Unit 11
Learning Outcomes
QVIRt To To be able to explain what current, current, charge, charge, voltage/potential i!erence an resistance are To To "now the e#uations that lin" these these To To "now the correct correct units to be use in each Omega
Defnitions Current, I $lectrical current is the rate o% &ow o% charge in a circuit' circuit ' $lectrons are charge particles that move aroun the circuit' (o we can thin" o% the electrical current is the rate o% the &ow o% electrons' Current is measured in Amperes (or Amps), A $#uation o% current Q nALq I= t = =nAvq t
•
)here n*number o% charges per unit volume #/e*the charge o% each particle carr+ing the current v* ri%t velocit+ o% the charges * cross-sectional area o% a wire
Charge, Q The amount o% electrical charge is a ph+sical #uantit+, similar to mass an length an time' .rom .rom the ata sheet we can see that the charge on one electron is actuall+ -1'0 x -1 10 2' This means that it ta"es '34 x 10 15 electrons to trans%er 12 o% charge' 6e7ne charge8 Q*It One coulomb is e7ne as the charge passing a point in a circuit in a time o% 1s when • the current is 1' Charge is measured in Coulombs, C Voltage/Potential Dierence, V Voltage, or potential i!erence, is the wor" one per unit charge' 9otential i!erence is a measure o% the i!erence in electrical potential energ+ between two points in a circuit' one volt is the potential i!erence between two points when one :oule o% energ+ is trans%erre b+ one coulomb passing %rom one point to the other' Voltage and p.d. are measured in Volts, V Resistance, R The resistance o% a material tells us how eas+ or i;cult it is to ma"e a current current &ow through it' The resistance R o% a wire wire is e7ne as the ratio o% the potential i!erence V across across the wire to the current I in it' Resistance Resistance * potential i!erence/current R*V/I The ohm is the resistance resistance o% a component when a potential i!erence i!erence o% 1 volt rives a current o% 1 ampere through it' •
•
• •
Resistance is measured in Ohms, Ω
Equations There are three three e#uations that we nee to be able able to explain an substitute substitute numbers into'
This sa+s that the current is the rate o% change o% charge per secon an bac"s up or iea o% current as the rate at which electrons
V =
W Q
This sa+s that the voltage/p'' is e#ual to the energ+ per charge' The ‘push’ of the electrons is equal to the energy given to each charge (electron).
! V = IR This sa+s that increasing the p'' increases the current' Increasing the ‘push’ of the electrons makes more ow. It also shows us that %or constant V, i% R increases I gets smaller' Pushing the same strength if there is more !locking force less current will ow.
Unit 11
Learning Outcomes
Ohm>s Laws an I-V ?raphs To be able to s"etch an explain the I-V graphs o% a ioe, 7lament lamp an resistor To be able to escribe the experimental set up an measurements re#uire to obtain these graphs To "now how the resistance o% an L6R an Thermistor varies Omega
"h#$s %a& %ter the last lesson we "new that a voltage s law the current &owing is proportional to the p'' pushing it' V * IR so this means the resistance is constant' On a graph o% current against p'' this appears as a straight line'
'a(ing )easure#ents To 7n how the current through a component varies with the potential i!erence across it we must ta"e reaings' To measure the potential i!erence we use a voltmeter connecte in parallel an to measure the current we use an ammeter connecte in series'
I% we connect the component to a batter+ we woul now have one reaing %or the p'' an one %or the current' Aut what we re#uire is a range o% reaings' One wa+ aroun this woul be to use a range o% batteries to give i!erent p''s' better wa+ is to a a variable resistor to the circuit, this allows us to use one batter+ an get a range o% reaings %or current an p'' To obtain values %or current in the negative irection we can reverse either the batter+ or the component'
I*V +rahs Resistor This shows that when p'' is @ero so is the current' )hen we increase the p'' in one irection the current increases in that irection' I% we appl+ a p'' in the reverse irection a current &ows in the reverse irection' The straight line shows that current is proportional to p'' an it obe+s Ohm>s law' ?raph a has a lower resistance than graph b because %or the same p'' less current &ows through b' -ila#ent %a# t low values the current is proportional to p'' an so, obe+s Ohm>s law' s the potential i!erence an current increase so oes the temperature' This increases the resistance an the graph curves, since resistance changes it no longer obe+s Ohm>s law' Dio.e This shows us that in one irection increasing the p'' increases the current but in the reverse irection the p'' oes not ma"e a current &ow' )e sa+ that it is %orwar biase' (ince resistance changes it oes not obe+ Ohm>s law'
'hree ecial Resistors
("lso seen in #$%& Physics ')
Varia0le Resistor variable resistor is a resistor whose value can be change' 'her#istor The resistance o% a thermistor varie with temperature' t low temperatures the resistance is high, at high temperatures the resistance is low' %ight Deen.ant Resistor %2D2R3 The resistance o% a thermistor varie with light intensit+' In im light the resistance is high an in bright light the resistance is low'
Unit 11
Learning Outcomes
Resistivit+ an Resistance To be able to state what a!ects resistance o% a wire an explain how the+ a!ect it To be able to escribe the experimental set up re#uire to calculate resistivit+ an e7ne it Omega
Resistance The resistance o% a wire is cause b+ %ree electrons colliing with the positive ions that ma"e up the structure o% the metal' The resistance epens upon several %actors8 %ength, l Length increases – resistance increases
'e#erature Temperature increases – resistance increases s temperature increases the ions are given more energ+ an vibrate more, the el ectrons are more li"el+ to collie with the ions' )aterial The structure o% an+ two metals is similar but not the same, some metal ions are closer together, others have bigger ions'
Resistivit4, 5 R
= ρ
l A
The resistance o% a material can be calculate using where is the resistivit+ o% the material' Resistivit+ is a %actor that accounts %or the structure o% the metal an the temperature' $ach metal has its own value o% resistivit+ %or each temperature' .or example, the resistivit+ o% copper is 1'Bx10 -5 Cm an carbon is Dx10 -4 Cm at room temperature' )hen both are heate to 100E2 their resistivities increase' Resistiit! is measured in Ohm metres , Ωm
)easuring Resistivit4 In orer to measure resistivit+ o% a wire we nee to measure the length, cross-sectional area
)e then rearrange the e#uation to
Unit 11
Learning Outcomes
RA l
an substitute values in
(eries an 9arallel 2ircuits To be able to calculate total current in series an parallel circuits To be able to calculate total potential i!erence in series an parallel circuits To be able to calculate total resistance in series an parallel circuits Omega
eries Circuits In a series circuit all the components are in one circuit or loop' I% resistor 1 in the iagram was remove this woul brea" the whole circuit'
The total voltage o% the circuit is e#ual to the sum o% the p''s across each resistor' V TOTAL = V 1 + V 2 + V 3 The total resistance o% the circuit is e#ual to the sum o% the resistance o% each resistor' RTOTAL = R1 + R2 + R3
Parallel Circuits 2omponents in parallel have their own separate circuit or loop' I% resistor 1 in the iagram was remove this woul onl+ brea" that circuit, a current woul still &ow through resistors 3 an D' The total current is e#ual to the sum o% the currents through each resistor' I TOTAL = I 1 + I 2 + I 3 The total potential i!erence is e#ual to the p''s across each resistor'
V TOTAL
=
V 1
=
V 2
=
V 3
The total resistance can be calculate using the e#uation8 1
RTOTAL
1 =
R1
1 +
R2
1 +
R3
6ater li.e Analog4 Imagine instea o% getting a potential i!erence we get a height i!erence b+ reaching the top o% a slie' This series circuit has three connecte slies an the parallel circuit below has three separate slies that reach the bottom'
Voltages/P2D2s In series we can see that the total height loss is e#ual to how much +ou %all on slie 1, slie 3 an slie D ae together' This means that the total p'' lost must be the p'' given b+ the batter+' I% the resistors have e#ual values this rop in potential i!erence will be e#ual' In parallel we see each slie will rop b+ the same height meaning the potential i!erence is e#ual to the total potential i!erence o% the batter+' Currents I% we imagine 100 people on the water slie, in series we can see that 100 people get to the top' ll 100 must go own slie 1 then slie 3 an 7nal slie D, there is no other option' (o the current in a series circuit is the same ever+where' In parallel we see there is a choice in the slie we ta"e' 100 people get to the top o% the slie but some ma+ go own slie 1, some own slie 3 an some own slie D' The total number o% people is e#ual to the number o% people going own each slie ae together, an the total current is e#ual to the currents in each circuit/loop'
Unit 11 Lesson 1B
$nerg+ an 9ower To "now what power is an how to calculate the power o% an electrical circuit
Po&er 9ower is a measure o% how #uic"l+ something can trans%er energ+' 9ower is lin"e to energ+ b+ the e#uation8 Energy Power = time
"o#er is measured in $atts, $ %nerg! is measured in &oules, & Time is measured in seconds, s
7e& Equations I% we loo" at the e#uations %rom the QVIRt lesson we can erive some new e#uations %or energ+ an power' Energ4 E V = E = VQ Q = It Q can be rearrange into an we "now that so combining these e#uations we get a new one to calculate the energ+ in an electric circuit8 E = VQ Q = It E = VIt G---------------------so <1=
Po&er I% we loo" at the top e#uation, to wor" out power we ivie energ+ b+ time8 E VIt = P = VI t t which cancels out to become <3=
V = IR I% we substitute into the last e#uation we get another e#uation %or power8 V = IR P = IV P = I 2 R G---------------------so
)e can also rearrange e#uation %or power8
P = VI
V = IR
I = G----------------------
I = into
V R
an substitute this into
V R
P = so
P = VI
to get our last
V 2 R
Energ4 again Two more e#uations %or energ+ can be erive %rom the e#uation at the top an e#uations D an H $nerg+ * 9ower x time Pt = I 2 Rt
Pt =
V 2 R
$#uation D becomes
t
E = I 2 Rt
E = $#uation H becomes
V 2 R
<4=
t <
Unit 13
Learning Outcomes
$. an Internal Resistance To "now what em% an internal resistance are To "now how to measure internal resistance To be able s"etch an interpret a V-I graph, labelling the graient an +-intercept
Omega
Energ4 in Circuits In circuits there are two %unamental t+pes o% component8 energ+ givers an energ+ takers'
Electro#otive -orce e#83, 9 $nerg+ givers provie an electromotive %orce, the+ %orce electrons aroun the circuit which trans%er energ+' E ε
=
Q
The si@e o% the em% can be calculate using8 This is similar to the e#uation we use to 7n voltage/potential i!erence an means the energ+ given to each unit o% charge' )e can thin" o% this as the energ+ given to each electron' The emf of a supply is the p.. across its terminals the energy is transferre from other forms to electrical %' is measured in &oules per Coulomb, &C * or Volts, V $nerg+ ta"ers have a potential i!erence across them, trans%erring energ+ %rom the circuit to the component' em% * energ+ giver p'' * energ+ ta"er $nerg+ is conserve in a circuit so energ+ in * energ+ out, or8 The total of the emfs = The total of the potential differences arond the whole circit
Internal Resistance, r The chemicals insie a cell o!er a resistance to the &ow o% current, this is the internal resistance on the cell' +nternal Resistance is measured in Ohms, Ω
%in(ing e#8 an. r
I% we loo" at the statement in the box above an appl+ it to the circuit below, we can reach an e#uation that lin"s em% an r ' Total em%s * total potential i!erences * *
The terminal p.. is the p.. across the terminals of the cell when a current is owing * * internal p'
)easuring e#8 an. r )e can measure the em% an internal resistance o% a cell b+ measuring the current an voltage as shown on the right, the variable resistor allows us to get a range o% values' I% we plot the results onto a graph o% voltmeter reaing against ammeter reaing we get a graph that loo"s li"e the one below'
?raphs have the general e#uation o% + * mxJc, where + is the vertical
V = %Ir # !
V = %r I # !
+ *m x Jc (o we can see that the8 !intercept represents the em an gradient represents (–)internal resistance
Unit 13
Learning Outcomes
Mirchho! an 9otential 6iviers To "now Mirchho!>s laws an be able to appl+ them to #uestions To "now what a potential iviers is an be able to calculate the output voltage To be able to explain an application o% a potential ivier Omega
:irchho$s %a&s Mirchho! came up with two
Catain "0vious$ -irst %a& $lectric charge is conserve in all circuits, all the charge that arrives at a point must leave it'
2urrent going in * current going out'
In the iagram we can sa+ that8 I 1 = I 2 # I 3 # I 4
Catain "0vious$ econ. %a& $nerg+ is conserve in all circuits, %or an+ complete circuit the sum o% the em%s is e#ual to the sum o% the potential i!erences' $nerg+ givers * energ+ ta"ers' In the iagram we can sa+ that8 * * p1 J p3 J pD J pH'
Potential Divi.ers potential ivier is use to prouce a esire potential i!erence, it can be thought o% as a potential selector' t+pical potential ivier consists o% two or more resistors that share the em% %rom the batter+/cell' The p''s across R1 an R3 can be calculate using the %ollowing e#uations8
V 1
=
V 0
R1 R1
+
R 2
V 2
=
V 0
R2 R1
+
R2
This actuall+ shows us that the si@e o% the potential i!erence is e#ual to the input potential multiplie b+ what proportion o% R1 is o% the total resistance' I% R1 is 10 C an R3 is 0 C, R1 contributes a tenth o% the total resistance so R1 has a tenth o% the available potential' This can be represente using8 R1 V 1
R 2
=
V 2 The ratio o% the resistances is e#ual to the ratio o% the output voltages'
;ses In this potential ivier the secon resistor is a thermistor' )hen the temperature is low the resistance < R3= is high, this ma"es the output voltage high' )hen the temperature is high the resistance < R3= is low, this ma"es the output voltage low' use o% this woul be a cooling %an that wor"s harer when it is warm'
In the secon potential ivier the secon resistor is a Light 6epenant Resisitor' )hen the light levels are low the resistance < R3= is high, ma"ing the output voltage high' )hen the light levels increase the resistance < R3= ecreases, this ma"es the output voltage ecrease' use o% this coul be a street light sensor that lights up when the surrouning are ar"'