Lesson 29 Electric Circuits

8/11/2019 Lesson 29 Electric Circuits

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Physics 30 Lesson 29 Electric Circuits

I. Circuit symbols

The various paths from source to load and back may be extremely complicated andmay contain many different types of electrical devices and connectors. Circuitdiagramsor schematic diagramsare drawn using the following symbols to showexactly how each device is connected to other devices. The components of an electriccircuit are called elements and their symbols are displayed below.

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!" generator

battery#power supply

resistor

variable resistor$rheostat%

lamp $resistor%

closed switch

open switch

fuse $! fuse is a device that meltswhen the current in a circuit goesabove a designed level. Themelting fuse breaks the circuit.%

ground

ammeter voltmeter


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II. Kirchhoff's rules

&hen components are connected together to create circuits they may be connected intwo different ways' in series or in parallel. &hen components are in series# currentflows consecutively through those components. &hen connected in parallel# currentflows concurrently through the components.

!s electrons move through a circuit# they lose energy in the various loads they passthrough# and they often have to part company and go different ways when they reach a

(unction of more than two wires. Two basic )uestions about how electric circuitsoperate are*

&hen electrons have several loads to pass through# what governs the amount ofelectric potential energy they will lose in each load+

&hen electrons have a choice of several possible paths to follow# what governs thenumber of electrons that will take each path+

!n understanding of the operation of simple series and parallel circuits depends on theanswers to these )uestions. ,ustav obert irchhoff $1/2-1//%# a ,erman physicist#proposed two rules that are used to analyse electric circuits. These rules are simplythe logical conse)uences that arise in circuits as a result of conservation laws.

1. The current rule$or point rule% states that the total current reaching any point in acircuit is e)ual to the total current leaving that point. $This is a conse)uence ofthe 3aw of "onservation of "harge.%

2. The oltage rule$or loop rule% states that the total change in potential around anyclosed loop in a circuit is e)ual to 4ero. $This is a conse)uence of the 3aw of"onservation of 5nergy.%

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series circuit parallel circuit


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The total resistance of the components in a circuit can be determined by making use ofirchhoff6s rules and 7hm6s law. "onsider a series circuit containing a battery andthree resistors. This circuit can be considered as a simple circuit containing only oneresistor that is e)uivalent to the total resistance of the three individual resistors.

8sing irchhoffs voltage rule# the magnitude of the potential difference of the battery$:o% e)uals the total potential difference across the three resistors.

:o; :1< :2< :=

!lso# considering the simpler circuit# 7hm6s law states that

:o; > ?series

so > ?series ; :1< :2< :=

!pplying 7hm6s law to each resistor

> ?series ; >11< >22< >==

@ut according to irchhoffs current rule the current flowing through all the resistorsmust be e)ual and that the same current $>o% would flow in the total resistance#

so >o; >1; >2; >=

Therefore the currents drop out of the e)uation

T; 1< 2< =

Aow consider a circuit containing a battery and three resistors in parallel. !gain thiscircuit can be considered as a simple circuit containing one resistor e)uivalent to thetotal resistance of the three individual resistors.

8sing irchhoffs current rule# the current through the circuit is e)ual to the sum of thecurrents on the three resistors.

>o; >1< >2< >=

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1

=

2

:o

T

; ?series

:o

1

2

=

:o

T; ?parallel

:o


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1.

Cou are given three resistors*

! 20

@ =0

" D0

&hat is the total resistance if they are connected*

a% in series+b% in parallel+c% ! and @ are in parallel which are then in series with "

a% T ; 1< 2< = ; 20 < =0 EFE< D0 ; %00

b% 1 ; 1 < 1 < 1 ; 1 < 1 < 1 ; 0.010=

T 1 2 = 20 =0 D0

T ; 1 ; 9.&

0.010=

c% first we calculate the resistance of the parallel resistors $!%

1 ; 1 < 1 ; 1 < 1 ; 0.0/=== ! ; 1G0.0/== ; 12

! 1 2 20 =0

now we can add the parallel part with resistor " in series

T ; 12 < D0 = &2

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I(. Electric circuit analysis

To solve problems concerning different kinds and combinations of circuits# we apply therules for series and parallel circuits given above. The main thing to do is to takeadvantage of those things that are constant for a circuit*

Hor series circuits# the currentis the same throughout.

Hor parallel elements# the voltageis the same for each resistor.

2.

Hor the given circuit# what is the current and potential drop across each resistor+

Hor a series circuit the current in each resistor is the same as the total current $>o%.Therefore# if we find >o# we know >1# >2# and >=.

>o ; :o ; D0 : ; 0.=== ! I% ) I2 ) I3 ) Io ) 0.333 *

T 1D0

:1 ; >11 ; 0.=== ! $0 % ; %3.3 (:2 ; >22 ; 0.=== ! $20 % ; &.&+ (

:= ; >== ; 0.=== ! $90 % ; 30.0 (

!s a check* :1 < :2 < := ; 1=.= : < I.I : < =0.0 : ;D0 : $expected value%

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T ;

1 <

2 <

= ; 0 + 20 + 90 = 1D0

1; 0

=; 90

2; 20

D0 :


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3.

Hor the given circuit# what is the current and potential drop across each resistor+ &hatis the current at the power supply $> o%+

Hor a parallel circuit the potential difference is the same for each resistor.

(% ) (2 ) (3 ) (o ) 30 (

>1 ; :1 ; =0 : ; 3.0 *

1 10

>2 ; :2 ; =0 : ; &.0 *

2 D

>= ; := ; =0 : ; 2.0 *

= 1D

>o ; >1 < >2 < >= ; =.0 ! < I.0 ! < 2.0 ! ; %%.0 *

4.

Hor the given circuit# what is the current and potential drop across each resistor+ &hatis the current at the power supply $> o%+

This circuit contains both a parallel part $1is parallel to 2< =% and a series part $2and =are in series%. >f we think of 2< =as one resistor !# we know that thepotential drop across 1and !is 10 :.

(% ; :! ; %0 ( I% ; :1G 1 ; 10 : G 10 ; %.0 *

Jince 2and =are in series# the current will be the same in both and will be >!.

>! ; :! ; 10 : ; 0.D0 ! Io; >1< >! ; 1.0 ! < 0.D0 ! ; %., *

! 1D < D

(2 ; >22 ; 0.D0 ! $1D % ; +., (

(3 ; >== ; 0.D0 ! $D % ; 2., (

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10 D 1D =0 :

10

1D

D 10 :


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(. -easuring current and $otential difference

>n order to measure the currentin different parts of a circuit# we use an ammeter.ecall that the current is the same for two resistors in series. Jince we want to havethe same current through the resistor and the ammeter# an ammeter is hooked up inserieswith the resistor. >n order that the ammeter does not substantially change theactual current of the circuit# the resistance in an ammeter must be as small as possible .Hor example# in the circuit shown below we temporarily break the circuit and hook up

the ammeter in series at that point.

>n order to measure thepotential differenceor voltage dropin different parts of a circuit#we use a voltmeter. ecall that the voltage drop is the same for two resistors inparallel. Therefore# a voltmeter is hooked up in parallelwith the resistor. >n order that

the voltmeter does not substantially change the actual current of the circuit# theresistance in a voltmeter must be as large as possible . Aote that unlike the ammeterwe do not break the circuit to measure the voltage drop.

5.

Hor the circuit above# draw in an ammeter to measure the current through the 10

resistor. Kraw a voltmeter that would measure the potential across the D resistor.

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!

:

!

10

1D

D 10 :

:

10

1D

D 10 :


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(I. Practice $roblems

1. Three resistors 20 # 10 and D are to be connected in a circuit with a 20 :

power supply. Hor each configuration below*

Kraw the circuit.

"alculate the total resistance.

Hor each circuit# calculate the current and the potential difference across

each resistor.

!. all in series

@. all in parallel

". The 10 and 20 are parallel and in series with the D resistor.

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(II. and/in assignment

1. ! string of / "hristmas tree lights connected in series to a 120 : source draws acurrent of 0.D !. Hind*

a% the total resistance of the string of lights $1I0 )

b% the resistance of each light $20 )

c% the potential difference across each light $1D :%

2.a% &hat is the resistance of a toaster that draws a current of I.0 ! from a 120

: source+ $20 )

b% &hat resistance would have to be added in a series with the same toaster

to reduce its current to .0 !+ $10 )

=. "alculate the total resistance in each of these cases*

a% 10 , =0 # and D0 in a series $90 )

b% I , D # and =0 in a parallel $2.D )

c% 9 in a series with a combination consisting of # and 12 connected in

parallel with each other. $12 )

. Bow many 1I0 resistors must be connected in parallel to draw a current of I.0

! from a 120 : source+ $/%

D. The potential difference across a heating coil is I.0 : when the current through itis =.0 !. &hat resistance must be added in series with the coil to reduce the

current through it to 2.0 !+ $1.0 )

I. ! portable radio is designed to operate at a potential difference of I.0 : and acurrent of 2D0 m!# but the only source available has a potential of 10.0 :. &hat

resistance must be added in series with the radio to make it operate properly+$1I )

. 5xamine these circuits and find the values indicated.

a% Hind* 2# =# and :=.

b% Hind* :0# 1# and total.

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c% Hind* =# >1# >2# ># >D# and >I.

d% Hind* >1# 1# and 2.

e% >0# >1# >2# >=# ># :1# :2# :=# and :.

a% $10 , 12 # ./ :)

b% $=I :# 1/ , I )

c% $ # 1 !# D !# =.= !# 1.I !# 1.1 !)

d% $=.0 !# 2.0 , I.0 )

e% $1.D !# 1.D !# 0.=0 !# 0.90 !# 0.=0 !# .2 :# 1./ :# 1./ :# 1./ :)

/. Hor the combination of resistors shown in the drawings below# determine the

e)uivalent resistance between points ! and @. $100 ,I.I )

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i ii


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9. The current in the /.00 resistor in the drawing is

0.D00 !. Hind the current in the 20.0 resistor and

in the 9.00 resistor. $0.D0 !# 2.10 !%

10. Hind the current across each resistor in the circuit in the diagram. $2.0 !# 1.0 !#1.0 !%

11. Two batteries# each with an internal resistance of 0.01D # are connected asshown in the diagram. >n effect# the 9.0 : battery is being used to charge the /.0: battery . &hat is the current in the circuit+ $==.= !%

12. &hat would be the current in the circuit above if the /.0 : battery was turned

around and reconnected+ $DI !%

1=. >n the circuit below L# C and M are switches. Kescribe which lights are on andwhich lights are off in the following situations.

!. L open# C closed# M closed@. L closed# C open# M closed". L closed# C open# M openK. L closed# C closed# M closed

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Circuit *nalysis *ctiity

eries Circuits

Problem#&hat is the relationship between electric potential and electric current in seriescircuits+

-aterials#low-voltage K" power supply $set at the maximum setting%

two resistors $1=0 # 220 %

K" ammeterK" voltmetervarious connecting wires

Procedure#

1. 8se the value on the block for each resistor in your calculations.

2. Jet up the series circuit shown in the diagram.

=. et and measurethe voltage $:o% from the power supply to .0 :.

. @efore proceeding with other measurements# calculate T# >o# >1# >2# :1and :2.Jhow all calculations.

D. 8sing the appropriate meters# measure the values of >o# >1# >2# :1and :2in thecircuit.

1uestions#1. Bow many different paths are there for an electron to take through a series

circuit+ 5xplain.

2. 8tili4ing an appropriate table of results# comment on how the measured and

calculated values compare.

=. >f the resistors were rearranged in this circuit# how would the total resistance#potential drop and current values change+

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1$1=0 )

2$220 )

:o


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Parallel Circuits

Problem#&hat is the relationship between electric potential and electric current in parallelcircuits+

Procedure#1. 8sing the same resistors and power supply from the series circuit activity# set up

the parallel circuit shown in the diagram.


. @efore proceeding with other measurements# calculate T# >o# >1# >2# :1and :2.

Jhow all calculations.

D. 8sing the appropriate meters# measure the values of >o# >1# >2# :1and :2in thecircuit.

1uestions#1. Bow many different paths are there for an electron to take through a parallel

circuit+ 5xplain.

2. 8tili4ing an appropriate table of results# comment on how the measured andcalculated values compare.


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1

2:

o


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Combination Circuit

Problem#&hat is the relationship between electric potential and electric current in combinationcircuits+

Procedure#1. Jet up the circuit shown in the diagram. $Hor this set up you re)uire another 1=0

resistor.%


. @efore proceeding with other measurements# calculate T# >o# >1# >2# >=# :1# :2# and:=. Jhow all calculations.

D. 8sing the appropriate meters# measure the values of >o# >1# >2# >=# :1# :2# and :=inthe circuit.

1uestions#1. Bow many different paths are there for an electron to take through a combination

circuit+ 5xplain.

2. 8tili4ing an appropriate table of results# comment on how the measured andcalculated values compare.


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1

2

=$1=0 )

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Lesson 29 Electric Circuits

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