Tutorial Potencial

8/18/2019 Tutorial Potencial

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In this tutorial, you will use the volt-ampere method to calculate power

parameters of an AC circuit.

A factory uses 220V commercial power. The principal users of power are a lighting load

of 200 kW and an injection molding machine rated at 600 kVA. The molding machine

has a power factor of .63 lagging and is !" efficient in con#erting electrical energy tomechanical energy.

$raw the power triangle for the molding machine.

What is % for this power triangle& 'ound your answer to the nearest kilowatt.

kW

That(s right. % ) * + pf ) 600 + .63 ) 3! kW

A factory uses 220V commercial power. The principal users of power are a lighting load of 200 ,W andan injection molding machine rated at 600 ,VA. The molding machine has a power factor of .63 laggingand is !" efficient in con#erting electrical energy to mechanical energy.

-elow is the resulting power triangle for the machine.

What is for this power triangle& 'ound your answer to the nearest kilo#ar.

k#ar

That(s right. ) s/rt*2 1 %2 ) s/rt6002 1 3!2 ) 66 k#ars


-elow is the resulting power triangle for the machine.

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What is the mechanical output power of the machine& 'ound your answer to the nearest

horse power.

5%

That(s right. %mech ) %elec + efficiency ) 3! + .! ) 32. kW

%in horse power ) %in watts7!6 ) 32007!6 ) 0. 8 4 5%


*ketch the circuit diagram for the factory.

$o you show the lighting load A in series or - in parallel with the molding

machine&

The lighting load must 9e in parallel with the machine. This is necessary to maintain

220V across all loads.

A factory uses 220V commercial power. The principal users of power are a lighting load of 200 ,W andan injection molding machine rated at 600 ,VA. The molding machine has a power factor of .63 lagging

and is !" efficient in con#erting electrical energy to mechanical energy.

The lighting load consumes k#ars. 5ow much is &

k#ars

That(s right. :or practical purposes; the lighting load consumes no reacti#e power.


The power triangle for the molding machine is shown 9elow.

The circuit 9elow represents the factory.

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$raw the power triangle for the factory.

What is the #alue of for this power triangle&

k#ars

That(s right. *ince the lighting load adds nothing to the reacti#e power of the factory; thereacti#e power of the factory is the same as the reacti#e power of the molding machine.


The power triangle for the molding machine is shown 9elow.


The power triangle for the factory is shown 9elow.

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What is the #alue of % for this power triangle&

kW

That(s right. -y conser#ation of energy; the total power used 9y the factory must 9e the

sum of the powers used 9y the indi#idual loads.

%total ) %machine < %lights ) 3! < 200 ) =! kW




* was found 9y taking the s/uare root of the sum of %2 and 2.

What is the power factor for the factory& 'ound your answer to 2 decimal places.

That(s right. %ower factor is the ratio of real power to apparent power.

pf ) %7* ) =!7!2 ) .!!0 8 .!


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What is the current in the power lines feeding the factory& 'ound your answer to the

nearest amp.

A

That(s right. Apparent power is just the product of the #oltage and current; taking

magnitude only; and ignoring phase angle.

> ) *7? ) !20007220 ) 33!2.! 8 33!3 A

A factory uses 220V commercial power. The principal users of power are a lighting load of 200 ,W and

an injection molding machine rated at 600 ,VA. The molding machine has a power factor of .63 laggingand is !" efficient in con#erting electrical energy to mechanical energy.



=


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A capacitor is to 9e connected to help correct the power factor. This capacitor should 9e

connected@

A. >n series with the molding machine.

-. >n series with the lighting load.

. >n series with the parallel com9ination of the lights and the molding machine.

$. >n parallel with the lights and the machine.

That(s right. The capacitor must 9e installed in parallel with all loads.

A factory uses 220V commercial power. The principal users of power are a lighting load of 200 ,W and

an injection molding machine rated at 600 ,VA. The molding machine has a power factor of .63 laggingand is !" efficient in con#erting electrical energy to mechanical energy.

The circuit 9elow represents the factory; with the capacitor installed.

The power triangle for the factory; 9efore the installation of the capacitor; is shown

9elow.

An industrial catalog lists the following capacitors a#aila9le.

A. 400 k#ars B 420V.

-. 300 k#ars B 220V.

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. 00 k#ars B 420V.

$. !00 k#ars B 220V.

>n order to get the 9est power factor correction; which one should you choose&

That(s right. Cou must pick a capacitor rated at at least 220 V. :or 9est power factor

correction; you want one rated at close to 66 k#ars. The closest one a#aila9le was the

220V; 300 k#ar model.

A factory uses 220V commercial power. The principal users of power are a lighting load of 200 ,W andan injection molding machine rated at 600 ,VA. The molding machine has a power factor of .63 lagging

and is !" efficient in con#erting electrical energy to mechanical energy.

The power triangle for the factory; 9efore the installation of the capacitor; is shown 9elow.


$raw the power triangle for all the loads of the factory.

What is the new &

k#ars

That(s right. All capacitors ha#e negati#e reactance; and thus they use negati#e reacti#e

power. >ndustrial catalogs may not use the minus sign; so you will ha#e to remem9er to

apply it.

total ) machine < lights < capacitor ) 66 < 0 < 1300 ) 466 k#ars


!


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The power triangle for the factory; after the installation of the capacitor; is shown 9elow.

The new power factor is .6.

What is the current in the power lines feeding the factory& 'ound your answer to the

nearest amp.

A

That(s right.

> ) *7? ) 6040007220 ) 2!34. 8 2!32 A

Adding the capacitor caused the line current to 9e reduced from 33!3A to 2!32A. The

real power consumed remained unchanged.

Dany industrial users of commercial power correct their power factors to near unity;

9ecause power companies charge a fee for poor power factors


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In this tutorial, you will transform a network, calculate a

phasor current, and then do the reverse transform to find the

time domain current. Consider the circuit below.

The ultimate task is to find it. To 9egin; we want to transform the circuit into the

phasor domain. -ut there(s a pro9lem. Ene of the sources contains a sine; not a cosine.

>t will therefore 9e necessary to con#ert the sine to a cosine.

420sin200t 1 F ) 420cos200t 1 θ

What is θ&

F

The circuit 9elow now shows 9oth sources as cosines.

Gow let(s con#ert the #oltage source to its phasor #alue. >t looks like this@ D7θ.

What is D&

420cos200t 1 F V con#erts to 42071F V.cos200t 1 60F A con#erts to 7160F A.


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Gow let(s con#ert the inductor to the phasor domain. >t looks like this@ jH.

What is H&

Ω

That(s right. IJ ) jωJ ) j200.! ) j40 Ω

420cos200t 1 F V con#erts to 42071F V.

cos200t 1 60F A con#erts to 7160F A.

>t is not necessary to con#ert the resistor #alue.

Gow let(s con#ert the capacitor to the phasor domain. >t looks like this@ jH.

What is H&

Ω

That(s right. I ) 47jω ) 1j7ω ) 1j7200 + 0 + 4016 ) 1j42= Ω

420cos200t 1 F V con#erts to 42071F V.cos200t 1 60F A con#erts to 7160F A.

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!00m5 con#erts to j40 Ω.

>t is not necessary to con#ert the resistor #alue.

it con#erts to the phasor I.

The transformed circuit 9elow reflects these #alues.

There are se#eral ways to sol#e for I. Ene of the 9est ways for this particular circuit is

the mesh current method. Gotice that mesh currents ha#e 9een entered on the a9o#e

circuit. >t is only necessary to write one mesh current e/uation; since the current in the

mesh on the right is already known. This e/uation for the mesh on the left is@

42071F ) j40I < HI 1 j42=KI 1 17160FL

What is H&

The mesh e/uation is then@

42071F ) j40I < 2=I 1 j42=KI 1 17160FL

After grouping terms; the mesh e/uation 9ecomes@

42071F ) j40 < 2= 1 j42=I 1 j=007160F

The j=007160F term is a 9it incon#enient; as it mi+es rectangular coordinates with polar

coordinates. Ge#ertheless; the term can easily 9e con#erted to straight polar form. Thiswill 9e@

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=007θ

What is θ&

F

That(s right. j=007160F ) =007160F < 0F ) =00730F

Eur mesh e/uation is now@

42071F ) j40 < 2= 1 j42=I 1 =00730F

After simplifying; the mesh e/uation 9ecomes@

I ) 42071F < =00730F72= < j4=

I can 9e resol#ed into polar form; D7θ.

What is Μ& 'ound your answer to the nearest integer.

That(s right. I ) 42071F < =00730F72= < j4= ) 46.3 < j434.272= < j4=

I ) 36.=74!.F72.4=730.6F ) 4.!7143.!F 8 4=7143.=F amps

This is the phasor current I.

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The phasor I must now 9e con#erted 9ack into the time domain.

it ) 4=cosωt 1 43.=F A

What is ω&

radians7second

it ) 4=cos200t 1 43.=F A. Therefore the pro9lem is sol#ed.

This tutorial will give you practice in calculating values

associated with sinusoidal quantities. ou will need your

calculator to work these problems.

onsider the time1#arying #oltage 9elow.

#t ) 220cos=0t 1 6!F #olts

What is #0& 'ound your answer to the nearest #olt.

#olts

That(s right. At t ) 0;

#t ) 220cos=0t 1 6!F ) 220cos1 6!F ) 6 #olts

What is #0.4 second& 'ound your answer to the nearest #olt.

#olts

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That(s right. At t ) 0.4;

#t ) 220cos=0t 1 6!F ) 220cos0.= radians 1 6!F ) 220cos26.=F 1 6!F ) 14!0 #olts

What is the ma+imum #alue of #t&

#olts

That(s right. The cosine function has a ma+imum #alue of one. Therefore the ma+imum

#alue of #t ) 220cos=0t 1 6!F is 220 #olts.

At what time t does #t reach its first ma+imum& Mi#e your answer in milliseconds; and

round to the nearest millisecond.

ms

That(s right. To find the time when the #t reaches its first ma+imum; we must sol#e

this e/uation@

#t ) 220cos=0t 1 6!F ) 220

cos=0t 1 6!F ) 4

=0t 1 6!F ) 0; where =0t is in radians.

on#erting to radians@

=0t 1 6!π740 ) =0t 1 4.46 ) 0

t ) 4.467=0 ) .023 s ) 23 ms

What is the minimum #alue of #t&

#olts

That(s right. The minimum #alue of #t will occur when the cosine is minimum.

#t ) 220cos=0t 1 6!F ) 22014 ) 1220 #olts

What is the peak1to1peak #alue of #t&

#olts

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That(s right. The peak1to1peak #alue is just twice the ma+imum #alue.

# peak1to1peak ) 2+220 ) 0 #olts

Jet(s ha#e a look at the /uantities we(#e calculated on a plot.

Gotice that #t starts at 6 V; rises to its first ma+imum at 23 ms; then 9ecomes 14!6 V

when t ) 0.4 seconds. The ma+ #alue of the wa#e is 220 V; the min #alue is 1220 V; and

the peak1to1peak #alue is 0 V.

#t ) 220cos=0t 1 6!F #olts

What is the a#erage #alue of #t&

#olts

That(s right. The a#erage #alue is Nero 9ecause the wa#e is centered around 0.

4=


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#t ) 220cos=0t 1 6!F #olts

What is the rms #alue of #t& 'ound your answer to the nearest #olt.

#olts

That(s right. Vrms ) Vm72472 ) 22074.4 ) 4=6 #olts

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#t ) 220cos=0t 1 6!F #olts

As a phasor; this #oltage would 9e ! ) 220 7θ. What is θ&

degrees

That(s right. :or #t ) 220cos=0t 1 6!F #olts; the magnitude of the phasor is 220 and

the phase angle is 16!F.

! ) 220 716!F #olts

#t ) 220cos=0t 1 6!F #olts

What is the angular fre/uency ω of the #oltage in radians7second&

radians7second

That(s right. :or #t ) 220cos=0t 1 6!F #olts; the =0 is ω; the angular fre/uency in

radians per second.

4!


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#t ) 220cos=0t 1 6!F #olts

What is the fre/uency f of the #oltage in hertN& 'ound your answer to an integer.

5N

That(s right. :or #t ) 220cos=0t 1 6!F #olts;

ω ) 2πf

f ) ω72π ) =072π ) !.6 8 5N

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#t ) 220cos=0t 1 6!F #olts

What is the period T of the #oltage wa#e in milliseconds& 'ound your answer to the

nearest millisecond.

ms

That(s right. We calculated f to 9e !.6 5N. T; the period; is calculated as follows.

T ) 47f ) 47!.6 ) 0.426 s ) 426 ms.

T is now shown in the figure 9elow.

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#t ) 220cos=0t 1 6!F #olts

>f this #oltage were connected to a 4000 ohm resistor; what would 9e the a#erage power

consumption of the resistor in watts& 'ound your answer to the nearest watt.

W

That(s right.

%a#g ) Vrms27' ) 4=6274000 ) 2.2 W 8 2 W

>t should now 9e clear why rms is called Oeffecti#e #alueO. The 4=6 V rms #alue of #t

has the same effecti#e #alue as a 4=6 V 9attery so far as powering the resistor is

concerned.

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ou will need a calculator for this tutorial "#$

#t ) 300 cos00t < 6=F V

:ind the root1mean1s/uare rms #alue of #t. 'ound your answer to the nearest #olt.

Vrms ) V

That(s right Vrms ) Vma+72472 ) 30074.4 ) 242.43 V. This rounds off to 242 V.

onsider the s/uare wa#e 9elow.

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:ind the root1mean1s/uare rms #alue of #t. 'ound your answer to the nearest #olt.

Vrms ) V

That(s right Vrms is sqrtmean #2.

Vrms ) PK=02 < 1=02L72Q472 ) PK2=00 < 2=00L72Q472 ) P2=00Q472 ) =0 V

Gow let(s find the rms #alue of a more comple+ #oltage.

To find the rms #oltage; we must analyNe this wa#eform in se#eral stages. Jet us start

with the formal definition of rms@

>n this e/uation; t0 is some ar9itrary time; and time t0 < T is the time one period later.

Although the e/uation is written for a #oltage v; it works for current i also. The wa#e

must 9e periodic repetiti#e; or it is not possi9le to find an rms #alue.

What is the period of this #oltage&

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T ) s

That(s right; the wa#e repeats e#ery 0 seconds; so the period T is 0 s.

The a9o#e e/uation can now 9e replaced 9y@

Although any #alue of t0 will work; what is pro9a9ly the most appropriate #alue&

t0 ) s

That(s right; the simplest thing is to just let t 0 ) 0. The integral will 9e done from t ) 0 s

to t ) 0 s.

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*ince t0 ) 0; this e/uation can now 9e replaced with@

#t is not a continuous function in the range t ) 0 s to t ) 0 s. Therefore; the function

and the integral must 9e 9roken up into parts for which #t is continuous. 5ow many

parts should it 9e 9roken into&

parts )

That(s right; there are 3 parts to #t; each part corresponding to a straight line on the

graph.

We can now e+press Vrms as follows@

What is the #alue of t4 &

t4 ) s

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That(s right; t4 ) 40 s; 9ecause the first flat part of the cur#e; the part where #t ) =0 V;

stretches from t ) 0 s to t ) 40 s. We also note that t2 ) 20 s; 9ecause the part 9etween t4

and t2 corresponds to the ne+t flat part of #t.


What is the #alue of #2 in the range t ) 0 to t ) 40&

#2 ) V2

That(s right; #t ) =0 V in the range from t ) 0 to t ) 40 s; so we just s/uare this #alue

to get #2; which is 2=00 V2.

2=


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What is the #alue of #2 in the range t ) 40 to t ) 20&

#2 ) V2

That(s right; #t ) 0 V in the range from t ) 40 to t ) 20 s; so we just s/uare this #alue

to get #2; which is 0.


:inding #2 in the range t ) 20 to t ) 0 is a 9it more tricky. We must first come up with

an e+pression for #t in this range. *ince the #oltage is a straight line in this region; it

o9eys the e/uation y ) m+ < 9; which; using our #aria9les; is # ) mt < 9.

What is the #alue of m in this e/uation&

m ) V7s

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That(s right; in the range from t ) 20 to t ) 0 s; the slope is

m ) K=0 1 1=0L70 1 20 ) 400720 ) = V7s

>n the range t ) 20 to 0 we now ha#e;

# ) mt < 9 ) =t < 9

What is the #alue of 9 in this e/uation&

9 ) V

That(s right; in the range from t ) 20 to t ) 0 s; the # intercept is found as follows.

# ) mt < 9 ) =t < 9

0 ) =30 < 9

9 ) 14=0

Therefore; the e/uation for #t in the range from 20 to 0 is@

# ) =t 14=0

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This e/uation can now 9e rewritten as@

Gow let(s sol#e the three integrals.

What is the #alue of the first integral 9etween 0 and 40&

:irst integral )

That(s rightR the integral 0f 2=00 dt is 2=00t. ?#aluated 9etween 0 and 40; this 9ecomes

:irst integral ) 2=0040 1 0 ) 2=000

We also note that the second integral e#aluates to Nero.

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This e/uation can now 9e rewritten as@

The third integral is the most difficult to sol#e. Gote that

=t 1 4=02 ) 2=t2 1 4=00t < 22=00

This makes it possi9le to di#ide the third integral into three parts@

What is the #alue of the integral of 22=00 dt&

>ntegral )

That(s rightR the integral 0f 22=00 dt is 22=00t. ?#aluated 9etween 20 and 0; this

9ecomes

>ntegral ) 22=000 1 20 ) =0000

What is the #alue of the integral of 4=00t dt&

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>ntegral )

That(s rightR the integral 0f 4=00t dt is 4=00t272. ?#aluated 9etween 20 and 0; this

9ecomes

>ntegral ) 4=007202 1 202 ) 00000

What is the #alue of the integral of 2=t2 dt& 'ound your answer to the nearest integer.

>ntegral )

That(s rightR the integral 0f 2=t2 dt is 2=t373. ?#aluated 9etween 20 and 0; this 9ecomes

>ntegral ) 2=7303 1 203 ) 66666.6!. This rounds off to 6666!.

'ecall that we had 9roken up the integral 9etween 20 and 0 into 3 integrals@

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Gow we can su9stitute the #alues we found@

Gow recall that

We can now su9stitute in the 4666! for the final integral@

What is Vrms& 'ound your answer to the nearest #olt.

Vrms ) V

That(s right.

Vrms ) 32.2! V

This rounds off to Vrms ) 32 V

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ou will need a calculator for this tutorial, preferably one with the

capability to manipulate comple% numbers. It is possible to use a

simpler calculator, but the calculations will be more difficult.

-rowser note@ Cour We9 9rowser should 9e capa9le of displaying sym9ols. 5ere is the

Mreek letter theta@ θ.€ This looks like a q instead of a theta on some primiti#e 9rowsers

without the Symbol character set loaded. >f you don(t see the theta as a Mreek letter;

consider updating your 9rowser; or else plan to mentally translate q into theta in the

following pages.

>n electrical engineering notation; j is the s/uare root of what&

j ) s/rt

That(s right.

j ) s/rt14; which is an imaginary num9er.

What is j + j&

j + j )

That(s right.

j + j ) j2 ) 14

What is 47j&

47j )

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That(s right.

47j ) 47jj7j ) j7j2 ) j714 ) 1j

omple+ num9ers can 9e e+pressed in three forms@ e+ponential form; polar form; and

rectangular form; as shown 9elow@

# ) De jθ ) D7θ ) Dcosθ < jDsinθ

When the comple+ num9er is gi#en in e+ponential form; the angle is usually gi#en in

radians. >f it(s in polar form; the angle is usually gi#en in degrees.

:or the comple+ num9er =e j4.=; the polar form is =7θ . What is θ& 'ound your answer to

the nearest degree.

θ ) F

That(s right.

4.= radians ) 4.= + 407π ) =.F; which is appro+imately 6F.

This means that =e j4.= ) =76F.

Gow let(s practice con#erting num9ers from polar to rectangular form.

=00740F ) A < j-

:ind A. 'ound your answer to the nearest integer.

A )

That(s right.

=00740F ) A < j-

A ) =00 cos0F ) 33.02.

This rounds off to 33.

Gow we ha#e@

=00740F ) 33 < j-

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:ind -. 'ound your answer to the nearest integer.

- )

That(s right.

=00740F ) 33 < j-

- ) =00 sin0F ) 324.3

This rounds off to 324.

Gow we ha#e@

=00740F ) 33 < j324

Jet(s try another con#ersion.

1007445F ) A < j-

:ind A. 'ound your answer to the nearest integer.

A )

That(s right.

1007445F ) A < j-

A ) 100 cos4=F ) 6==.32

This rounds off to 6==.

Gow we ha#e@

1007445F ) 6== < j-

:ind -. 'ound your answer to the nearest integer.

- )

That(s right.

10074=F ) 6== < j-

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- ) 100 sin4=F ) 1=.6

This rounds off to 1=.

Gow we ha#e@

1007445F ) 6== 1 j=

Ge+t; we(ll practice con#erting from rectangular form to polar form.

=! 1 j63 ) D7θ. :ind D. 'ound your answer to the nearest integer.

D )

That(s right.

=! 1 j63 ) D7θ

En a good calculator; you can enter =!;163; then do a rectangular to polar con#ersion

to get D ) .6; which rounds off to =. >f your calculator doesn(t ha#e comple+

num9er features; you can still calculate the magnitude D as follows@

D2 ) A2 < -2 ) =!2 < 1632 ) !24

D ) s/rt!24 ) .6; which rounds off to =.

Gow we ha#e@

=! 1 j63 ) =7θ. :ind θ. 'ound your answer to the nearest degree.

θ ) F

That(s right.

=! 1 j63 ) =7θ

En a good calculator; you can enter =!;163; then do a rectangular to polar con#ersion

to get θ ) 1!.6F; which rounds off to 1F. >f your calculator doesn(t ha#e comple+

num9er features; you can still calculate the angle θ as follows@

θ ) arctan-7A ) arctan1637=! ) 1!.6F; which rounds off to 1F.

Gow we ha#e@

3=


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=! 1 j63 ) =7−48F

Jet(s try another rectangular to polar con#ersion@

140 < j6 ) D7θ

:ind D. 'ound your answer to the nearest integer.

D )

That(s right.

140 < j6 ) D7θ

En a good calculator; you can enter 140;6; then do a rectangular to polar con#ersion

to get D ) 43.06; which rounds off to 43. >f your calculator doesn(t ha#e comple+

num9er features; you can still calculate the magnitude D as follows@

D2 ) A2 < -2 ) 1402 < 62 ) 4060.

s/rt4060 ) 43.06; which rounds off to 43.

Gow we ha#e@

140 < j6 ) 437θ

:ind θ. 'ound your answer to the nearest degree.

θ ) F

That(s right.

140 < j6 ) 437θ

En a good calculator; you can enter 140;6; then do a rectangular to polar con#ersion

to get θ ) 44.!F; which rounds off to 44F. >f your calculator doesn(t ha#e comple+

num9er features; you can still calculate the angle θ as follows@

θ ) arctan-7A ) arctan67140

Cour calculator will gi#e 13.=3F for this arctan; 9ut this angle will 9e wrong.

alculators always calculate arctangents in the right half plane. These #alues will 9e off

9y 40F if the real part of the comple+ num9er is negati#e; as it is in this case. To

correct the #alue; add 40F@

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θ ) 40F 1 3.=3F ) 44.!F; which rounds off to 44F.

Gow we ha#e@

140 < j6 ) 437141F

Gow we(ll try some math with comple+ num9ers.

A < j- ) =0 < j!= < 2 1 j. :ind A.

A )

That(s right.

A < j- ) =0 < j!= < 2 1 j ) 2 1 j4

Cou just add up the real parts; then add up the imaginary parts.

D7θ ) =741F!7−56F. :ind D.

D )

That(s right.

D7θ ) =741F!7−56F ) = + !7θ ) 3=7θ

When multiplying comple+ num9ers in polar form; you can multiply the indi#idual

magnitudes to find the net magnitude.

D7θ ) =741F!7−56F. :ind θ.

θ ) F

That(s right.

D7θ ) =741F!7−56F ) = + !74F1=6F ) 3=714=F

When multiplying comple+ num9ers in polar form; you can add the indi#idual angles to

find the net angle.

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D7θ ) 4207135F7762F. :ind θ.

θ ) F

That(s right.

D7θ ) 4207135F7762F ) 4207743=F162F ) 4=773F

When di#iding comple+ num9ers in polar form; you can di#ide the indi#idual

magnitudes to find the net magnitude and su9tract the indi#idual angles to find the net

angle.

D7θ ) =0763F < 2= 1 j30. :ind θ. 'ound your answer to the nearest degree.

θ ) F

That(s right.

D7θ ) =0763F < 2= 1 j30 ) 22.! < j.= < 2= 1 j30 ) !.! < j4.= ) .716.9F

This rounds off to =0717F.

A < j- ) 407−12F7= 1 j!. :ind -. 'ound your answer to the nearest integer.

- )

That(s right.

A < j- ) 407−12F7= 1 j! ) 407−12F7.67−54.5F ) 20.742.5F ) 4=. < j4.4

This rounds off to 4= < j4.

D7θ ) 1= < j407−152F 1 4=0 1 j40072712F

:ind D. 'ound your answer to the nearest integer.

D )

That(s right.

D7θ ) 1= < j407−152F 1 4=0 1 j40072712F

3


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D7θ ) 40.37119.1F407−152F 1 40.37133.7F72712F

D7θ ) 4037−32.9F 1 0.27145.7F

D7θ ) 6. 1 j=6.0 1 63.0 1 j6.=

D7θ ) 23. < j.= ) 2.720.0F

This rounds off to 2=720F.

111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

In this tutorial, you will study a typical three-phase circuit, and

calculate voltages, currents, and power. Consider the problem below.

A three1phase C1connected generator with a phase #oltage of =00 V and a fre/uency of60 5N is connected through power lines to a 9alanced delta1connected load that

consumes 00 kW of power at 0.!! pf lagging. alculate the power consumed in one

phase of the load; the phase #oltage of the load; and the line current.

To approach this pro9lem; it is wise to a draw a diagram; as shown 9elow.

The OCO of the C1connected generator is drawn upside down for con#enience. >t was

specified that the phase #oltage of the generator is =00 V. This is shown on the diagram.

'ecall that this is the rms #oltage. >t is shown as a magnitude only. The phase angle can

9e ignored in 9alanced three1phase pro9lems.

>t is gi#en that the load consumes 00 kW of power at a pf of 0.!! lagging.

What is the power consumed in one phase of the load& 'ound your answer to the

nearest kilowatt.

kW

3


40/44

That(s right.

%φ ) %T73

This is 9ecause one third of the power is consumed in each phase.

The ne+t thing the pro9lem calls for is to calculate the phase #oltage of the load. -efore

we can do this; we must first calculate the line #oltage.

What is the line #oltage& Mi#e your answer to the nearest #olt.

V

That(s right.

for a C1connected source or load; so VJ ) 66 V. This #oltage is now shown on the

diagram 9elow.

0


41/44

Gow we(re ready to calculate the phase #oltage of the load.

What is the phase #oltage of the load& Mi#e your answer to the nearest #olt.

V

That(s right. Vφ ) VJ for a delta1connected source or load; so Vφ ) 66 V. This is not

shown e+plicitly on the diagram; 9ecause it should 9e clear that Vφ and VJ are the same.

The ne+t thing that is asked for is the phase current of the load. -efore you can calculate

that; you(ll need to calculate the apparent power; *; in one phase of the load. 'ecall thatthe power per phase was calculated to 9e 433 kW; and the pf was gi#en as 0.!!.

What is the apparent power consumed in one phase of the load& Mi#e your answer to the

nearest ,VA.

,VA

That(s right.

* ) %7pf

* ) 43370.!! ) 4!3 ,VA

4


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Gow we(re ready to calculate the phase current.

What is the current in one phase of the load& Mi#e your answer to the nearest amp.

A

That(s right.

> ) *7V ) 4!3000766 ) 200 A

Jet(s re#iew what we know@

%hase #oltage of source ) =00 V

Jine #oltage ) 66 V

%hase #oltage of load ) 66 V

%ower consumed 9y load ) 00 kW

%ower consumed in one phase of the load ) 433 kW

%ower factor ) 0.!!

Apparent power in one phase of the load ) 4!3 ,VA

%hase current of the load ) 200 A

The diagram has 9een updated to show some of these #alues.

2


43/44


44/44

Jet us now check our work. Ssing the #alues of VJ and >J a9o#e; calculate the total power consumed 9y the load. Mi#e your answer to the nearest watt.

W

That(s right.

for all types of three1phase loads; so %T total power ) 364 W. There is a small error

from the e+pected 00000 W 9ecause of round1off in the calculations for VJ and >J.

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