Transcript Solution

Chapter 13
Inductance and
Inductors
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13.0
Preview
Inductor [Fig. 13-1, page 493]
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coil of wire [Fig. 13-1a]
effect on circuit operation is to oppose changes in current
inductance: property of inductor to oppose current change
may referred to as chokes or reactors.
Sysmbol [Fig. 13-1b]
unit: Henry
non-ideal inductor [Fig. 13-1c]
applications :tuning circuit in radio receiver
ballast circuit that limit current in fluorescent lamp
protection circuit used to control short-circuit current
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13.1
Electromagnetic Induction
[page 494]
Motional emf [Fig. 13-2a]
• the magnet is thrust into the coil
• the meter deflects upscale
• the magnet is withdrawn
• the meter deflects downscale
• voltage magnitude is proportional to how fast the magnet is
moved
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Motional emf [Fig 13-2b, page 495]
• voltage is induced if a conductor is moved through the magnetic field
• if the conductor is moved to the right, its far end is positive
• if it is mvoed to the left, the polarity reverses and its far end becomes
-ve
• voltage magnitude is proportional to how fast the wire is moved
Mutually induced voltage [Fig. 13-2c, page 495]
• voltage is induced in coil 2 due to the magnetic field created by the
current in coil 1
• meter kick upscale at the instance the switch is closed
• meter kick downscale at the instance the switch is opened
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Self-inducted voltage [Fig. 13-2d, page 495]
• voltage is induced in a coil by its own current
• the top end of the coil becomes +ve at the isntance the switch is
closed
• the top end of the coil becomes -ve at the instance the switch is
opened
Faraday’s Law [page 495]
• voltage is induced in a circuit whenever the flux linking(i.e. passing
through) the circuit is changing.
• the magnitude of the induced voltage is proportional to the rate of
change of the flux linkages
• also stated in terms of the rate of cutting flux lines
Lenz’s Law
[page 496]
• the polarity of the induced voltage is such as to oppose the cause
producing it
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13.2
Induced Voltage and Induction
[page 496]
Steady DC current [Fig. 13-3a]
• current is constant, magnetic field is constant
• induced voltage is zero
• inductance has no effect on steady, unchanging dc current
Increasing current [Fig. 13-3b]
• increasing current causes expanding field
• the expanding field induces a voltage that opposes the current buildup
• the faster the current increases, the larger the opposing voltage
Decreasing current [Fig. 13-3c]
• decreasing current causes collapsing field
• the collapsing field induces a reversed voltage that opposes the curret
decrease
• the faster the current decreases, the larger the induced voltage
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Counter emf [Fig. 13-4, page 497]
• the induced voltage tries to counter changes in current, it is called a
counter emf or back voltage
Note:
• this voltage does not oppose current, it oppose only changes of current
• it does not prevent the current from changing
• it only prevents it from changing abruptly
Result of back voltage:
• current in an inductor changes gradually and smoothly from one value
to another
Refer to Figure 13-4 [page 497] - Current in an inductance
(a) Currennt cannot jump from one value to another suddenly.
(b) Current must change smoothly with no abrupt jumps.
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Iron-core and air-core inductors
• voltage induced in a coil depends on flux linkages
• flux linkages depend on core materials
Iron-core coils
• coil with ferromagnetic core
• flux almost entirely confined to their cores pass through all windings
flux linkage = N 
[Fig. 13-5]
induced voltage
e = N x rate of change of 
= N (d/dt)
Refer to Figure 13-5 [page 497] - When flux  passes through all N
turns, the flux linking the coil is N .
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Example 13-1 [page 497]
If the flux through a 200-turn coil changes steadily from 1 Wb to 4 Wb
in one second, what is the voltage induced?
Solution
The flux changes by 3 Wb in one second. Thus, its rate of change is 3
Wb/s.
e = N x rate of change of flux
= (300 turns) (3 Wb/s) = 600 volts
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Air-core coils
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coil with air or nonferromagnetic core
flux not confined to their cores [Fig. 13-6]
the flux linking the coil is proportional to current
Self-inductance, L (unit: Henry)
e = L x rate of change of current
= L (di/dt)
Refer to Figure 13-6 [page 498], it should be noted that not all flux
links all turns. Nonetheless, the flux linking the coil is
proportional to current.
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13.3
Self-inductance [page 499]
• the inductance of a coil is one henry if the voltage created by its
changing current is one volt when its current changes at the rate of
one ampere per second [Fig. 13-8, page 499]
• voltage across an inductor is denoted by VL rather than by e
EXAMPLE 13-2 [page 499]
If the current through a 5-mH inductance changes at the rate of 1000 A/s,
what is the voltage induced?
Solution
VL = L x rate of change of current
= (5 x 10-3 H) (1000 A/s) = 5 volts
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Inductance formula
approximate inductance fo the coil [Fig. 13-9, page 499]
L = μN2 A / l
(H)
where l is in meters, A is in square meters, N is the number of turns, and
 is the permeability of the core.
EXAMPLE 13-3 [page 500]
A 0.15 m long air-core coil has a radius of 0.006 m and 120 turns.
Compute its inductance.
Solution
A = r2 = 1.131 x 10-4m2
 = O = 4 x 10-7
Thus,
L = 4 x 10-7 (120)2 (1.131 x 10-4)/0.15 = 13.6H
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• To provide geater inductance in smaller spaces, iron cores are
used
(before saturation)
• to solve variable permeability problem after iron core saturated,
air gap may be used to control saturation
• if the gap is wide enough, coil inductance is approximately
(Note: equation break down for small 1)
L = oN2 Ag / 1g
Refer to Fig. 13-10 [page 500] for the iron-core coil with an air gap
to control saturation
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Example 13-4 [page 500]
The inductor of Figure 13-10 has 1000 turns, a 5-mm gap, and a
cross-sectional area at the gap of 5 X 10-4m2. What is its
inductance?
Solution
L = (4 X 10-7)(1000)2(5 X 10-4)/(5 X 10-3) = 0.126 H
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Practical notes
1. Since inductance is due to a conductor‘s magnetic field, it
depends on the same factors that the magnetic field depends on.
The stronger the field for a given current, the greater the
inductance. Thus, a coil of many turns will have more inductance
than a coil of a few turns (L is proportional to N2) and a coil
wound on a magnetic core will have greater inductance than a coil
wound on a nonmagnetic form.
2. However, if a coil is wound on a magnetic core, the core‘s
permeability  may change with flux density. Since flux density
depends on current and inductance depends on permeability,
inductance in this case depends on current, I.e., it is not constant,
but is a non-linear function of current. For example, the inductor
of Figure 13-11 has a nonlinear inductance due to core saturation.
All inductors encountered in this book are assumed to be linear,
I.e., of constant value.
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13.4
Computing Induced Voltage [page 501]
• polarity of induced voltage depends on current increasing or
decreasing
• current increases, induced voltage is +ve
• current decreases, induced voltage is -ve
Example 13-5 [page 501]
Figure 13-12 is the current through a 10-mH inductance.
Determine voltage L and sketch it.
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Example 13-5 (Solution)
[page 502]
Solution
Break the problem into intervals over which the
slope is constant, determine the slope for each segment, then
compute voltage using L = L x slope for that interval:
0 to 1 ms:
Slope = 0. Thus, L = 0V.
1 ms to 2 ms:
Thus,
Slope = i / t = 4 A/(1 x 10-3 s) = 4 x 103 A/s.
L = L i / t = (0.010H)(4 x 103A/s) = 40V.
2 ms to 2 ms:
Thus,
Slope = i / t =  8 A/(2 x 10-3 s) =  4 x 103 A/s.
L = L i / t = (0.010H)( 4 x 103A/s) =  40V.
4 ms to 5 ms:
Slope = 0. Thus, L = 0V.
5 ms to 6 ms:
Thus,
Same slope as from 1 ms to 2 ms.
L = 40V.
The voltage waveform is shown in Figure 13-13 [page 502].
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13.5
Inductances in series and parallel [page 503]
• same rules as resistance
• for series case (Fig. 13-15, page 503)
LT = L1 + L2 + L3 + …. + LN
• for parallel case (Fig. 13-16, page 504)
1/LT = 1/L1 + 1/L2 + 1/L3 + …….+ 1/LN
For two inductances,
LT = L1．L2 / (L1 + L2)
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Example 13-6 [page 502]
Find LT for the circuit of Figure 13-17 [page 504].
Solution
The parallel combination of L2 an L3 is
L2 L3
6 2
Leq 
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 1.5
L2  L3 6  2
This is in series with L1 and L4.
Thus, LT = 2.5 + 1.5 + 11 = 15H.
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13.6
Practical considerations [page 504]
Core types
• depends on its intended use and frequency range
• iron core for audio or power supply applications as they need large L
• air or ferrite cores RF and TV circuits as iron has large power loss ar
high frequency
Variable inductors
• changing coil spacing with a screwdriver adjustment
• threaded ferrite slug is moved in or out of coil’s winding
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Refer to Fig. 13-20 [page 505] for Circuit symbols:
(a) Iron-core
(b) Ferrite-core
(c) Variable
(d) Air-core
Stray capacitance [Fig. 13-21(a), page 506]
• stray capacitance exists between inductors’s windings
• effect depends on frequency - neglected on low frequency while take
into account on high frequency.
In Figure 13-21(a), Cw is stray capacitance.
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Stray inductance [page 506]
• all current-carrying conductors have inductancne
• leads on circuit components and traces on PCB all have inductance
• stray inductance is so small that it can be neglected
13.7
Inductance and steady state DC [page 507]
• ideal iductor looks like short circuit
• dc equivalent of non-ideal inductor is coil resistance
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Example 13.8
[page 507]
In Figure 13-24(a), the resistance of the coil is 14.4. What is I?
Solution
Reduce the circuit as in (b),
ETh = (9/15)(120) = 72V
RTh = 6 9 = 3.6
Now replace the coil by its dc equivalent circuit as in (b). Thus,
I = ETh/RT = 72/(3.6 + 14.4) = 4A
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Example 13-9 [page 508]
The resistance of coil 1 in Figure 13-25(a) is 30  and that of coil 2 is
15  . Find the voltage across the capacitor assuming steady state dc.
Solution
Replace each coil inductance with a short circuit and the capacitor with
an open circuit. As you can see from (b), the voltage across C is the
same as the voltage across RL2. Thus,
VC = [Rl2/(Rl1 + Rl2)]E = (15/45)(60) = 20 V
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13.8
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Eergy stored by an inductance [page 509]
energy is stored in its magnetic field
when the field collapses, this energy will return to the circuit
for ideal inductor with no internal resistance, no power is dissipated
power to inductor
P = VL i
, where VL = L di/dt
W = (1/2) L i2 , where i = instantanneous value of current
W = (1/2) L I2 , where I = steady state current
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Example 13-10
[page 509]
The coil of Figure 13-28(a) has a resistance of 15 . When the current
reaches its steady state value, the energy stored is 12J. What is the
inductance of the coil?
Solution [page 510]
I = 100 V/25 = 4A
W = 1/2LI2 J
12J = 1/2L(4A)2
Thus,
L = 2(12)/42 = 1.5 H
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13.9
Inductor troubleshooting hints
[page 510]
Open coil
• causes: result of poor solder joints or broken connections
• check: visual inspection
disconnect the inductor and check it with ohmmeter
(open-circuited coil has infinite resistance)
Shorts
• causes: Shorts occur betweenn windings
or between coil and its core
• check: Result in excessive currennt and overheating
Look for burned insulation, discolored component, etc.
Use ohmmeter to check for shorts between windings and core
Subsitute a known good inductor for the suspected one
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