electric power notes

7/23/2019 Electric Power Notes

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ELECTRIC POWER

Apply a field model to magnetic phenomena including shapes and directions produced by bar

magnets, and by currnets in ires, coils and solenoidsThe direction of a magnetic field is determined by the direction the north end of a magnetic compass

points. The magnetic field can be represented as continuous loops that begin at the North end of a

magnet travel in space around to the South end of the magnet and then move through the inside of themagnet back to the North end.

Electric currents were shown by Oersted to have a magnetic effect. That is they make the compass

needle change direction. The electric current had a magnetic field associated with it, which also is in acontinuous loop. Electric current can be reversed in direction by changing the connections to a battery.When this is done the compass needle reverses direction. f you don!t have a compass needle, then youwill need a rule to determine the direction of the field. The rule is called the "ight #and $rip "ule.

The "ight #and $rip "ule$rab the wire in your right hand with your thumb pointing in the direction of the conventional current.%our fingers will then wrap around the wire in the direction of the magnetic field.

When you do the same thing for a coil of wire, you find that the magnetic field is pointing in the samedirection for all parts of the coil, i.e. the magnetic field is coming into one side of the coil and leavingthe other side. This means that the coil has a North side &or end' where the magnetic field is leavingthe coil, and a South side &or end'.

(ircle with a dot means coming out of page. (ircle with a cross means going into the page.

) solenoid is merely a series of coils lined up parallel to each other so that each of the individual coil!smagnetic fields add together to produce a stronger field.

The physical *uantity, magnetic field, is a vector with si+e and direction. The unit for the strength ofthe magnetic field is Tesla, abbreviated T-. or e/ample, the Earth!s magnetic field at the E*uator is0.1 mT, North.

(opyright )2 Ed (omm 3004


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!uantify magnetic forces on current current carrying ires using " # Il$ here the directions of

I and $ are either perpendicular to, or parallel to, each other%When a wire carrying a current is placed in a magnetic field it e/periences a force. What determinesthe si+e of this force5 6ogic would tell you that the stronger the magnetic field or the larger the currentin the wire, the bigger the force.

The other factor that determines the si+e of the force is the length of wire actually in the magnetic field.)ll these combine to give the e*uation for 78the orce, , on a (urrent8carrying wire7 9 :l,

where : is the symbol for the magnetic field, for current and l for the length of wire.

or the orce to be in Newtons, the ;agnetic ield must be in Tesla, the (urrent in )mps and the6ength of wire in ;etres.

n general, the orce, , on a (urrent8carrying wire7 9 n:l, where n 9 number of coils.

n this phenomenon, we have a vector, the magnetic field, acting on another vector, the current, to

produce a third vector, the force. What direction does the force act on the wire5

Once again, a rule has been devised to remember how to determine the force!s direction. n fact thereare two different ways of remembering the rule. Teachers will have their preferences depending ontheir own schooling. The two versions are7

6eft #and "ulende/ &or pointing' finger, pointing ahead ;agnetic fieldSecond &or longest' finger, at right angles to the inde/ finger (onventional (urrentThumb, upright at right angles to both fingers orce

6ock the three fingers in place so that they are all at right angles to each other. Now rotate your handat the wrist so that the field and current line up with the directions in your problem. The thumb willnow point in the direction of the force.

"ight #and Slap "uleingers &out straight' ;agnetic field

Thumb &out to the side of the hand' (onventional (urrent2alm of hand orce

#old the hand flat with the fingers outstretched and the thumb out to the side, right angles to thefingers. Now rotate your hand so that the field and current line up with the direction in your problem.The palm of your hand now gives the direction of the force, hence the name.

What happens if the wire is not at right angles to the magnetic field5

n general, the length of the wire is really the length of the wire as seen- by the magnetic field, i.e. thecomponent of the wire cutting across the wire.



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f the wire is at right angles to the field, cutting across the magnetic field, then the force is thema/imum, 9 :l.

f the wire and the current are running parallel to the field, then the force is +ero. )s the angle with thefield increases from +ero to


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n this position there is a turning effect, but it is opposing the rotation and brings the coil back to theperpendicular position. The coil will now permanently stay in this position unless there is a change inthe design.

>esign Options) meterThe turning effect is proportional to the current. ) helical or watch spring can be attached to the coilto oppose the magnetic force, and a needle attached to the coil. The larger the current, the larger themagnetic force, the more the spring is unbent, the larger the needle deflection.

) motorTo continue the rotation, the turning effect needs to be reversed each half turn. This is achieved bereversing the direction of the current in the coil with a commutator.

t is most important that the turning effect and current are re'ersedwhen the coil is at right anglestothe field, when the turning effect is momentarily +ero. The current reverses tice in each turn.

Apply a field model to define magnetic flu(, , using # $A and the )ualitati'e effect of

differing angles beteen the coil and the field) current in a wire in a magnetic field can result in a force and hence movement of the wire. (an amoving wire in a magnetic field produce a current5 %es. This is called Electromagnetic nduction.

The movement of a magnet in and out of a coil or solenoid easily demonstrates the generation of acurrent, which re*uires a voltage. Simple e/periments indicate that78

the si+e of the current or voltage is to the speed at which the magnet is moved,

the direction of the current is reversed if the action of the magnet is reversed,

there is no current or voltage when there is no relative movement between the magnet and coil.

Similar results are found when the magnet is replaced by an electromagnet, that is a coil with a >(current passing through it.

#ow can this phenomenon be e/plained5 and how can the si+e of the induced voltage be determined5

One e/planation uses the ideas from the previous section on magnetic forces. )nother introduces anew concept called ;agnetic flu/-.

?sing ;agnetic orce on ) (urrent or (harge


N S

N S


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magine a metal rod, not connected to anything, being pulled across a magnetic field. The movingpositive and negative charges in the rod each constitutes a current and will e/perience a magnetic force.

What will be the direction of the force on the positive and negative charges5

The forces are in opposite directions. This will give the top of the rod a negative voltage, and thebottom a positive voltage.

f the top and bottom of the rod are connected by a wire outside the magnetic field, then the negativelycharged electrons will move from the top of the rod through the wire to the bottom of the rod. Themagnetic force, due to the moving rod, will then take them to the top of the rod, giving them theenergy to go around the circuit again.

The moving rod in the magnetic field- is a source of electrical energy like a battery.

Once a current is established, the magnetic field will act on it. The direction of this force will be to theleft, slowing down the rod. To maintain the rod!s velocity, and the flow of current, a constant push tothe right is needed. This is the source of energy that supplies the electrical energy.

The force on the current must be to the left because if it was to the right, the rod would accelerate,producing a larger voltage, and a larger current, which produces a larger force, which produces agreater acceleration, ..... n this case energy is being created from nothing. So, from energyconservation principle alone, the magnetic force of the current must be to the left.


vel

vel

(onventionalcurrent


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?sing ;agnetic lu/;agnetic lu/ is defined as the amount of magnetic field passing through the area of a coil. Thestronger the field, the more flu/@ the larger the area, the more flu/.

This leads to the formula =B A. where the units of are weber.

Explain the generation of voltage, including AC voltage, in terms of the rate of change ofmagnetic flux,(Faradays Law), the direction of the induced current (Lenzs Law), and number

of loops through which the flux passes, including calculations using induced

= -

/

t

araday used the idea of magnetic flu/ to e/plain the observation of the generation of electricitydescribed at the beginning of this section. #e summarised some of the observations with thestatement78The magnitude, or si+e, of the induced emf, or voltage, is proportional to the rate of change ofmagnetic flu/-

t6en+ e/tended araday!s e/planation to include the direction of the induced emf. #e stated78The direction of the magnetic flu/ of the induced current was such that it opposed the change in flu/of the applied magnetic field.- This is incorporated as a minus- sign.

= Nt

, where N is the number of coils.

Note the minus- sign does not mean a negative voltage.

Describe the production of voltage in generators and AC voltage in alternators, including the

use of commutators and slip rings;

The magnetic flu/ through a rotating coil in a magnetic field follows a sine curve with the ma/imumflu/ when the coil is at right angles to the coil. The magnetic flu/ is changing most rapidly when thecoil is parallel to the field. n this case the flu/ is changing from going through one side to goingthrough the opposite side.

$iven that the graph of magnetic flu/ with time is a sine curve, the graph of the induced emf will alsobe a sinusoidal curve, but ( voltage.



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Compare sinusoidal AC voltages produced as a result of the uniform rotation of a loop in a

constant magnetic flux in terms of frequency, period, amplitude, peak-to-peak voltage (V p-p) and

peak-to-peak current (Ip-p);

The fre*uency of the )( voltage will be the same as the fre*uency of the turning coil. The fre*uency,which is the number of cycles in one second, of 3=0A)(, is 10 #+. This is set by the generators in the

power stations. The period is the time for one complete cycle and is e*ual to 0.03 seconds.

The amplitude of )( voltage, or the peak voltage, is the ma/imum value of the voltage. The peakvalue of the )( voltage is proportional to the si+e of the magnetic field, :

the area of the coils, )the number of coils, Nthe fre*uency or how *uickly the flu/ changes.

The peak to peak voltage, written as p pV , is e*ual to twice the peak voltage.

Use rmsvalues for a sinusoidal AC voltage, V RMS= Vpeak /2 and IRMS= Ipeak /2, and interpret

rmsin terms of the DC supply that delivers the same power as the AC supply;

)(, or Alternating Current, voltage varies as a sine wave. The domestic )( voltage supplied by theSE( is described at 3=0 A )(. The 3=0 volts specifies the ";S or root mean s*uare value, which is away of averaging a sine wave. The ";S value of an )( voltage indicates the >( voltage that wouldhave the same heating effect, that is provide the same power.

This relationship is given by peak RMSV xV= 2 9 BB( motors use a source of >( voltage to produce rotation. $enerators and alternators use a source ofrotation to produce a voltage, either )( or >c in the case of a generator, only )( in the case of analternator. )ll three devices have a coil of wire rotation in a magnetic field.

) >( motor must use a split ring commutator, an alternator must use slip rings, while a generator

could used either depending on whether )( or >> voltage was the re*uired output.

Explain transformer action, modelled in terms of electromagnetic induction for an ideal

transformer, qualitatively; and quantitatively using number of turns in primary and secondary

coils, voltage and current;

( 1

2

1

2

2

1

NN

VV

II

= = for an ideal transformer)

One of the advantages of )( voltage and current is that it produces a constantly changing magneticfield. This property is used in the design of a transformer. ) transformer is another e/ample of anapplication of electromagnetic induction.

magine a soft iron core shaped as a s*uare ring. )round two sides are coils of wire. f an )( voltageis applied to one of the coils, called the primary coil, an alternating magnetic field will be set up in the


Time &sec'

A

BB< A


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iron core. This alternating magnetic field will propagate through the iron core to the other coil, calledthe secondary coil. #ere the alternating field will induce an alternating voltage in this coil.

f the primary coil has many turns, a strong field will be set up in the iron core. )t the secondary coil,this strong field will set up a high voltage in each turn of the coil. f the secondary coil has many turns,a large )( voltage will be across the terminals of the secondary coil.

n summary ps

p

sVV

NN

= , where Apand Asare the primary and secondary voltages, and Npand Nsare the

number of turns in the primary and secondary coils. This relationship is true for an ideal transformerwhere there is no energy loss either in the coils, or in the iron core.

f there is no energy loss, then the 2ower n- to the 2rimary coil will e*ual the 2ower Out- of theSecondary coil. This can be written as A229 ASS

Model mathematically power supplied as P = VI and transmission losses using voltage drop V =

IR) and power loss (P= I2R);

Transformers are essential for the transmission of electrical power. They enable the transforming of

relatively low generating voltages of CC,000 volts up to considerably higher voltages of 100kA forwhich the energy or power loss is less.

) typical power output from a power station would be B00 ;W, i.e. B.0 / C0 4 watts. )t this voltage of100kA the current is D00 amps. This large current will flow down and back through long, very lowresistance cables to ;elbourne. ?nfortunately, even though their resistance is low, the large currentmeans that some energy or power is lost.

The e/pression for power lost in a resistance is 3". or a cable of total resistance both ways ofresistance of 1.0 ohms, this gives power loss of D00 / D00 / 1.0 9 C.4;W, i.e. 0.D of the poweroutput.

Whereas, if the same power were transmitted at C00kA, the current would be 1 times bigger at B000

amps, and the power loss would be 31 times larger at =.1 / C0F watts or C1 of the power outputwhich is a significant amount.

The effect of the s*uared term in the 2ower 6oss e/pression is a significant factor.

t is important to also be aware of the conse*uences on the voltage that is available at the other end ofthe cables. n the second case the voltage drop across the cables is B000 amp / 1 ohms 9 C1 kA,leaving 41 kA.

Whereas in the case of the higher step8up transformer, the voltage drop is D00 amp / 1 ohm 9 B kA,leaving =

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