# 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 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
# 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
# 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'
# 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 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'
# 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
# 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
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'
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
# 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
# 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
# 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
# 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
# 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'
# 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
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
# 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'
# 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'
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
# 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