10**-3#   # Inductor 2=8 mH

a = 1./l1#
b = 1./l2#

Leq = 10/(a+b)#
print 'The Equivalent Inductance = %0.2f Henry'%Leq
print 'i.e 4 mH'
The Equivalent Inductance = 0.04 Henry
i.e 4 mH

Example No. 19_20 Page No. 598

In [35]:
from math import sqrt
# Two series coils, each with an L of 250 uH, have a total inductance of 550 uH connected series-aiding and 450 uH series-opposing. (a) How much is the mutual inductance Lm between the two coils? (b) How much is the coupling coefficient k?

# Given data

l1 = 250*10**-6#     # Coil Inductance 1=250 uH
l2 = 250*10**-6#     # Coil Inductance 2=250 uH
Lts = 550*10**-6#    # Inductance series-aiding=550 uH
Lto = 450*10**-6#    # Inductance series-opposing=450 uH

Lm = (Lts-Lto)/4.
print 'The Mutual Inductance = %0.2e Henry'%Lm
print 'i.e 25 uH'

lt = sqrt(l1*l2)#

k = Lm/lt#
print 'The Coupling coefficient k = %0.2f'%k
The Mutual Inductance = 2.50e-05 Henry
i.e 25 uH
The Coupling coefficient k = 0.10

Example No. 19_21 Page No. 608

In [36]:
# A current of 1.2 A flows in a coil with an inductance of 0.4 H. How much energy is stored in the magnetic field?

# Given data

l1 = 0.4#     # Coil Inductance 1=0.4 H
I = 1.2#      # Current=1.2 Amps

E = (l1*I*I)/2#
print 'The Energy Stored in the Magnetic Field = %0.2f Joules'%E
The Energy Stored in the Magnetic Field = 0.29 Joules