tutorial de electricidad

8/18/2019 Tutorial de Electricidad

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This tutorial reviews topics from physics that relate to electricalnetworks. Specifically, the concepts of voltage, current, charge, power,

and energy will be reviewed. You are advised to have a calculatorhandy.

Consider the circuit below.

How much charge flows through the headlight in 2 hours?

Q = coulombs

Charge flowing through the headlight in 2 hours = 5 amps x 36 seconds!hour x 2hours = 36 coulombs.

How long does it ta"e before one million electrons flow through the headlight? #ound$our answer to the nearest femtosecond %one femtosecond is & '&5 seconds(.

&


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t = femtoseconds

)he time it ta"es for one million electron to flow through the headlight = Q!* = &.6 2 x& '&+ coulombs per electron x & 6 electrons!5 amps = 32. , x & '&5 seconds- which isapproximatel$ 32 femtoseconds.

How much power is consumed b$ the headlight?

= watts

*n /C circuits- power is 0ust 1oltage times current. = * = &2 x 5 = 6

4lectrical engineers express energ$ either in "ilowatt'hours %" 'hr( or 0oules % (-depending on the application.

2


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How much energ$ is deli1ered to the headlight if it is left on for 3 da$s? #ound $ouranswer to the nearest " 'hr.

= " 'hr

or" %energ$( is power times time. = t = 6 watts x 3 da$s x 2, hours!da$ = ,32 watt'hours = ,.32 " 'hr

)his rounds to , " 'hr.

ow let7s do a problem using 0oules as the unit of energ$.

How much energ$ is used b$ the headlight in 5 minutes?

=

or" %energ$( is power times time. = t = 6 watts x 5 minutes x 6 seconds!minute = &8- 0oules

This tutorial will test your understanding of Kirchhoff's Voltage aw !KV ",Kirchhoff's #urrent aw !K# ", and $hm's aw. %ll the circuits in this tutorialcan be solved &in your head&, but writing down a few loop e uations and node

e uations couldn't hurt. There are a lot of &distracters& (( circuit elements that donot contribute to the solution.

3


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9ind the current *. )$pe the 1alue in the box below- then clic" on the Submit Answer button. *f $ou don7t get it right- $ou7ll be gi1en a hint and as"ed to tr$ again.

:

)he formal solution for this problem loo"s li"e this;

Consider the loop shown in red- consisting onl$ of the 25 Ω resistor and the sol1ed around the red loop;' 25


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9ind the current *.

:

)he formal solution for this problem loo"s li"e this;

>et7s sol1e > around the red loop- the one consisting of the 25 Ω resistor- the &25 batter$- and the


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9ind the 1oltage .

*n the figure below- it can be seen that the 5: current from the current source also goesthrough the &8 batter$ and the & Ω resistor.

*n this wa$- C> is sol1ed b$ inspection. ow all $ou need to do is appl$ @hm7s >aw; = *# = 5 x & = 5

ow let7s add another circuit element and sol1e again.

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9ind the 1oltage .

)he traditional wa$ to sol1e this problem is shown below. 9irst- the un"nown currents*&8 and *& are defined- then two nodes are identified.

C> can then be sol1ed at ode :;'5 8 * &8 = *&8 = 5 ' 8 = '3:

ow we sol1e C> at ode A;'8 ' * &8 *& = '8 ' %'3( * & = *& = 8 ' 3 = 5:

)hen we appl$ @hm7s >aw; = *# = 5 x & = 5

)he abo1e is a perfectl$ good wa$ to sol1e the problem- but there is a faster wa$. )hisin1ol1es the use of a supernode .


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:n$thing can be treated as a node- so we choose to lump the 8: source and the &8 batter$ into a single node. *t can be seen that onl$ the current from the 5: source enters

the supernode- and onl$ * & lea1es. C> is then eas$ to write;'5 * & = *& = 5:

Bou will find that using supernodes can greatl$ simplif$ some problems.

*t should be clear that the 1alue of the 8: current source ma"es no difference to thesolution.

ow let7s add $et another circuit element and sol1e again.

9ind the 1oltage .

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)he circuit is shown below- with * & and the supernode defined.

C> can then be sol1ed at the supernode;'5 && *& = *& = 5 ' && = '6:

)hen we appl$ @hm7s >aw; = *# = '6 x & = '6

e could ha1e used the 2 ohm resistor and the &2 batter$ in a supernode instead ofusing the 8: source and the &8 batter$. )he results would ha1e been exactl$ the same.

hat is the resistance between : and A?

ohms

+


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@rdinaril$- two resistors are not in series if there is a node between them- but in this problem- the & Ω and 3 Ω resistors beha1e as though the$ are in series- because nocurrent can flow through the 2 Ω resistor. )he result is a total resistance of , Ω.

ow let7s loo" at another problem.


ohms

)he concept of shorting out components is important. *f a resistor has both endsconnected to the same point- the resistor is shorted out and contributes nothing to thecircuit. Aoth ends of the 2 Ω resistor are connected to point A. )he total resistance is

0ust the & Ω resistance.

Here7s another problem.

&


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@ddl$ enough- the 2 Ω and 3 Ω resistors are in parallel. ote that both resistors areconnected to the node 0oining all 3 resistors on one end- and point A on the other end.)his ma"es the resistance of the parallel combination &2 Ω- and the net resistance &< Ω.

Here7s a new one;


ohms

&2


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)he < Ω resistor is shorted out because it7s connected to the same point on both ends. *tdoesn7t figure in the solution. )he , Ω and 2 Ω resistors are in series- ma"ing 6 Ω. )his6 Ω resistance is then in parallel with the 3 Ω resistor- ma"ing 2 Ω. 9inall$- the 2 Ω resistance is in series with the + Ω resistor- ma"ing && Ω.

)he 1oltage di1ider and current di1ider are useful concepts in anal$Ding circuits. >et7sstart with the 1oltage di1ider- as shown in the circuit below.

Ese the 1oltage di1ider formula to find 2.

2 = 1olts

)hat7s right.

2 = F# 2!%# & # 2( = &2 G& !%2 & ( = ,

ow let7s tr$ another circuit.

&3


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Ese the 1oltage di1ider formula to find 2.

2 = 1olts

)hat7s right.

2 = F# 2!%# & # 2( = 2, G& !%5 & ( = &6

ow let7s mo1e on to the current di1ider.

Ese the current di1ider formula to find * 2.

&,


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*2 = amps

)hat7s right.

*2 = *F# &!%# & # 2( = 3 G,!%, 6( = &2 :

)he current di1ider e uation can be used for more complex circuits.

Ese the current di1ider formula to find * 2.

*2 = amps

)hat7s right.

&5


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*2 = *F# &!%# & # 2( = ,5G, !%, 5 ( = 2 :


How man$ essential nodes are there?

nodes

)here are 3 essential nodes. )he$ are shown below- circled in blue.

How man$ irchhoff7s Current >aw e uations are need to completel$ sol1e the circuit?

&6


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e uations

)here are 3 essential nodes. )his means that 2 C> e uations are needed.

How man$ irchhoff7s oltage >aw e uations are need to completel$ sol1e the circuit?

e uations

2 C> and 3 > e uations are needed. )he meshes for the > e uations are shownon the diagram in red.

&


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How man$ @hm7s >aw e uations are needed?

e uations

2 C>- 3 >- and 5 @hm7s >aw e uations are needed- for a total of & .

How man$ un"nown currents should be s"etched on the circuit?

currents

&8


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2 C>- 3 >- and 5 @hm7s >aw e uations are needed- for a total of & . )he fourre uired currents are shown in the s"etch below. )he directions for the currents arearbitrar$. ote that it is not necessar$ to define a current in the branch on the left-

because the current %2 :( is alread$ "nown.

How man$ un"nown 1oltages should be s"etched on the circuit?

1oltages

2 C>- 3 >- and 5 @hm7s >aw e uations are needed- for a total of & e uations.)here are , re uired currents and 6 re uired 1oltages- for a total of & un"nowns.

otice that the number of e uations e uals the number of un"nowns.

)he re uired 1oltages are shown in the circuit below. )he sign of the 1oltage of eachresistor is determined b$ the current direction. )he sign of the 1oltage for the currentsource is arbitrar$.

&+


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Aelow are the 5 @hm7s >aw e uations needed.3 = 2 x 3, = ,* ,< = e uations;)op ode; '2 ' * 5 ' *< = Aottom ode; 2 * , II =

9ill in the blan" in the abo1e e uation.

ut the missing current here.


Aelow are the 5 @hm7s >aw e uations needed.3 = 2 x 3, = ,* ,< = e uations;)op ode; '2 ' * 5 ' *< = Aottom ode; 2 * , *6 =

)hree > e uations must be written- one for each of the simple loops- as shown below.>eft >oop; ' * 3 5 ' , = )op #ight >oop; '5 ' 8 ' < = Aottom #ight >oop; , 6 ' II =

2


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9ill in the blan" in the abo1e e uation.

ut the missing 1oltage here.


Aelow are the 5 @hm7s >aw e uations needed.3 = 2 x 3, = ,* ,< = e uations;)op ode; '2 ' * 5 ' *< = Aottom ode; 2 * , *6 =

)hree > e uations must be written- one for each of the simple loops- as shown below.>eft >oop; ' * 3 5 ' , = )op #ight >oop; '5 ' 8 ' < = Aottom #ight >oop; , 6 ' + =

Bou now ha1e & e uations and & un"nowns. *t is therefore possible to sol1e for all ofthe un"nowns.


2&


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Fe1eral points on the circuit are identified b$ the letters a through g. Fome of these areessential nodes- and some are nonessential nodes. :lthough an$ node can ser1e as areference node- some are more con1enient than others. )he intelligent engineer willselect the node that allows the fewest- simplest node 1oltage e uations to be written.

hich point % a through g( do $ou thin" is the best choice for the reference node?

node

ode a - the reference node- is now mar"ed with the ground s$mbol. ote that node ahas the most branches %,(- and that it has the most 1oltage sources connected to it %one'' no other node has more than one 1oltage source connected to it(.

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Bou will now ha1e to label most nodes with un"nown 1oltages. *f node c were such anode- $ou would label it as c. 9or one node- $ou don7t ha1e to specif$ an un"nown

because it is possible to specif$ the actual 1oltage.

hich node % b through g( is it?

node

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Fince node g is connected to node a onl$ through a 1oltage source- we can label thisnode with its actual numerical 1oltage- instead of labeling it with g.

hat 1alue should be used?

1olts

ode g is now labeled on the diagram as '6 . ode e is also labeled. Fince we don7t"now what 1oltage it has %with respect to the reference node(- we must assign it anun"nown % e(.

2,


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*n order to calculate the current flow in the branch with the 5 batter$ and the < ohmresistor- it will be helpful to define a 1oltage at the nonessential node f . *nstead oflabeling the node as f- we will label it with respect to the 1oltage at node e.

ode f should therefore be labeled as;

e

ode f - labeled as e 5- is now shown on the circuit.

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odes c and d are special. )hese two essential nodes are directl$ connected b$ a 1oltagesource. )his would be eas$ to deal with if either c or d were the reference node %recallhow we defined node g(.

)he correct wa$ to handle this situation is to define one of the nodes as an un"nown-then define the second node with respect to the first one. >et7s label node c as c.

ode d should therefore be labeled as;

c

)he circuit below is now full$ labeled with node 1oltages.

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e are now read$ to begin writing irchhoff7s Current >aw e uations at appropriatenodes. o e uation will be needed at either of the nonessential nodes % b and f (- nor willone be re uired at the reference node.

hat other node % c- d - e- or g( does not need a C> e uation?

node

ode g needs no C> e uation because its 1oltage is alread$ "nown.

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ow let7s write part of the C> e uation for node e. )he current from e to g is;J e ' %'6(K!# = % e 6(!#

# in the abo1e expression is;

ohms

Fo far- so good. )he current from e to g is % e 6(!+. )his is 0ust @hm7s law for thecurrent through the + Ω resistor.

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)he current from e to d is;JL ' % c ,(K!8 = %L ' c ' ,(!8

L in the abo1e expression is;

@hm7s law sa$s that the current from e to d will be J e ' % c ,(K!8. )his is found b$ ta"ing the difference in the 1oltages on the two sides of the 8 Ω resistor. e nowha1e 2 of the 3 currents we will need at node e.

2+


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ow we will see wh$ it was con1enient to define a 1oltage at node f . otice that thecurrent in the 5 batter$ is the same as the current in the < ohm resistor. )herefore- ifwe can find an expression for the leftward current through the < ohm resistor- that same

expression will appl$ to the leftward current out of node e. )his current is;% e 5 ' L(!<

L in the abo1e expression is;

e can now assemble these 3 currents to write the C> e uation at node e;% e 6(!+ % e ' c ' ,(!8 % e 5(!< =

3


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@ur next tas" will be to deal with nodes c and d . )his can be done in two wa$s; supernode or defined current . *n the first method- nodes c and d are treated as asupernode. *n the second- a current is defined in the batter$ connecting nodes c and d .

ote that a circle is drawn around nodes c and d - so that the$ can be treated assupernode cd . )he sum of the currents out of this supernode is Dero.

3&


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)here are , currents flowing out of this supernode- one for each of the 2 branches atnode c- and one for each of the 2 branches at node d . )he current through the &2 ohmresistor is c!&2.

hat is the current out of node c going through the && ohm resistor?

:

)hat ta"es care of the node c part of the supernode. >et7s mo1e on to node d .

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)he current from node d to node e is% c L ' e(!8

hat is L?

Bou can see that- when dealing with a supernode- all points on the supernode ma$not ha1e the same 1oltage. )he top end of the supernode has a 1oltage c. )he bottomend has a 1oltage c ,.

)he remaining current is the current from node d to node g- which is;J c , ' %'6(K!&3 = % c & (!&3

ow we ha1e all , currents out of the supernode- so we can write C> at thesupernode;

c!&2 ' 2 % c , ' e(!8 % c & (!&3 =

33


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#emember that we also had the C> e uation for node e;% e 6(!+ % e ' c ' ,(!8 % e 5(!< =

otice that this is 2 e uations and 2 un"nowns '' all we need to sol1e the circuit. @ncee and c are calculated- all 1oltages and currents in the circuit can be easil$ found.

ow let7s see how to approach nodes c and d using a defined current instead of asupernode.

e define a current %in an arbitrar$ direction( in the , batter$ as shown below.

3,


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Ha1ing done this- we now sol1e nodes c and d in the usual wa$. C> for node c is;

c!&2 ' 2 L =

hat is L?

9or node c we ha1e;c!&2 ' 2 *, =

ow let7s tr$ node d .

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C> for node d is;

% c , ' e(!8 J c , ' %'6(K!&3 ' L =

hat is L?

@ur 3 C> e uations are now; ode e; % e 6(!+ % e ' c ' ,(!8 % e 5(!< = ode c; c!&2 ' 2 *, = ode d ; % c , ' e(!8 % c & (!&3 ' *, =

36


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)here are 3 un"nowns % c- e- and *,( in the abo1e e uations. )he circuit can therefore be sol1ed %do so- if $ou wish.(


3


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)hree meshes are defined- a - b - and c. *t will be necessar$ to define a current in each ofthese meshes. )wo of them will be un"nowns- to be found b$ calculation. )he third can

be immediatel$ labeled with a numeric 1alue.

hich mesh % a through c( can be labeled numericall$?

mesh

Mesh c can be labeled numericall$- because it contains a current source that is shared

b$ no other mesh.

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hat numerical 1alue should be assigned to the current in mesh c?

amps

Mesh c is now labeled numericall$. )he other 2 meshes are labeled with un"nowncurrents. 9urthermore- the 3: source is now labeled with an un"nown 1oltage. *t is notnecessar$ to label the 2: source with an un"nown 1oltage- because no e uation will bewritten for mesh c.

3+


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e can now begin writing the mesh current e uations for the circuit. *t is traditional to put sources on the left of the e uation- and passi1e de1ices li"e resistors on the right.9or mesh a - the left side of the e uation will be;

L ' 3

hat is L?

)he , source is entered as negati1e in the expression because 1oltage drops areta"en as negati1e when the$ appear on the left side of the e uation.

,


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e now complete writing the mesh current e uation for mesh a ;

', ' 3 = +*a 8%*a ' L(

hat is L?

)he e uation for mesh a is now complete;

', ' 3 = +*a 8J*a ' %'2(K = +*a 8%*a 2( = &


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e now write the mesh current e uation for mesh b ;

3 5 = & L 6J*b ' %'2(K

hat is L?

)he e uations for the meshes are complete.

mesh a ; ', ' 3 = &


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otice that there are 2 un"nown currents and one un"nown 1oltage. )herefore we needan additional e uation.

)his additional e uation will in1ol1e the source

)he current in the 3: source has not $et been used in an$ e uation.

,3


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)he e uation for this current is;

3 = *b ' L

hat is L?

e now ha1e 3 e uations and 3 un"nowns. )he simplified 1ersions of the e uations are

shown below.

mesh a ; '2 ' 3 = &


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ith 3 e uations and 3 un"nowns- it is now possible to sol1e the circuit.

)his simulated laborator$ is intended to teach some aspects of )he1enin7s )heorem for/C circuits. Bou will need a calculator.

:n instructor in a circuits laborator$ uses a selection of resistors- /C 1oltage sources-and /C current sources %all of which ma$ be considered to be ideal( to build a complexcircuit. Fhe attaches a wire to each of two points in the circuit- then seals the rest of thecircuit inside a blac" box- with 0ust the two wires stic"ing out- as shown below.

___________________ | | | |-------o | Black Box | | |-------o |___________________|

Bou arri1e in the lab- and the lab instructor as"s $ou to find out how much current willflow between the two wires when the$ are tied together. )he onl$ things she lets $ouuse are a resistor and a digital 1oltmeter.

Clic" the button below to connect the 1oltmeter to the two wires- then loo" at the resultson the 1oltmeter.

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___________________ | Voltmeter | o-------| |

| | o-------| Volts | |___________________|

Bou examine the resistor- and $ou find that it has a 1alue of 6&6 ohms.

Bou now connect the resistor between the two wires coming out of the blac" box. Clic"the button below to connect the 1oltmeter across the resistor- then loo" at the results onthe 1oltmeter.

Aased on this information- $ou should be able to figure out the solution. hen $outhin" $ou "now the answer- enter it in the box below- and clic" the Fubmit :nswer

button. Ae sure $our answer is in milliamps.

)his is what $ou want to tell the lab instructor; *f * connect together the two wires from

the blac" box- the current will be milliamps

hat 1alue of # > will cause the most power to be consumed in # >?

# > =

)hat7s right. 9or maximum power transfer-# > = # )H = %&8 (%&2 (!%&8 &2 ( =


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ow we want to calculate the power consumed b$ # >. Aefore we do that- we shouldfirst find the )he1enin 1oltage of the circuit connected to # >.

)H = 1olts

)hat7s right. )he )he1enin 1oltage is found as follows;)H = %, (%&8 (!%&8 &2 ( = 2,

)he complete )he1enin e ui1alent circuit- with # > connected to it- is shown below.

hat is the power consumed b$ # >?

= watts

)hat7s right.> = )H 2!%,# >( = 2, 2!%, x


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Fuperposition is a powerful theorem that allows the engineer to brea" up a large-difficult problem into se1eral small- simple problems.


e want to find * . )his could be done b$ using such techni ues as the node 1oltagemethod or the mesh current method. *nstead- we will sol1e it using superposition.

Calculate the contribution of the ,8 source to * .

*,8 = :

)hat7s right. Bou start from the original circuit.

)hen $ou redraw the circuit with the &2 : source and the


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ow it7s clear that the & Ω and &, Ω resistors are in series- and the &5 Ω resistor isshorted out.*,8 = ,8 !%& &, ( = 2 :

>et7s mo1e on to the next source. Calculate the contribution of the &2 : source to * .

*&2 = :

)hat7s right. Bou start from the original circuit.

)hen $ou redraw the circuit with the two 1oltage sources remo1edN here7s what $ou get;

)he &5 Ω resistor is shorted out. * &2 can be found from the current di1ider e uation;*&2 = %&2(%& (!%& &, ( = 5 :

>et7s mo1e on to the final source. Calculate the contribution of the


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)hen $ou redraw the circuit with the ,8 source and the &2 : source remo1edN here7swhat $ou get;

)he mesh current e uation for the mesh on the left can then be written.'


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1%t( = 3 6t2 5e ',t 2sin%8t( 1olts

e will soon calculate i%t(- but for now- let7s 0ust concern oursel1es with the current inthe resistor.

:t t = - what will the current in the resistor be?

#esistor current = amps

)hat7s right.1%t( = 3 6t2 5e ',t 2sin%8t( 1olts- so1% ( = 3 6x 5x& 2x = 8resistor current = 1% (!# = 8!& = .8:

>et7s show all the branch currents on the circuit diagram;

5&


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e now "now that i # % ( = .8:.

*n general-

i# %t( = 1%t(!# = .3 .6t2 .5e ',t Lsin%8t( amps

9ind L.

L =

)hat7s right.i# %t( = 1%t(!# i# %t( = J3 6t2 5e ',t 2sin%8t(K!&i# %t( = .3 .6t2 .5e ',t .2sin%8t( amps

ow let7s tr$ to find i C%t( .

#ecall the general formula for calculating the current in a capacitor;iC%t( = C%d1C!dt(

otice that the 1oltage across the capacitor is the source 1oltage- 1%t(. )his gi1es us thefollowing;iC%t( = 2 x& '6 Jd1%t(!dtK

*t appears that we must calculate the time deri1ati1e of 1%t(;d1%t(!dt = dJ3 6t2 5e ',t 2sin%8t(K!dt

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>et7s do this one term at a time. hat7s the deri1ati1e of the 3?

/eri1ati1e =

)hat7s right. )he deri1ati1e of a constant is Dero.

d1%t(!dt = dJ3 6t2 5e ',t 2sin%8t(K!dt

>et7s do the next term. hat7s the deri1ati1e of 6t 2? 9ormat $our answer so that there areno spaces between characters.

/eri1ati1e =

)hat7s right.d%6t2(!dt = 6x2t = &2t


hat7s the deri1ati1e of 5e ',t ? Fince $ou can7t t$pe superscripts- 0ust t$pe the answer inone long string- with no spaces.

/eri1ati1e =

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)hat7s right.d%5e',t (!dt = %',(5e ',t = '2 e ',t


hat7s the deri1ati1e of 2sin%8t(? ust t$pe the answer in one long string- with nospaces.

/eri1ati1e =

)hat7s right.dJ2sin%8t(K!dt = %8(2cos%8t( = &6cos%8t(

utting this all together- we ha1e;d1%t(!dt = dJ3 6t2 5e ',t 2sin%8t(K!dtd1%t(!dt = &2t ' 2 e ',t &6cos%8t(

#ecall that we too" this deri1ati1e so that we could calculate i C%t(- sinceiC%t( = Cd1!dt

)his gi1es us the following e uation for i C%t(;iC%t( = %2 x&'6 (J&2t ' 2 e ',t &6cos%8t(KiC%t( = J2, t ' , e ',t 32 cos%8t(Kx&'6

iC%t( = 2, t ' , e ',t 32 cos%8t( microamps

hat is the capacitor current at t = ? Oi1e $our answer in microamps.

5,


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iC% ( = µ:

)hat7s right.iC% ( = 2, % ( ' , e',% ( 32 cos% (µ:iC% ( = ' , 32 µ:iC% ( = '8 µ:

e had pre1iousl$ found;i# % ( = .8:

ow let7s consider the current in the inductor. 9irst- let us assume that it is gi1en that theinductor current is + : when t = . Fome boundar$ condition on the inductor current isnecessar$ as a starting point for the integration we are about to do. *n an$ problem $ou

are gi1en- it will alwa$s be possible to figure out boundar$ conditions- if the$ are notexplicitl$ stated.

hat is i%t( at t = %the current in the 1oltage source(? #ound off $our answer to thenearest amp.

i% ( = :

)hat7s right. A$ C>;i% ( = i# % ( iC% ( i>% (i% ( = .8 : ' 8 µ: + :i% ( = +.


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Aelow is the formula for the current flow in an inductor;i>%t( = %&!>(Jintegral to tR 1%τ(dτ] i>% (%H)M> does not ha1e mechanisms for displa$ing the integral s$mbol.(

9or the abo1e circuit- this e uation becomes;

i>%t( = %&!. ,(Jintegral to tR 1%τ(dτ] +i>%t( = 25Jintegral to tR 1%τ(dτ] +where 1%τ( is gi1en b$;1%τ( = 3 6 τ2 5e ', τ 2sin%8τ( 1olts

i>%t( then becomes;i>%t( = 25Jintegral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ] +

>et us sol1e the integral one term at a time.integral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ

hat is the integral of the 3? Fince $ou can7t t$pe the Oree" letter τ- use t instead.

*ntegral =

)hat7s right.integral%3( = 3τ

>et7s find the next term in the integral.integral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ

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hat is the integral of 6 τ2? Ese t in place of τ in $our answer. :lso- $ou can7t t$pe asuperscript- so 0ust t$pe a string.

*ntegral =

)hat7s right.integral%6τ2( = 2 τ3

integral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ

hat is the integral of 5e ', τ? Ese t in place of τ in $our answer. Bou can7t t$pe asuperscript- so 0ust t$pe a string.

*ntegral =

)hat7s right.integral%5e', τ( = '&.25e ', τ

integral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ

hat is the integral of 2sinJ8 τK? Ese t in place of τ in $our answer. Ae sure to use

brac"ets- not parentheses- in $our answer. 9or 1alues less than &- put a Dero before thedecimal point.

5


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*ntegral =

)hat7s right.integral%2sinJ8τK( = ' .25cosJ8 τK

>et7s re1iew the results of the integration;integral%3 6τ2 5e ', τ 2sinJ8 τK(dτ = 3τ + 2τ 3 − 1.25 e ', τ − .25cosJ8 τK

Aut the 1alues of τ must still be e1aluated between the limits of and t. )o get the upper limit- we7ll 0ust substitute t for τ. )he lower limit will be a constant.integral to tR %3 6τ2 5e ', τ 2sinJ8 τK(dτ = 3 t + 2t3 − 1.25 e ',t − .25cosJ8tK where 3t + 2t3 − 1.25 e ',t − .25cosJ8tK is the upper limit of the integration- and is thelower limit.

hat is ?

>ower limit of integration =

)hat7s right. = &.5

)he net result of the integration is;integral%3 6τ2 5e ', τ 2sinJ8 τK(dτ = 3 t + 2t3 − 1.25 e ',t − .25cosJ8tK &.5

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#ecall wh$ we did this integration. e needed it to calculate i >%t(.i>%t( = %&!>(Jintegral to tR 1%τ(dτ] i>% (i>%t( = %&!. ,(Jintegral to tR 1%τ(dτ] +i>%t( = 25Jintegral to tR 1%τ(dτ] +

e can now plug in for the integral we 0ust calculated;i>%t( = 25J3t + 2t3 − 1.25 e ',t − .25cos%8t( &.5] +

i>%t( then becomes;i>%t( = ;i%t( = i# %t( iC%t( i>%t(i%t( = ,6.8


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)n this tutorial, you will use what you have learned about first ordersystems to solve three problems, each one progressively more comple*

than the last, and building on the results of the last. #onsider thecircuit below.

:ssume that the capacitor is initiall$ uncharged. )he switch is closed at t = .

9ind 1% ' (. )his is the 1oltage on the capacitor the instant before the switch is closed.

1% '( =

e now "now that 1% ' ( = .

9ind 1% (. )his is the 1oltage the instant after the switch is closed.

1% ( =

e now "now that 1%'

( = 1% ( = .

6


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9ind 1%infinit$(.

e now "now that 1% '( = 1% ( = - and 1%infinit$( = &2 .

9ind τ- the time constant. Oi1e $our answer in milliseconds.

ms

e now "now that 1% '( = 1% ( = - 1%infinit$( = &2 - and τ = 6 ms.

6&


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e can now appl$ the following formula to get a complete expression for the beha1iorof the 1oltage for t S ;

1%t( = 1f %1 ' 1 f (e%'t!τ(

)he result will be

1%t( = &2 ' IIIIe %'t!.6(

hat is the 1alue that7s supposed to be in the blan"?

e now ha1e the final expression for the capacitor 1oltage as a function of time;

1%t( = &2 ' &2 e%'t!.6(

Bou7ll need $our calculator for this part. 9ind 1%t( for t = 2 ms. #ound $our answer tothe nearest 1olt.

1%2 ms( =

ow let7s modif$ the circuit b$ adding another resistor.

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*t is no longer necessar$ to assume that the capacitor is initiall$ uncharged. *nstead- wesimpl$ sa$ that the switch has been open for a long time- then the switch is closed at t =

.

9ind 1% ' (.

e now "now that 1% ' ( = .

9ind 1% (.

e now "now that 1% ' ( = 1% ( = .

63


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9ind 1%infinit$(.

e now "now that 1% ' ( = 1% ( = - and 1%infinit$( = + .


ms

e now "now that 1% ' ( = 1% ( = - 1%infinit$( = + - and τ = ,5 ms.

6,


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1%t( = 1f %1 ' 1 f (e%'t!τ(

)he result will be

1%t( = + ' + e%'t! ___(


ow let7s modif$ the circuit b$ adding another 1oltage source.

)he switch has been open for a long time- then the switch is closed at t = .

9ind 1% ' (.

65


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e now "now that 1% ' ( = '2 .

9ind 1% (.

e now "now that 1% ' ( = 1% ( = '2 .

9ind 1%infinit$(.

e now "now that 1% ' ( = 1% ( = '2 - and 1%infinit$( = , .

66


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ms

e now "now that 1% ' ( = 1% ( = '2 - 1%infinit$( = , - and τ = ,5 ms.


1%t( = 1f %1 ' 1 f (e%'t!τ(

)he result will be

1%t( = , ' IIe %'t!.45(


6


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)he final result for the capacitor 1oltage is1%t( = , ' 2, e %'t!.45(

:t t= - this e1aluates to '2 . :t infinit$- it e1aluates to , . )he 1oltage is adeca$ing exponential that graduall$ approaches , .

)his will gi1e $ou practice in dealing with #>C circuits. Consider the circuit below.

)he switch has been closed for a long time. )he switch is opened at t = . *t isnecessar$ to find i%t( for t S . 9irst- let7s find initial and final conditions %the boundar$

1alues(.

hat is i% ' (? )his is the current 0ust before the switch is opened.

amps

e now "now that the current before the switch is opened is .

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ow let7s open the switch.

hat is i% (? )his is the current the instant after the switch is opened.

amps

)he initial current i% ( = i = .

*t will also be useful to determine initial conditions for the capacitor.

hat is 1 C% '(?

1olts

6+


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*f 1C% '( = '8 - then 1 C% ( = '8 - because $ou can7t change the 1oltage across a

instantaneousl$

)he initial current i% ( = i = . )he initial capacitor 1oltage is '8 . )he circuit isredrawn below for the situation after the switch is opened.

e will need to find i7% ( %the deri1ati1e of i at t = (. e can do this b$ first finding1>% (.

hat is 1 >% (?

1olts

)he initial current i% ( = i = . )he initial capacitor 1oltage is '8 . )he initialinductor 1oltage is 8 .


71/78

hat is di!dt e1aluated at t = ?

amps!second

)he initial current i% ( = i = . )he initial capacitor 1oltage is '8 . )he initialinductor 1oltage is 8 . i7% ( %di!dt e1aluated at t = ( = 2 amps!second.

hat is i f = i%infinit$(?

amps

)he initial current i% ( = i = . )he initial capacitor 1oltage is '8 . )he initialinductor 1oltage is 8 . i7% ( %di!dt e1aluated at t = ( = 2 amps!second. )he finalcurrent i f = .


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)his concludes the anal$sis of boundar$ conditions.

e now notice that the circuit abo1e is a #>C circuit

*t is a series #>C circuit. otice that the two resistors can be combined into a single,8 Ω resistor.

hat is α ?

radians!second

α is 6 radians!second.


73/78

hat is ω %the resonant fre uenc$(?

radians!second

α is 6 radians!second and ω is 5 radians!second. otice that these areangular fre uencies.

)his circuit is damped

α is 6 radians!second and ω is 5 radians!second. )he circuit is o1erdamped.9or o1erdamped circuits- two damping constants are important;s& = ' α s rt%α 2 ' ω 2(s2 = ' α ' s rt%α 2 ' ω 2(


74/78

hat is s &? #ound off $our answer to a whole number.

seconds '&.

α is 6 radians!second and ω is 5 radians!second. )he two damping constantsare;s& = ' α s rt%α 2 ' ω 2( = '2683!secondss2 = ' α ' s rt%α 2 ' ω 2( = '+3&

)he general form of the e uation for o1erdamped circuits is;

x%t( = xf

*n this particular case- we wish to find i%t(- so-

i%t( = if

*f this formula is e1aluated at t = - the result isi% ( = if L B-

where L is


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)he e1aluation of this e uation for t = gi1es us-: & = ': 2

*n order to get the one additional e uation we need- we must ta"e the deri1ati1e of thegeneral e uation abo1e. )his gi1es-

i7%t( = di!dt =

where L is

#eiterating what we "now about the boundar$ 1alues;)he initial current i% ( = i = . )he initial capacitor 1oltage is '8 . )he initial

inductor 1oltage is 8 . i7% ( %di!dt e1aluated at t = ( = 2 amps!second. )he finalcurrent i f = .

#eiterating what we "now about the series #>C circuit;α is 6 radians!second and ω is 5 radians!second. )he two damping constantsare;s& = '2683!secondss2 = '+3&

*n general-

i%t( = if

)he e1aluation of this e uation for t = gi1es us-: & = ': 2


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*n addition to the abo1e information- we now ha1e an expression for i7%t(;

i7%t( = di!dt =

>et us now e1aluate this e uation at t = . hat is the 1alue of : &? Oi1e the 1alue to 3significant figures.

: &= amps

)he complete e uation for the current in the circuit is now;i%t( = 3. &%e'2683t ' e '+3&


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i%t( reaches a maximum at t = &88 µs.

hat is the pea" current? Oi1e $our answer to 3 significant figures.

imax = amps

)he complete e uation for the current in the circuit is;i%t( = 3. &%e'2683t ' e '+3&

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