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EE 201

SUPPLEMENTARY COURSE NOTES

Denard Lynch, P.Eng. Electrical and Computer Engineering University of Saskatchewan © 2000 – 2010 Issue Date: Sep, 2011

Supplementary Course Notes EE201

©Denard Lynch Sep , 2011 of 126

Magnetic Circuits and Examples

Review of Magnetism

Originates at the molecular level: electrons orbiting a nucleus induce a magnetic “moment” each “orbit induces its own separate moment depending on particular molecule, can have a net magnetic moment (i.e. ferromagnetic) several molecules group together in “domains” that can also have a net magnetic moment (i.e. N-S), but these domains are typically randomly oriented, even in magnetic materials unless influenced by some magnetizing force.

This implies and illustrates a fundamental relationship between electricity and magnetism Two important and related facts:

a moving charge generates a magnetic field, and although not obvious from this fact… relative motion between a magnetic field and a charge exerts a force on the charge(s) (and on that which contains them; i.e. a conductor)

So what is this “magnetic field”?

a force field visualized as “lines of magnetic flux” that appear to follow rules: form closed loops from one “pole” to another directional (by convention, from north to south pole of magnet) repel each other * follow the path of least resistance* *these last two rules conflict with one another! Remember… the field consists of continuous lines, but “lines” are really a visual aid Magnetic flux can also be visualized as flowing, similar to water or electric current (symbol: Φ) We are often interested in its density (symbol B)

Depending on the material, the magnetic moments from each set of orbiting electrons in a molecule can reinforce or cancel each other to give the molecule a net magnetic moment. Various molecular arrangements lead to three types of materials:

Paramagnetic – only very slightly receptive to a magnetic field (e.g. air, glass, wood, paper, plastics etc.) Diamagnetic (anti-magnetic) – form opposite dipoles in response to an imposed magnetic field (bismuth, pyrolytic graphite), but only a weak response Ferromagnetic – net magnetism at the molecular level; get together into polarized “domains”. They are normally oriented randomly, but may be aligned temporarily or permanently. (E.g. iron, cobalt etc.)

We are interested in this last category because it is relatively easy* to establish magnetic flux, Φ, in these materials in response to a magnetic field.

Figure 1



this “relative ease” is somewhat temperature dependent. Each material has a “Currie temperature”, above which they behave like paramagnetic materials.

Why is this (ferromagnetic) ability of interest to us? So we can design “magnetic circuits” to route flux in certain ways and to certain places; the same way we route electric current or hydraulic fluid. We route this to certain places for one of two reasons:

The force it can provide Speakers, sound reproduction Motors, generators Door bells, electromagnets its magnetizing effect (i.e. channelizing/routing flux) HDDs, magnetic tapes Transformers, inductors

How can we create a magnetic field, or a magnet? Back to moving charges… Moving charges, for example a current flowing in a conductor, create a magnetic field. The induced “lines of flux” are formed in concentric circles in a plane perpendicular to the direction of current flow.

The direction of the flux, Φ, can be determined using a “right-hand-rule” (RHR), and Figure 2 illustrates the diagrammatic convention for the current: into page/board = ⊗; out of page/board = ¤. Looking at a current carrying conductor from one end (Figure 3), the density of the magnetic field, B, is given by:

B =

µ0 I2!r

, outside the

B =

µ0 Ir2!R2

conductor, and

inside the conductor.

B

r R

R

Charge-carrying conductor

⊗ ¤

Figure 2



If we route current through a cylindrical coil of wire, we will end up with a fairly uniform magnetic field inside the coil. The coil, in effect, becomes a magnet (electromagnet), which is a source of magnetomotive force (mmf), symbol: ℑ. A source of magnetomotive force (mmf), which could be an electromagnet or a permanent magnet, will cause an amount of magnetic flux, Φ, to ‘flow’. The amount of flux that floes depends on the magnetic ‘resistance’ of the circuit or path. Again, we can use a right-hand-rule (RHR) to determine the north pole of the source of mmf. (For an electromagnet consisting of a coil of wire, if you wrap the fingers of your right hand around the coil in the direction of current flow, your thumb points in the direction of the flux flow, and the north pole.) The ‘resistance’ of the path is proportional to the length, l, inversely proportional to the area, A, and is know as the reluctance, ℜ. Similar to electrical resistance where, the magnetic reluctance,

! =

lµA

, where µ is the permeability of the material and is the product of the relative permability,

µr, and the permability of free space, µ0 (µ=µrµ0). Also similar to the electric circuit case, there is a fundamental relationship between the driving force, ℑ, the reluctance of the path, ℜ, and the amount of flux, Φ, that flows in the circuit. This relationship is know as Ohm’s Law for Magnetic Circuits: ℑ=Φℜ, which will be a fundamental tool for analysis of magnetic circuits. Ω’s Law for Magnetic Circuits: ℑ = Φℜ, where: ℑ = magnetomotive force (mmf) = (number of turns)(current in Amperes) = NI Φ = total magnetic flux in Webers, Wb

! =

lµA

= reluctance in Amp-turns per Weber (A-t/Wb) or rels, and

l = average path length in meters A = cross-sectional area in m2 µ = µrµ0, and µr is unit-less and specific to the material,

and µ0 = 4π X 10-7 Wb/At-m. µr ~ 1 for paramagnetic materials,

and >>1 for ferromagnetic materials

Figure 3: Field Strength versus Distance from a Conductor



Substituting ℑ=NI and ! =

lµA

into Ω’s Law for Magnetic Circuits gives us:

! =

µAl

NI , with a slight rearrangement…

!A= µ NI

l, now defining two new quantities:

B =

!A

and H =

NIl

, where:

B is the flux density in Webers per square meter (Wb/m2) or Teslas, (T) (1T=10,000Gauss) H is the magnetizing force or magnetic field intensity in Amp-turns/m (A-t/m), and substituting these expressions for B and H in the last expression, we have:

B = µH

Take the expression for flux density: B =

!A

, or Φ = BA. Recalling Ω’s Law and that the mmf, ℑ

= NI = Φℜ, we can do another manipulation:

NI = !( ) "( ) = BA( ) l

µA#$%

&'(=

Bµ

l = Hl

or NI = Hl, which is a variation of Kirchhoff’s Voltage Law for electric circuits, which can also be stated as: NI = Hl!! , or the sum of the mmf rises around any loop in a magnetic circuit must equal the sum of the magnetizing force drops. This relationship will prove very useful in analyzing magnetic circuits! Let us re-examine the relationship between the flux density, B (which is proportional to Φ) and the magnetizing force, H (proportional to NI) given as B=µH. µ is the slope of the relationship between B and H and is typically not linear for magnetic materials. It is usually given in graphical form in a “B-H curve”. If the flux density is static, we can use the “normal” permeability. If the flux varies, we can use “incremental”, “average”, or “differential” permeability (Jackson). Unless otherwise stated, in this course we will assume a static flux and use the “normal” permeability. Hysteresis Hysteresis is a property of magnetic material that causes some residual magnetism to remain in a material after it has been exposed to an external source of mmf. Once a ferromagnetic material is subjected to a magnetic field, the domains align. If the magnetic field is removed, most of the domains will return to (approximately) their original orientation. Dependent on the type of material, some domains may stay oriented in the induced direction, giving the sample some residual magnetism of its own. The amount of (reverse) field strength needed to return the magnetic moment of the sample back to zero is called the coercive force.



Magnetic Circuit Problems

Magnetic circuit problems faced by a designer are generally one of two types: the NI (mmf) is known, and we need to find the flux, Φ or B, or the desired flux, Φ, or flux density, B, is known and we want to find the required NI.

Useful Circuit Comparisons: Fluids Electricity Magnetics

The parallel between magnetic circuits and (particularly) electric circuits is very useful for analyzing and designing magnetic circuits!! Channeling Flux: Gaps and Fringing; Laminations and Magnetic Force of Attraction Fringing: Recall that two opposing properties of magnetic ‘lines’ of flux is that they will “avoid” each other by spreading out as much as possible, but that they also take the path of least resistance, so they will “get together” to follow a lower reluctance path. This means that when they have been channeled through a ferromagnetic material (easy path) and then come to a paramagnetic (e.g. air) gap, they will start to “spread out” to avoid each other again over the length of the gap, which will affect the density, B. To estimate the effective area of the path in an air gap:

add the length of the gap, g, to the length and width of a rectangular or square cross-section:

AEff !gap = l + g( ) w + g( ) add the length of the gap, g, to the radius of a circular (or elliptical) cross-

section: AEff !gap = " r + g( )2

Laminations: Magnetic cores are often laminated to reduce heating and losses from eddy currents (those currents induced in a conductor by a varying magnetic field). For laminated cores, the effective area is the nominal area multiplied by the Stacking Factor (S.F.).

StackingFactor =

AEff

AOverall

For a rectangular cross-section, for example, AEff. = (S.F.)(length X width)

Pressure (pump)

Restriction (valve)

Flow (fluid) Flow (current)

Restriction (resistance)

Pressure (voltage) +

Pressure (mmf)

Flow (flux)

Restriction (reluctance)

S

N

l w

effective area

physical area

l+g

w+g

r+g

r

Figure 4: Ohm's Law Analogies

Figure 5: Fringing Effective Area



Note: the Stacking Factor is sometimes given as a percentage: e.g. S.F. = .95 or 95%. Overlapping Areas: Where two legs of a magnetic circuit come together, the areas may not be the same or oriented to exactly match. In these cases, the flux will almost entirely pass through the overlapping area. Although the lines of flux will have to make some path adjustment when transitioning from one piece to the other, this happens quickly enough so an assumption that it happens immediately at the transition is a reasonable one. Magnetic Force of Attraction: In any magnetic circuit, there is a ‘magnetic force of attraction’. This attractive force exists anywhere in the circuit, but is most useful where there is a physical break or separation between one piece and another. This makes it possible to lift or hold ferromagnetic loads with a magnet (or electromagnet). The magnetic force of attraction is given by:

FMag! Attr =

Bgap2 Agap

2µ0.

The designation ‘gap’ in this formula should really be interpreted as ‘interface’, as this force exists at any interface (or other point in the circuit), regardless of whether an actual gap exists or not. Also note, it depends only on the flux density, B, and the area of the contact, not on the length of any ‘gap’. The gap length, however, may indirectly influence the force by increasing the reluctance of the magnetic path and thus reducing the total flux. Remember, when dealing with overlap situations, use the smallest common area as the ‘Agap_’.

2cmX4cm

Effective overlap: 2cmX2cmX0.95

S.F.=0.95

Figure 6: Overlap Effective Area



Table 1: Summary of Electric and Magnetic Similarities Electric Magnetic

E = IR ℑ = Φℜ

R =

!lA

! =

lµA

Φ = BA KVL

ΣVloop = 0 Σℑloop = 0 ΣVloop = ΣIR ΣNIloop = ΣHl

KCL ΣInode = 0 ΣΦnode = 0 ΣIin = ΣIout ΣΦin = ΣΦout

Where R typically stays reasonably constant with varying current, ℜ (reluctance) does not usually

stay constant with varying levels of flux!B – H Curves are used to determine the relationship between the “force” and the “flow”

B = µH

µ =

BH

Pressure (mmf)

Flow (flux)

Restriction (reluctance)

S

N

Flow (current)

Restriction (resistance)

Pressure (voltage) +



Some Quick Examples: Fringing: Overlap/Laminations: Hysteresis:

Φ

Φ

gap

gap

gap

Φ

Φ

B

H



Ex. 1

a) What magnetomotive force, mmf in A-t, is needed to provide a flux of 0.35 mWb in this cast steel core? b) What is the direction of the flux in the core if the current flows into the top terminal as shown?

a)

leg A(m2) l(m) Φ(Wb) B(T) H(A-t/m) Hl(A-t) Cast S. .0005 .15π .35mWb Φ/A=.7 4501 212 Sheet S. “ “ “ “ 150 71

Note 1: find H from the B-H curves: ~450 b) using the RHR, flux is generated going clockwise

I

N 15 cm

Crossection = 2 cm X 2.5 cm



Ex. 2 N = 1000t; I = .25A Find the total flux, Φ, produced by the coil if the core is made of:

sheet steel cast steel cast iron

Leg A(m2) l(m) Φ(Wb) B(T) H(A-t/m) Hl(A-t) a) Sheet S. .0004 .4 53mWb 1.32 625 250 b) Cast S. .0004 .4 .38mWb .95 625 250 c) Cast Iron .0004 .4 .10mWb .25 625 250

10cm

10cm

2cm 2cm

I

N



Ex: 3 a) The core in the figure is made of Cast Steel. If the coil is 1000 turns and the current is 1.35A, determine the flux in the core. b) Given the same conditions as in a), determine the flux if a 2mm air gap is cut in the core as shown. Consider fringing at the gap. **Considering Fringing**

leg A(m2) l(m) Φ(Wb) B(T) H(A-t/m) Hl(A-t) a) .0004 .5 .000592 1.48 2700 1350 b) cast steel .0004 .498 .000308 .77 500 249 gap .000484 .002 .000308 .636 506,000 Guess 75%

1012 1261/1350=93.4%, too low =1261 cast steel .0004 .498 .000333 .83 540 269 gap .000484 .002 .000333 .687 547,000 Guess 81%

1094 1363/1350=101%, close enough! =1363 So the flux in the gap is .333mWb.

2mm gap – part b) L= .5m

2cmX2cm

I

N



Ex: Mag 4 N = 500t; I = 0.6A *Neglecting Fringing* a)Find the flux in the air gap, and b) the total flux produced by the oil.

leg A(m2) l(m) Φ(Wb) B(T) H(A-t/m) Hl(A-t) L-2 .0003 .1 .000465 1.55(sat!) 2440 244 L-3 .0003 .099 .000085 .283 195 19 L-gap .0003 .001 .000085 .283 225,000 Guess 75%

225 L-1(SS) .0009 .4 .000550 .611 75 30 =274 274/300=91%, too low! L-2 .0003 .1 .000459 1.53(sat!) 2750 275 L-3 .0003 .099 .000096 .320 205 20 L-gap .0003 .001 .000096 .320 255,000 Guess 85%

255 L-1(SS) .0009 .4 .000544 .604 75 30 =305 305/300=102%, close enough! So… a) the flux in the gap is 96µWb, and b) the total flux produced by the coil is .544mWb

I

N

L1 = 0.4m

L2 = 0.1m

1cmX3cm 3cmX3cm

Sheet steel

Cast steel

1mm gap

L3



Eg. 5: Consider the magnetic circuit shown in the figure. N1 = N2 = 200t I1 = 8A Stacking Factor for the [laminated] cast steel section is 0.93. The sheet steel sections (with the gap) are solid. Consider fringing effects at the air gap. Find the current, I2, required to establish a flux density, B, of 1.1T in the air gap.

First draw an Electric Equivalent… Note: You can combine the cast segments into one reluctance and also the 2 sheet steel segments into 1 ℜ because:

i) same material ii) same physical characteristics (dimensions) iii) same flux

Second, make a table, then put in what you know (bold), and finally calculate the rest.

leg A(m2) Φ(Wb) B(T) H(A-t/m) l(m) Hl(A-t) cast steel .0008371 .0011264 1.346 1700 .42 714 gap .0010242 .0011264 1.1 875352 .002 1750 sheet steel .0009 .0011264 1.25 460 .178 82 2546 Hl = 2546 = N1I1 +3 N2I2 = (200)(8) + (200)I2; I2 = 4.73A Notes: 1) cast steel area: .03 X .03 X .93 = .000837m2 to consider the stacking factor in the laminated section 2) air gap area: (.03 + .002) X (.03 + .002) = .001024m2 to consider fringing at the gap 3) the coils are additive, both producing a cw flux with the currents as shown.

S.S.NI 1

NI 2

C.S.

N1

N2

laminatedCastSteel

SheetSteel

I1

I2

2mm gap

3cm

3cm

all x-sections

15cm

15cm



Eg. 6: The electromagnet (EM) shown at right is made of solid sheet steel. The coil has 200 turns of 12ga insulated copper wire. If the EM is brought to within 1mm of a 20kg cast steel bar of the same cross-section, find the current required to just start lifting the bar. Consider the effects of fringing at the air gaps.

Draw Elec. Equiv. The total force required from both gaps together to lift the 20kg bar:

F = (20kg) X (9.81m/s2) = 196.2N Note we can use either the total area of both gaps or half the required force to find the required B in the air gap.

F = B2A2µo

; B = 2FµoAgaps

= 0.75

Now use this to do the rest of the mag ckt calculations:

leg A(m2) Φ(Wb) B(T) H(A-t/m) l(m) Hl(A-t) sheet steel

EM .0004 .000331 .827 110 .24 26

gap .000441 .000331 .75 596831 .002 1193 cast steel

bar .0004 .000331 .827 550 .10 55

1274 NI = Hl = 1274 to create enough magnetic force to just start lifting the bar. I = Hl/N = 1274/200 = 6.37A Notes: the length of the air gap is twice one gap since they are in series B = µoH in air, for metal we use the graphs (B-H curves) Agap = .021 X .021 = .000441m2 because of fringing

+ E.M.Gaps(2mm total)

NI

Bar

2cm

2cm

all x-sections

20 kg

1mm gap

N

I

l=24cm

8cm



Force on a moving charge in a magnetic field

(Note: Consult a refresher on cross-products if required.) Recall the two sides of the relationship between electricity and magnetism:

i) a moving charge creates a magnetic field, and ii) a magnetic field exerts a force on a a moving charge

We are now going to explore part ii) of this relationship. First, highlight the differences between electric and magnetic fields w.r.t. their effect on a charged particle:

Electric Fields Magnetic Fields Fe is in the direction of the field Fm is perpendicular to the field Fe will act on a particle regardless of velocity, and will accelerate a particle

Fm only affects a charge in motion, but will not change its velocity (i.e. no acceleration)

Fe does work to displace/move a charge, and the energy of the charge is affected

Fm does no work on the charge, and the particles energy is not affected

Recall: Work = Force acting through a distance. F•dS = F•vdt = 0, i.e. no work is done since there is no force in the direction of the motion. To explore Fm a little further, from physics we know:

Fm ∝ q, vel, B, where: B = strength of the magnetic field, vel is the velocity of the particle(s), and q is the magnitude of the charge.

Also, Fm is perpendicular to both vel and B.

Fm = q ! vel " B (1) Now consider a collection of charged particles in a medium in the presence of a magnetic field of concentration B. If the charges and the magnetic field are static, there is no force or effect on the charged particles. However, the charged particles are all ‘drifting’ with a velocity of vdrift, then the force on each one is given by (1) above, where q is the magnitude of the charge on one particle, vel is its velocity and B is the density of the field. The total number of charge carrying particles in the medium (e.g., a wire/conductor) is equal to nAl, where n = number of particles/m3, A = cross-sectional area in m2, and l is the length of the material in m.

FmB

vel



Then the total force (due to B) on the medium (wire) is:

Fm! total = q " vel # B( ) nAl( ) (2)

where q is the magnitude of charge per carrier in coulombs/carrier n is the density of the charge carriers per m3

A is the cross-sectional area in m2, and vdrift is the velocity of the charges in m/s.

and:

I = qnAvdrift (3) (Note: that all units in the expression cancel to leave coulombs/sec, and 1 c/s = 1 ampere.) Substituting (3) into (2) results in the expression for the force on a wire in a field, B:

Fmag = Il ! B (4) where I is the current in A, l is the length of the wire in the filed in meters, m, and B is the magnetic field density in Teslas, T.

Let’s look at the implication of the cross-product or directionality. The force on a small segment of wire, ds, carrying a current, I, in a field of strength, B is:

dFmag = Ids ! B Integrating both sides:

dFmag! = Fmag = I ds! " B

where ds! is simply the vector sum of all the “ds”s along the current’s path. Let’s look at 2 special cases: Case 1) a wire passing through a field:

Fmag = I dsa

b

! " B

where the result of dsa

b

! is simply the straight line a-b = l’, and:

Fmag = Il' ! B

as indicated by (4) above.

ds BI

ds B

Ia

bl’



Case 2) a wire loop in a field Similar to above, the force is:

Fmag = I ds!! " B = 0!

because the vector sum around the loop, ds!! ,

is 0! But although the net force on the whole loop is 0, the force on any part of the loop is not. This has some significant implications for rotating electric machines! Let’s examine the top and edge views of a rectangular loop of wire in a uniform magnetic field of strength B.

First, calculate the force on the side legs of length b: F = Il ! B = i b( )Bsin900 (since the current is flowing 900 to the filed, B). Looking at the edge view, we can see that the resulting force is acting downwards on the left leg, and upwards on the right leg. The force on the end legs (length a): F = Il ! B = i a( )Bsin00 = 0, and F = Il ! B = i a( )Bsin1800 which also = 0. Although the net force on the loop is 0, we can see there is a torque tending to rotate the loop about the “axle” placed through the centre. The total torque on the loop is:

! = IbB( ) a2+ IbB( ) a

2= IbBa , or IB(area of the loop). (Note: the torque is ccw in this case.)

Top View Edge View

B B

a

bI

I

I I

Fm

Fm

a/2 a/2

dsB

I



If the loop happened to have N turns of wire, the total torque would be:

! = NIAloopB In this particular configuration, the magnetic force will always be pointing straight up or down, and varies sinusoidally (in magnitude) as the loop rotates. An equal, sinusoidal force would have to be applied to cause the loop to rotate at a constant angular velocity. In motors and generators, the windings and magnetic (flux) path are designed to provide a more uniform torque (either in or out).



A quick review of cross products:

Take an example of a cross-product expression: F = Il X B. It is also correct, and more convenient, to think of the expression as: Il X B = F. In this form, we can think of the cross-product operation as:

(first vector) X (second vector) = (resultant vector) e.g. Il X B = F

RHR 1 (thumb) X 2 (index finger) = 3 (second finger) The Right Hand Rule (RHR) gives the general direction of the resultant given the direction of the input vectors. Mathematically, the cross product expression may be expressed as:

F = |Il| |B| sinθ , where θ is the angle formed by rotating the first vector clock-wise into the second vector. If you do this on a desktop, for example, and both the input vectors are in the plane of the desktop, then a positive result indicates the resultant is perpendicular to, and into the desktop, and a negative result indicates the result is out of the desktop. A couple of simple examples will illustrate this process: 1) If the vectors both fall in quadrant 1 or 2, then θ will be between 0 and 1800, and sinθ will always be positive, meaning the resultant, F will be into the page.

2) If the vectors both fall in quadrant 3 or 4, then θ will be between 180 and 3600, and sinθ will always be negative, meaning the resultant, F will be out of the page.

In most cases, we can use the simple RHR using the component of I (i.e. Isinθ) that is perpendicular to B. (Satisfy yourself, in the above examples, that at 900 and 2700 the RHR gives exactly the same result regarding the direction of the resultant.)

B

I

I

Fm

!

!

II

BFm!

!



Faraday’s/Lenz’s Laws of Electromagnetic Induction and Motional emf

Faraday’s Law:

! = " d#mdt

, where Φm = total “flux linkages” =

BdA! .

For a coil or winding with N turns,

! = "N d#mdt

Note: the ‘-‘ is there mainly to indicate that the induced voltage opposes the change in flux. Consider a loop of wire moving through a magnetic field of density, B, at a constant velocity, vel:

a) dΦ/dt = 0, so E = 0

b) dΦ/dt = +, so E = –ve value

c) dΦ/dt = 0, so E = 0

d) dΦ/dt = -, so E = +ve value

e) dΦ/dt = 0, so E = 0

E.g.: If the loop was 1m2, and was moving through a field B = 1T at a velocity, vel = 1m/s, then there would be a change in flux of 1Weber per second as the loop entered and exited the field (not while it was totally outside or inside the field). During these transition periods, there would be an emf of ±1V induced in the loop. Lenz’s Law tells us that the direction (polarity) of the induced voltage, Em, will be such that it will create a current that produces a flux that opposes the change in flux which induced the original Em. Two interpretations:

1) Considering b) above, the flux inside the loop is increasing, so the induced voltage will create a ccw current that will create opposing flux. In d), where it is decreasing, it will cause a cw current which will produce a flux to reinforce the field, thus opposing its reduction.



2) In b) a upward current in the right leg (ccw) will cause a force on the wire that points to the left, opposing the force that is pushing the loop into the field. In d), the current must flow upward in the left leg (cw) to create a force to the left to oppose that force which is trying to pull it to the right.

Of course the idea of a potential (E) in a closed loop of, theoretically, zero resistance is problematic. Even with an infinite current flowing there will never be any voltage. Of course an infinite current means an infinite magnetic force opposing the movement in the first place, so nothing would move(!). If nothing moves, there are no changes in flux and thus no induced voltage(?). This apparent paradox can be partially explained by considering what happens as things just start to move. An incremental movement would cause an incremental change in flux which would cause an incremental induced voltage which would cause some minute current to flow that would produce some minute flux that would oppose the change caused by the motion. In a practical sense however, loops are not resistance-less and loops are not usually closed (short circuited). In these practical situations, the tendencies predicted by Faraday’s and Lenz’s Laws create voltages and currents that also must satisfy other laws and ultimately force an appropriate balance in the system. It is important to remember that Faraday’s Law will hold, and a motional emf will be induced whenever a conductor full of charges moves in a magnetic field, regardless of whether any current flows or not. We will soon see that the amount of current can be determined by forcing a balance of other circuit and physical laws. Consider yet another explanation of Faraday’s law with respect to motional emf; moving a conducting bar (full of charges) through a magnetic field, B, with a velocity, vel. Recall the force on a moving charge in a magnetic field is:

F = q ! vel " B . Also recall, that the force on a charge in an electric field is:

F = qE For a natural balance to occur without a “current” flow, the force due to an electric field must balance the force due to the the magnetic field:

qE = q ! vel " B , and since the voltage, V, observed over a length, l, is El:

qV = q El( ) = ql ! vel " B , and removing the common “q” from each side we are left with the general expression for motional emf:

V = !motional = l " vel # B

As in the case of the force on a moving charge, this is a cross product expression, indicating that the vectors involved must be orthogonal to each other. If we apply the Right Hand Rule (RHR) for cross products to this situation, the resultant “vector”, Em, can be thought of as a “battery” or “source” where the +ve direction of the vector points towards the positive terminal (i.e. indicates the +ve polarity).

vel

B

l

++++

-- --

+Em

l vel

B

1

2

3



E.g. (similar to the loop above) If the bar in the sketch above was a 1m long conductor moving through a field of density 1T pointing into the page as shown, with a velocity (perpendicular) of 1m/s, the motional emf, Em would be 1V, with the positive end on the top as indicated. Now if an electric circuit was connected to this “battery” or generator, current would flow according to Ω’s Law. However, once a current is flowing, we have our other famous equation to contend with:

F = IlXB! In nature, nothing comes for free. If we have a current flowing that is doing some work, some energy must be put into the system. Using these two equations (

V = !motional = l " vel # B , and F = IlXB), we can explore the basic principles of how motors and generators work. But first, we need to explore another characteristic of an electromagnetic circuit: inductance.



INDUCTANCE

Resistance opposes current; Capacitance opposes changes in voltage; Inductance opposes changes in current.

Recall Faraday’s Law: EM = !d"dt

for the voltage

induce in a loop of wire. (Which resulted in: Em = lvel X B) If we have N turns, it’s like multiplying the flux linkages by N, so:

dtNd

dtdNEM

)( !=!= , and since BA=!

dtBAdNEM)(= ; and A is constant, so

!)(

, define ;)(

,constant ,, ;)(

, ;)(

;

22

dtIdLE

lANL

dtId

lANE

lNdtlNId

NAE

lNIHHlNI

dtHdNAE

HBdtdBNAE

M

M

M

M

M

=

==

=

===

==

µµ

µ

µ

µ

Where the letter ‘L’ is used to represent “self inductance”, usually referred to as just “inductance”, and the unit is the Henry (H). (E.g. A rate of change of current of 1 ampere per second in a 1 Henry inductor will produce a voltage of 1 Volt across the inductor.) Now looking at the expression for Inductance,

) valid!(always (2) that so ; that recall and ;(1) 22

!==!= NL

Al

lANL

µµ

For an air-core coil, the reluctance “outside” the coil is small compared to the reluctance “inside” the coil (Ri >> Ro). Thus, for an air-core inductor where the length is >> than the diameter (i.e. l/d > 10), the inductance can be estimated using:

4%. typicallyiserror the,10 where;2

0 <!=dl

lANL µ

(Note: for l/d < 10, you can apply Nagoaka’s correction factor)

V

V

!

RO

RI



Now go back for a moment to:

L = N 2

!; and " = NI = #!, (similar to V = IR) and substituting ! = NI

#

we are left with L = N#I

(also, always valid!).

Example 1: An air-core coil… find the inductance, L. Check: l/d = 12.5 – OK.

Example 2: A Laminated sheet steel core (S.F. = .93) with dimensions as shown. Calculate the inductance, L. (Note: Assume we are operating in the linear region of the B-H curve).

(Use a ‘normal’ approximation from the B-H curves to a B~=1T. Jackson’s ~ .0025. Boylestad is always around .001 for B up to about .8)

L = µAN 2

l

L =.0025( ) .01m( )2 100t( )2

.1m= 25mH

12mm

15cm

120t

1cm X 1cm

10cm

S.F. = .93

100t

L =µoAN

2

l

L =µo( ) .0062( ) !( ) 1202( )

.15( )L = 13.64µH



Example 3: A laminated Sheet Steel ‘C’ core inductor (S.F. = .96) with an air gap cut to prevent saturation. Again, use estimate for permeability assuming it is linear. .001 fr Boylestad up to B~.8T Assume total reluctance is in air gap: Consider fringing, so A = (1cm+1mm)2

Calculate the total reluctance first, including the metal

1cm X 1cm

1mm

6cm

6cm

100t

L =µoAN

2

l

L =µo( ) .0112( ) 1002( )

.001( )L = 1.52mH

! =lcoreµoA

+lgapµA

! =.239( )

.001( ) .012 .96( )( ) +.001( )

µo( ) .0112( )! = 2.489X106 + 6.577X106

! = 9.066X106 rels

L =N 2

!=

1002( )9.066X106

= 1.103mH



Back to the expression for voltage across an inductor:

dtdiLV = .

For nice “straight line” (constant slope) changes in current, the calculation of voltage across an inductor is straight forward. Example 4: Determine the voltage across a 50mH inductor for the current shown in the following graph. Graph the voltage on a comparable time-scale.

T(start) T(end) I(start) I(end) L V=L(di/dt)

0.000 0.002 0 4 0.050 100.0 0.002 0.003 4 4 0.050 0.0 0.003 0.005 4 -2 0.050 -150.0 0.005 0.006 -2 4 0.050 300.0 0.006 0.007 4 2 0.050 -100.0 0.007 0.008 2 2 0.050 0.0 0.008 0.010 2 0 0.050 -50.0

i(A)

T (msec)0 2 4 6 8 10

4

2

0

-2



V

T (msec)0 2 4 6 8 10

(X100)

3 2

1

-1 -2



Transient Response

Up until now we have considered inductors excited by direct current (D.C.) in situations where the circuit conditions have reached “steady state”. Now let’s look at the transient response of inductors to step changes in current (still using a D.C. source). (Note: the derivations and expressions below apply to a Thevenin Equivalent circuit specifically as shown at right. To apply most of these formulas, the circuit should first be reduced to its Thevenin Equivalent.) Using Kirchhoff’s Voltage Law:

! !

!

" "

#=$%&'

() #

==$%&'

() +=

+#=

$%&'

() #

=$%&'

() #

$%&'

() #

$%&'

() #=

#=

+=

===

I

0 0

LR

1

-E/Rb 1,a ,ln1

baudu

sidesboth integrate now ,

1,-by sideeach multiply and ,

,dt

by sidesboth multiply and ,

side...right thefrom LR

an out take,

..arranging.-re ,

,V and ,V that recalling ;

t

dropsrises

dtLR

di

RE

i

baua

dtLR

RE

i

di

dtLR

iREdi

iRE

iRE

LR

dtdi

LiRE

dtdi

dtdi

LiRE

dtdi

LiRVV

R

EL

@ t=0

i

+ -+

-



[]

( )

!! 1I

constant, time the, as RL

define note ,1I

sides...both toRE

- add and ,

,RE

-by sidesboth multiply ,

get... we, that recall and ,

e'' ofexponent an as sideeach use now ,ln

this...reverse ,)ln(lnln recall ,lnln

,ln

)ln(

ln

00

!!"

#$$%

&'=

!!"

#$$%

&'='=

'='

='

'

==

'=!!!!

"

#

$$$$

%

&

'

'

'=!"#$

%&'=!

"#$

%&''!

"#$

%& '

'=()

*+,

-!"#$

%& '

'

''

'

'

'!!!!

"

#

$$$$

%

&

'

'

.

.

t

tLR

tLR

tLR

tLR

xtLR

RERE

I

tI

eRE

eRE

eRE

RE

eRE

RE

I

e

RERE

I

xeee

tLR

RERE

I

baba

tLR

RE

RE

I

tLR

RE

i



A similar derivation can be used to determine VL as a function of time:

VL = L didt

, and using the expression for current derived above...

VL = L

d ER

1! e!t"

#

$ % %

&

' ( (

#

$

% %

&

'

( (

dt, taking out the constants...

VL = ELR

d 1! e!t"

#

$ % %

&

' ( (

#

$

% %

&

'

( (

dt, and d(1) = 0, d e

!t"

#

$ % %

&

' ( ( = !1

"e!t" , and L

RTh= ",

VL = E" 0 ! !1"e!t"

#

$ % %

&

' ( (

#

$

% %

&

'

( ( ,

VL = Ee!t" !!

It can also be shown that for circuits under D.C. excitation, the decaying transient can be described by:

iL = ISteady!State e!t"

#

$ % %

&

' ( ( , where " again = L/RTh,

and

vL = !I0RTh e!t"

#

$ % %

&

' ( ( , where I0 is the steady state current,

and RTh is the effective resistance "seen" by the inductor.

Note: In such exponentially rising/decaying circuits, the varying quantity (i or v) is generally assumed to have reached steady state after 5 time constants have elapsed (i.e. 99.3% of ultimate value).



Initial Conditions:

In the above derivations, we have ignored the possibility of any pre-existing constant terms in order to simplify the derivation. The implied assumption is that the current was at 0 at time t 0 (i.e. when the circuit conditions for the inductor changed). Of course, in many situations the initial current is not 0, and this pre-existing condition must be taken into consideration. A more general expression for the behaviour of exponentially rising or decaying quantities is given by:

( ) t

f f ix( t ) X X X e !"

= " " ; where Xi is the initial value of the quantity, and Xf is the final value. You can satisfy yourself that this expression correctly reduces to the expressions derived above for both voltage or current by substituting 0 for either the initial or final values. These expressions along with other circuit analysis techniques can be used to determine the initial and final values of the parameter in question. Example 1: Consider the circuit at right. a) find vL, i, i1, and i2 at t=0. b) Find all the above at t=>>5τ (=>> .418s). c) Find iL and vL as a function of time before and after the switch is opened. Find the maximum voltage across the switch contacts after the switch is re-opened.

a) at t=0, before the switch is closed, .



Energy stored in an Inductor:

Consider now the energy stored in an inductor. This can be thought of as the energy or work required to establish a magnetic field, and since it is not lost in heating (in an ideal inductor at least), it is available to be returned to the system as the magnetic field collapses.

, substituting the expressions previously derived for V and I,

, taking the constants out of the integration,

, and integrating the two terms separately,

, recalling that ,

, evaluating the integral,

recalling that: , and (anything)0 = 1,

,

, then 2

2LIW =

W (work,= energy) = (Power)(Time) = Pdt = VLILdtt1

t 2!

0

"!

W = E e! t

"# $ %

& ' ( ER1! e

! t"#

$ %

& ' ( dt

0

)*

W =E 2

Re!t"#

$ %

& ' ( 1! e

! t"#

$ %

& ' ( dt

0

)*+

, -

.

/ 0 =

E 2

Re!t" !e

!2t"#

$ %

& ' (

0

)*+

, -

.

/ 0

W =E 2

Re!t

"# $ %

& ' ( dt ! e

!2t"#

$ %

& ' ( dt

0

)*

0

)*+

, -

.

/ 0

eax =1a

! eax,a ="1#,and

"2#

W = E 2

R!"e

!t" ! ! "

2e!2t

"# $ %

& ' (

)

* +

,

- . 0

/

= E 2

R!"e

!t" + "

2e!2t

"# $ %

& ' (

)

* +

,

- . 0

/

W =E 2

R!"e

!#" ! !

"2e!2#

"$ % &

' ( )

$

% &

'

( ) ! !"e

!0" ! !

"2e! 2( )0

"$

% & &

'

( ) )

$

% & &

'

( ) )

*

+ , ,

-

. / /

a!" =1a"

=1"

= 0

W =E 2

R!" 0( ) ! !

"2

0( )# $ %

& ' (

# $ %

& ' ( ! !" 1( ) ! !

"2

1( )# $ %

& ' (

# $ %

& ' (

)

* +

,

- . =

E 2

R0 + " !

"2

) * +

, - . =

E 2

R"2

, and " = LR

W =E 2

2RLR

=L2E 2

R2 , and since ER

= I



Consider as an example the inductor and the circuit from the previous example. At “steady state”, the current in the Inductor was 0.8196A. This means that the energy stored in the inductor would be:

W =I 2L2

=.8196A2( ) 1H( )

2= 0.336watt ! sec( joules, J )

This energy is “discharged” as the current decays (i.e. the magnetic field collapses) and in this case was dissipated as heat in the resistances in the circuit. This concept of energy temporarily stored in the magnetic field of an inductor is also important in understanding how the “Conservation of Energy” law is true in circuits with reactive components like capacitors and inductors. Energy supplied by the source would not equal the amount dissipated in other elements of the circuit without considering the energy temporarily stored in these devices.

Inductors in Series or Parallel:

The inductance of a combination of inductors in series or parallel can be determined in a manner very similar to that of resistors. For inductors in series: LT = L1 + L2 + L3 + ...

For inductors in Parallel: ...LLLLT

+++=321

1111

Example: Find the total equivalent inductance of the circuit show below:

(ANS: 3.83H)



Example 2: For the circuit at right, the switches have been open for a long time (i.e. @ S.S.), and then both switches are closed at the same time. Find the values of vL and iL before the switches are closed. Find the energy stored in the inductor before the switches are closed What is the time constant, τ, in the circuit after the switches are closed For the period following the switches being closed, what is the initial current ( iL(t = 0+)) and the final current (iL(t = ∞)) Write the expression for, and graph the current through the inductor. Write the expression for vL(t) and graph its value for at least 5 time constants after the switches are closed. Find the time, t, when the current through the inductor is 0A (assuming t = 0 when the switches are closed).

Solution: Before the switches are closed, the only path for the current is from the 10V battery, through the 100Ω resistor, through the inductor, through the 33Ω resistor and back. We will have a positive current, (in the direction indicated (⇒)):

IL = 10V/133Ω = 75.188mA. At steady state, the voltage across the inductor VL = 0V.

The energy stored in the inductor at this point:

W =

LI 2

2. Using our value for current from above:

W =

.3H( ) .0752A( )2

= 848.0µJ

With the switches closed, the left side of the circuit is effectively eliminated and we are left with the modified circuit show at right:

RTh = 33/470 = 30.835Ω

ETh =

33!33+ 470( )! 12V( ) = 0.787V

The resultant equivalent circuit is also show at right.

! = L

RTh= .3H

30.8" = 9.729ms

!"1 = 102.8s"1

Figure 7: Thévènin Equivalent

Circuit



The initial current just after the switches are both closed is the same as the previous steady state value:

iL(t=0+) = +75.2mA The final current can easily be determined by the Thevenin Equivalent representing the circuit after the switches are closed.

iL(t=∞) = .787V/30.8Ω = -25.53mA Note that it is in the opposite direction to the original current through the inductor (which was called positive).

In general, iL( t ) = I f ! I f ! Ii( )e!t

" . We have all the values required to write the

expression:

iL( t ) = !25.5mA! !25.5mA! 75.2mA( )e!t

9.73ms = !25.5 +100.7e!102.8t( )mA

Graphing:

t t=0

IL

75mA

-25mA

τ=7.93ms



Use the initial current (75.2mA), the Equivalent circuit, and KVL to find the initial voltage across the inductor during this period. We need to adopt a convention for the polarity of the voltage; for convenience, we can assume that +ve is upward in the Thevenin Equivalent circuit (would mean that the right side was +ve in the original circuit).

VL -(75.2mA)(30.8Ω) -0.787V = 0; VL-initial = 3.11V

vL( t ) = 3.105e!t9.73msV

The final voltage, Vf = 0V, as usual.

Using our expression for iL(t):

iL( t ) = 0A = !25.5 +100.7e!102.8t( )mA ;

e!102.8t =

25.575.7

= .2535

t = 13.35ms As a check, plug this value back into our expression for voltage across the inductor:

vL(13.35ms ) = 3.105 e!102.8 13.35ms( )( )mA = .787V ,

which, as it must, satisfies Kirchhoff’s Voltage Law at this point in time!

t t=0

VL

τ=7.93ms

3.1V



Mutual Inductance:

The “opposition to changes in current” we have considered previously have been due to a device’s reaction to magnetic lines of flux produced by a current flowing in its own windings and the resulting induced voltage. This characteristic or parameter was called self-inductance, commonly referred to a just “inductance”. In a similar fashion, a coil of wire can react to the field established as a result of a current flowing in another coil adjacent to, or at least in the vicinity of, the first (and vice-versa). This reaction is referred to as “mutual inductance” M, and is measured in the same units as self inductance, the Henry(H). Mutual inductance is defined as:

where M21 is the mutual inductance of coil 2 with respect to coil 1; N2 is the number of turns in coil 2; Φ21 is the flux produce by coil 1 that links coil 2; and I1 is the current in coil 1. Similarly, M12 is the mutual inductance of coi1 1 with respect to coil 2:

The voltage induced in coil 2 as a result of flux created by coil 1:

and similarly for the voltage induced in coil 1 as a result of coil 2 is given by:

. It can be shown that M12 = M21, and can be called simply M. Using this simplification, the induced voltages are:

If the coefficient of coupling between the two coils is 100% (i.e. φ12 = φ21), we can also write a simplified expression for the mutual inductance, M:

While mutual inductance can be a nuisance in some circumstances (e.g. stray interference), it is exploited in devices called transformers, where the mutual coupling between two coils is, by design, virtually 100%. This is usually accomplished by winding both coils on one continuous core so that all the flux produced by one coil links the other. By manipulating the above expressions, it can also be shown that under these circumstances, the ratio of the “induced or output” voltage to the “input” voltage is the same as the “turns ratio”; a very useful result for electrical engineers.

M21 = N2!21I1

M12 = N1!12I2

V2 = !N1d"21dt

= !M21dI1dt,

V1 = !N2d"12dt

= !M12dI2dt,

V1 = !M dI2dt

, and V2 = !M dI1dt

M = N2!I1

= N1!I2



Capacitors

Recall from PHY155 that the capacitance of an ideal parallel-plate capacitor, C = !r!oAd

, where A

= the plate area (m2), d = the distance between the plates (m), εo = permittivity of free space (8.854 X 10-12 C2/N-m2), and εr is the relative permittivity of the dielectric between the two capacitor plates (a unit-less number giving the ratio of the permittivity of the dielectric to that of a vacuum, with a vacuum considered essentially the same as space). Also recall that the voltage across an ideal capacitor is proportional to the charge held on the plates

and inversely proportional to the amount of the capacitance: VC =QC

(Q in coulombs and C in

farads will give V in volts). Since the amount of charge is a function of the current and how long

it has been flowing, we can also write: VC =i ! tC

. An incremental increase in voltage can be

expressed: dVC =i !dtC

, and re-arranging, we get an expression for the current: iC = C dVCdt

In a “steady-state” DC circuit, a capacitor appears like an open circuit, with no current flowing through it and a voltage across it determined by the other circuit components and their configuration.

Redraw the schematic VC1 = 72V( ) 2!

2 + 7!( ) = 16V

VC2 = 72V( ) 7!

2 + 7!( ) = 56V

Alternatively, we could have figured out the current flowing through R1 and R2 (8A) and determined the voltages across the capacitors from that.



Transient Response

Up ‘til now we’ve considered capacitors excited by direct current (D.C.) in situations where the circuit conditions have reached “steady state”. Now let’s look at the transient response of capacitors to step changes in voltage (still using a D.C. source). (Note: the derivations and expressions below apply to a Thevenin Equivalent circuit specifically as shown at right. To apply most of these formulas, the circuit should first be reduced to its Thevenin Equivalent.) To mathematically derive the response, first convert the circuit above to an equivalent with a current source as shown immediately below. Note that from the capacitors perspective, the I-R’ combination looks exactly like the E-R equivalent in the first diagram. Also note that I = E/R and R’ = R. Assume there is no charge on the capacitor (i.e. no voltage) prior to the switch being closed

Using Kirchhoff’s Current Law for the top rail:

Iin = Iout ; recalling that IR =!! vR '

, and IC = Cdvdt

,

I = IR ' + IC =vR '

+ C dvdt

, re-arranging...

dvdt

=I ! v R 'C

, take out an 1R'C

from the right side...

dvdt

=1R 'C

IR '! v( ), and multiply both sides by dtIR '! v( ) ,

diIR '! v( ) =

1R 'C

dt, and multiply each side by -1,

div ! IR '( ) =

!1R 'C

dt, now integrate both sides duau + b

=1a

ln au + b"#$%&'(

, a = 1, b = -IR'

dvv ! IR '( ) =

!1R 'C

dt0

t

"0

v

"

ln v ! IR '( )"# $%0V =

!1R 'C

t[ ]0t ,

ln V ! IR '( ) ! ln !IR '( ) = !1R 'C

t, recall ln ab

"#$

%&'= ln a( ) ! ln(b), reverse this...

I (=E/R)

Sw1

RÕ (=R) C

vi iRÕ iC

+E

Sw1 R

Ci



ln I ! IR '!IR '

"#$

%&'=

!tR 'C

, now use each side as an exponent of the constant 'e'

eln V ! IR '

! IR '"#$

%&' = e

! tR 'C , and recall that eln(x ) = x, we get...

V ! IR '!IR '

= e! tR 'C , multiply both sides by -IR',

V ! IR ' = !IR 'e! tR 'C , and add -IR' to both sides...

V = IR '! IR 'e! tR 'C = IR ' 1! e

! tR 'C

"#$

%&'

, now define R'C as ( , the time constant,

V = IR ' 1! e! t"

#$%

&'(

Finally, recalling that R’ = R and I = E/R from the original series equivalent circuit,

and VC = E 1! e! t"

#$%

&'(!! , where we note that E is the value of the original voltage source and is the

voltage present across the capacitor at t = ∞ (i.e. at “steady-state”).

Now, recalling that Cdvi Cdt

= , and using the expression for voltage across the capacitor derived

above, we can also develop an expression for IC for this same circuit as a function of time:

1 1t t

C

d E e d edvi C C CEdt dt dt

τ τ− −⎛ ⎞⎛ ⎞ ⎛ ⎞− −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠= = =

and since d(1) = 0, and d e

! t"

#

$%&

'(=!1"

e! t" , and " = RC, we are left with:

10t t

CCEi CE e e

RC RCτ τ− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞−= − = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

!!t

CEi eR

τ−

=

where we observe that E/R is the initial, or maximum current that will flow in the circuit at time, t = 0+ , just after the switch is closed.



It can also be shown that for circuits under D.C. excitation, the decaying transient can be described by:

''Th'

'0

'Th

, where ' is the time constant of the decaying circuit R

and

, where V is the initial voltage accross the capacitor.

and R is the effective resistance "se

to

CTh

t

C o

Vi e CR

v V e

τ

τ

τ−

−

⎛ ⎞−= =⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟

⎝ ⎠

en" by the capacitor in the discharging circuit.

Consider the following diagrams showing the charging and discharging phases for the voltage and current in a capacitor (from Boylestad’s Circuit Analysis, 3rd Canadina Ed., CDROM)

Note: In such exponentially rising/decaying circuits, the varying quantity (i or v) is generally assumed to have reached steady state after 5 time constants have elapsed (i.e. 5τ = 99.3% of ultimate value). It is important to note that the expressions given above apply only to a simple E-R-C circuit, and that complex circuits must be reduced to a simple equivalent to use these formulas.



Initial Conditions: In the above derivations, we have ignored the possibility of any pre-existing constant terms in order to simplify the derivation. The implied assumption is that the voltage was at 0 at time t 0 (i.e. when the circuit conditions for the capacitor changed). Of course, in many situations the initial voltage is not 0, and this pre-existing condition must be taken into consideration. A more general expression for the behaviour of exponentially rising or decaying quantities is given by:

( ) t

f f ix( t ) X X X e !"

= " " ; where Xi is the initial value of the quantity, and Xf is the final (steady-state) value. You can satisfy yourself that this expression correctly reduces to the expressions derived above for both voltage or current by substituting 0 for either the initial or final values. These expressions along with other circuit analysis techniques can be used to determine the initial and final values of the parameter in question.

Capacitors in series and in parallel

Capacitors in parallel add simply (like resistors or inductors in series): CTotal = C1 + C2 + C3 + ...

CTotal = 800µF + 60µF +1200µF = 2050µF Capacitors in series add like resistors or inductors in parallel:

CTotal =1

1C1

+ 1C2

+ 1C3

+ ...

CTotal = 200µF( )!1 + 50µF( )!1 + 10µF( )!1"

#$%

!1= 8µF



Energy stored in a Capacitor: In a derivation very similar to that used to determine the energy stored in an inductor (using the expressions previously derived for the voltage across and current through a capacitor), we can determine that the energy stored in a capacitor will be:

WC = Poweritime = iCvC dt =

CV 2

20

t

!



Examples Eg. 1: (based on Robbins & Miller, 3rd Ed., Q 11-11)

1. Determine the expressions for the capacitor voltage and current when the switch is connected to terminal 3,

2. and then again when it is instantly connected to terminal 2.

3. Graph both the parameters over time until they finally reach a steady-state.

Assume there was no initial charge on the capacitor before the switch was first operated. Also assume that the system is allowed to reach a steady-state while connected to terminal 3 (before it is switched to 1).



Eg 2: (based on Robbins & Miller, 3rd Ed., Q 11-12) Consider the capacitor circuit at right. The switch has not been activated for a long time, and there are no initial conditions to consider. The switch is then moved to terminal 3 for 5 ms, then switch instantly to terminal 1 where it is left to reach steady-state. 1. determine the expressions for iC(t) and vC(t)

while connected to terminal 3 2. determine iC and vC at t = 5ms. 3. determine the expressions for iC(t) and vC(t)

while connected to terminal 1 4. determine iC and vC at t = 10ms. 5. sketch iC and vC from t = 0 through both switch operations until they reach steady-state.



Inductive and Capacitive Transients – SUMMARY

(Jackson: Ch 13(C), 17(L). Boylestad Ch 10(C), 11(L)) Changing current, i, and voltage, v, in an inductor or capacitor:

i v

L

C

Instantaneous: Exponential: Transient Response timeframes:

Use x(t) = X f ! X f ! Xi( )e! t " , where Xf and Xi are the final and initial values during the period of

interest, and ! is the time constant ( L RTh,RThC ) and RTh is the resistance “seen” by the inductor

or capacitor again, during the period of interest.

t = 0 (time when the circuit changes from L’s or C’s perspective

i/v

t

Time period of interest t = 0+ to t (∞)

Previous conditions t = -t to 0-



Determining Initial and Final Values:

initi

al For L iL(t=0+) = iL(t=0-)

vL(t=0+): use old current(t=0-) and new circuit(t=0+) and KVL

For C vC(t=0+) = vC(t=0-) iC(t=0+): use old voltage(t=0-) and new circuit(t=0+) and KVL

final

At steady-state, the voltage across and inductor or the current through a capacitor is always 0 (vL(t=∞)=0V, iC(t=∞)=0A). For the final current in an inductor or the final voltage across a capacitor, use basic circuit analysis, remembering that at steady-state(SS, t=∞), an inductor looks like a short and a capacitor looks like an open

LSS: CSS:



Thevenin “101”

Summary: The purpose of this note is to provide a brief guideline on when and how to use Thevenin and Norton equivalents to help with circuit analysis. It provides some simple steps for determining the equivalents as well as an example to demonstrate the process. Finally, it provides a couple of cautions about the limitations of using equivalents. Why use equivalents? Simply stated, an equivalent circuit can be used to simplify determination of the voltage across, or current through a specific element. In order to do this, we replace a rather complicated network of components with a simple 2-element replacement that acts exactly the same as far as the target component is

concerned. Using the equivalent network to find the voltage or current through the target component is usually not a problem. The challenge is determining the equivalent itself, and sometimes when it will be helpful or necessary to do so.

When to use equivalents: There are many techniques that can be used for circuit analysis, and they will all give the same result. While there is often some advantages in using one method versus another in a specific circuit situation, which one you use will depend on the questions you are trying to answer and your personal familiarity and expertise at applying them. This note will focus on the analysis you will typically have to do for transient D.C. circuits. In these cases, a Thevenin or Norton equivalent is often very useful. You will typically require an equivalent if the network between a source(s) and the target component (i.e. the component of interest) has a combination of series and parallel components. Another, but less obvious way to tell is if the circuit is such that varying the current or voltage in the target component will affect the voltage or current at the component. For example, if we’re interested in the current through the 470kΩ resistor in the circuit above, we can see that changing the current in it will change the voltage across it and thus affect the contribution for the various sources in the network. This is a good hint that an equivalent circuit will be helpful! At worst, you can’t go wrong by using an equivalent circuit (it will always perform the same as what it replaces), you may just do a bit more work than necessary. Of course this assumes that you determine the equivalent correctly! Definitions: A Thévenin or Norton equivalent circuit consists of a source (voltage or current respectively) and one resistor, forming a two-terminal network, that will behave exactly the same from the target component’s perspective as the more complex network it replaces. It is very important to understand that the source and the resistor in the equivalent network are generally not the same as any original component, even though they may have the same values! This means, for example, you cannot infer that the current through an original resistor is the same as the current going through RTh, even though it may have the same value. In some cases some relevant information can be deduced from the equivalent, but this should be done cautiously!



How to Determine Equivalents: A simple series of steps will help derive an equivalent circuit: 1. define two terminals isolating the component of interest (target component), 2. “remove” the target component, and determine the resistance “looking back” into the terminals

with the sources replaced by their internal resistance (a short for voltage sources and an open for current sources),

3. determine the “open circuit” voltage (Thévenin) or short circuit current (Norton) at the terminals,

4. use the source and resistance to build a series network (voltage source, resistor =, Thévenin) or parallel network (current source, resistor = Norton) and attach it to the two terminals in place of the original network.

Admittedly, steps two and three can cause some challenges with some circuits. In these cases, some practice with circuit analysis will help determine the necessary component values. Finding the equivalent resistance (2), RTh or RN, is usually just a matter of reducing resistor networks. This topic is well covered elsewhere and not the prime target of this discussion. Finding the open circuit voltage or short circuit current (3) can often be aided by using Thévenin<->Norton source conversion themselves. A brief example will illustrate how. Consider the circuit example shown above. Let’s say we’re interested in the current through the 4.7kΩ resistor. We first define two terminals around the component. (Note: I’ve also “flipped” the 3V source so that it is a +3V facing the other way.)

Next we remove the component and replace the sources by their “equivalent impedances”. Looking back into the terminals, we “see”

(100k + 200k)//1MΩ, so RTh = 230.7kΩ Next we need to determine the open circuit voltage at the terminals. Here we should recognize the conversion relation between Thévenin and Norton equivalents:

Note this is the same as between all equivalents: same VOC, same ISC! First, notice a “Norton Equivalent” on the right that could be converted to a Thévenin Eq. and combined in series with the other “Thévenin Eq. on the right side:

+RTh = RNETh = INRN

IN = EThRTh

RTh

RNRTh = RN

ETh IN

1M!

100k!4.7k!

200k!

6µA9V

3V

1M!

100k!

200k!

6µA9V

3V

RTh/RN



Also note that the polarities between the equivalent’s terminals must also match! At this point we have two choices: use the voltage diver rule to determine VOC; or convert both the remaining Thévenin Eq. to Norton Eq., combine them into 1 Norton Eq. and convert back to 1 Thévenin Eq. Voltage divider first: Note the two sources are opposing in polarity which leaves (9-3.6) 5.4V to drop across the two resistors. Using simple voltage divider rules, and observing appropriate polarities, we can determine that 4.15V drops across the 1MΩ resistor, and the rest across the 300kΩ resistor. This leaves us with a voltage between our terminals of interest of -4.85V (assuming the bottom terminal is the reference), and the following equivalent circuit..

Using Source Conversions:

combining:

and converting back: ! We can now simply calculate the required value: -20.6µA (assuming the positive direction is defined as down through the resistor) Final Summary and Cautions: The steps suggested above will help you derive Thévenin and Norton equivalents for D.C. circuits containing resistors and independent sources, and use them to simplify circuit analysis. The most important word of caution, as stated above, is that once the equivalent substitution has been done, the original components are not there. In general, you cannot use the equivalent components to deduce any parameters (current through, voltage across) for the original components that were replaced. This is also true of simple Thévenin <-> Norton source conversions; even though the resistances are the same value, they are not the same resistors!!

100k! 6µA 100k!

200k! 3V .6V 300k! 3.6V

+

230.8k!

-4.85V4.7k!

1M! 300k!12µA9µA 230.8k!21µA

+

230.8k!

-4.85V4.7k!

1M! 300k!

9V

3.6V

4.15V 1.25V+ +- -



Bar and Rail Example 1 B = 0.8T, R = 2Ω, Rbar = 0.1Ω, lbar = 1.0m, lr = 0.8m, µ = 0.15, M = m = 1kg Assume the system has reached steady-state:

a Draw a FBD for the bar b What is the magnitude and direction of the current? c What is the velocity and direction of the bar? d Do a complete power balance on the system. e Is the bar acting like a motor or generator? f What is the overall efficiency, η?

m

M

B µ

lr

R

6V

+



Bar and Rail Example 2 Rail Gun Example (from the movie ”Eraser”)

fires aluminum rounds at “near the speed of light” using an EM (electromagnetic) pulse. Let’s check and see if the “physics were [really] impossible”!

Some useful constants:

speed of light, c = 3 X 108 m/s specific mass of aluminum: 2.443 g/cm3 specific heat of aluminum: .89 J/g-C0 or .89 W-s/g-C0 resistivity, ρAl = 5 X 10-8Ω-m so .5cm X .5cm X 1cm round = 2 X 10-5Ω

Some assumptions:

near speed of light, let’s say 1/3 C = 108m/s is really fast, lets say 1/10 of that or 107m/s (Note: high-powered rifle in order of 800 -900 m/s?) assume rails are 1m long and 1cm apart A large, strong lab magnet might have a field density of 2.5 – 5T; let’s assume a 5T magnet in the gun Assume pellets (rounds) are .5 X .5 X 1cm: Volume: (.005)(.005)(.01) = 2.5 X 10-7m3 or 0.25cm3 Mass: (2.443g/m3)(.25cm3) = 0.6075g Resistance: ρl/A = (5 X 10-8)(.01)/(.0052) = 2 X 10-5Ω Specific heat for one round: (.6075g)(.89W-s/g-C) = .54W-s/0C

B +

1m

1cm V



Bar and Rail Example 3 (2006 midterm) Consider the bar&rail system shown in the diagram below to answer the following questions. Assume that the mass of the bar is negligible and that it is resistance-less. The rails are 1m apart, and the field density is 1T.

a) The switch has been in position 1 (as shown) for long enough for the system to reach steady state. i) Determine the velocity and direction of the bar. ii) Is it acting as a motor or a generator?

Note that an initial current of 20A (20V/1Ω) flowing downward through the bar will cause an Fmag to the right. This motion will induce a voltage in the bar in opposition to the battery. The system will reach a steady velocity when the velocity (EM) is enough to limit the current so that Fmag = Mg.

Mg = Fmag = Il X B; I = 1 9 81

9 811 1

mkg( . )Mg sI . Al B ( m )( T )

= = =!

Using KVL: Em + (9.81A)(1Ω) -20V = 0; EM = 10.19V EM = l vel X B; 10 19M

ME mE l vel B; vel . sl B

= ! " = =!

to the right.

Since the electrical source is delivering power to the bar and the mass is being lifted (mechanical work being done), the system is acting at a motor. b) The switch is now moved to position 2 and the system has once again reached a steady-state. i) What is the direction and velocity of the bar? ii) Is it acting as a motor or a generator? iii) Do a complete power balance to prove that energy is conserved. (Be sure to specific all the mechanical sources/sinks.) With no external source of current, the mass will pull the bar to the left until the induced EM is large enough to drive enough current through the other 1Ω resisitor and give a balancing Fmag.

Again, at steady state, Mg = Fmag = Il X B; I = 1 9 81

9 811 1

mkg( . )Mg sI . Al B ( m )( T )

= = =!

Using KVL: Em + (9.81A)(1Ω) = 0; EM = -9.81V

9 81MM

E mE l vel B; vel . sl B= ! " = =

! to the left.

1kg

+

1

2

20V

1!

1!



Since there is only a source of mechanical power (the mass dropping), the system is now acting as a generator. Power:

Mechanical: Source: Mass falling providing force Mg = 9.81N at 9.81m/s = 96.2W Sink: the bar being pulled with a force, Fmag = 9.81N @ 9.81m/s = 96.2W

Electrical: Source: Bar (generator) (Em)(I) = (9.81V)(9.81A) = 96.2W Sink: Resistor heating I2R = (9.81)2(1Ω) = 96.2W



Bar & Rail Problem Flowchart



Nothing moves!

B

M

R

V

+

Generator Motor

Right Left

Yes

Yes Yes

No

Find vel using Em=lvelXB

Use KVL and I from above to find Ebar

Find I when Mg =

Fmag+Ffr

Find I when Fmag_ = Ffr+Mg

Will bar accelerate left

or right?

Is net Force > friction?

Calc. net

force

Is there a mechanical

force? F=Mg

Is there a current? F=IlXB

Begin assuming the bar is at rest.

Begin assuming th4



Motor – Generator: Energy Sources & Sinks

mechanical

losses

mechanical power out

Conversion to

mechanical power

electrical power to convert

electrical losses

total electrical

power in

electrical losses

electrical power out

Conversion to electrical

power

mechanical power to convert

mechanical losses

total m

echa

nical

power in

MOTOR

GENERATOR sources sinks

Figure 9: Energy Sources & Sinks in Motors and Generators

m-s

inks

m-s

ourc

e

e-si

nks

e-so

urce

ee201 notesDC full - College of Engineering · nominal area multiplied by the Stacking Factor...

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