The sum of electromotive force = the voltage around a closed system.
\nIn a series circuit Kirchoff's first law tells us that the voltage is shared between the components, if each component has the same resistance then the voltage is shared equally between them.
\nIn a parallel circuit Kirchoff's second law tells us that the current in will be the same out. So a path with a component with more resistance will take less current and there will be more current in the other path.
\nIf one branch has half the resistance of the other branch then it will have double the current and take 2/3 of the total current.
\n\n\nResistors in series is: R = R1 + R2
\nDerived from: V = V1 + V2
\n\nResistors in parallel
\nI⁄V = I⁄V + I⁄V
\nUsing v = IR, I⁄V = 1⁄V
\nso
\n1⁄R = 1⁄R + 1⁄R
" }, { "subject": "AS Level Physics", "revisioncardtitle": "10 4internalresistance", "revisioncard": "From Kirchoff's law, the relationship between the e.m.f the terminal p.d and the lost volts:
\nelectromotive force = terminal p.d + lost volts
\nE = V + lost volts
\n\nelectromotive force = terminal p.d + current × internal resistance
\nE = V + Ir
\n\nelectromotive force = current × resistance + current × internal resistance
\nE = IR + Ir
\nE = I(R + r)
\n\n\nPotential Dividers allow voltages to be varied.
\n\nBy considering the total p.d and the fraction of the total resistance \n provided by R, we can determine the value of Vout
\n\nThe equation is:
\nVout = (R2⁄R1 + R2) × Vin
\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "10 6sensingcircuits", "revisioncard": "Using a pair of fixed resistors in series in a potential divider has the effect of splitting the p.d, but what if you want to vary Vout
\n\nTo vary Vout is to replace one of the fixed resistors with a variable resistor.
\n\nA potentiometer is a variable resistor with three terminals. Adjusting this contact varies the p.d between the terminals giving a variable Vout
\nWhen moved to R2 the Vout increases, when moved R1 the Vout decreases.
" }, { "subject": "AS Level Physics", "revisioncardtitle": "6 1springsandhookslaw", "revisioncard": "A pair of equal and opposite forces are required in order to apply a tensile force
\n\nThe Force and the extension are directly proportional as the material being the k constant
\n\nIt is proportional until it hits the limit of proportionality which will lead to plastic deformation
\n\nF = k × Δx
\n\nDirectly proportional -> elastic deformation
\n\nLimit of proportionality -> plastic deformation
" }, { "subject": "AS Level Physics", "revisioncardtitle": "6 2elasticpotentialenergy", "revisioncard": "A material that is elastically deformed stores the energy and can be recovered/conserved when under the elastic limit
\n\nWhen the material passes the limit of proportionality energy cannot be recovered because energy goes into the atoms moving
\n\n\nΔW = F × Δx
\n\nArea under a Force against extension graph is the Elastic Potential Energy
\n\nE = 1/2Fx
\n\nF = kx
\n\nE = 1/2 × kx × x
\n\nE = 1/2 × kx^2
\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "6 3deformingmaterials", "revisioncard": "When jumping with a bungee cord the Elastic Potential Energy is turned into Kinetic Energy and Gravitational Potential Energy when fully stretched. Then they accelerate back up.
\n\nDifferent Material react differently under tensile forces
\n\nDuctile materials like metal wire obeys hookes law and has a small elastic region(steep gradient) and a large plastic region(less steep gradient)
\n\nBrittle materials obeys hookes law, have small elastic region(steep gradient) but no plastic region because they break when they hit the limit of proportionality
\n\nPolymeric materials don't obey either
\n\nA ductile material that obeys Hooke's law when unloading the distance the material has extended is larger when it has passed the limit of proportionality, and is parallel to it.
\n\nIn rubber the graph produces a hysteresis loop. The area under the loading curve is the useless Thermal energy and the area under the unloading curve, is recovered Elastic Potential Energy
\n\n\n\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "6 4stressstrainandyoungmodulus", "revisioncard": "Stress is the measure of the force over an are with units of (N/m^2).
\nStrain is the ratio between the extension and the original length.
\nYoung's Modulus is the gradient of the two and determines the constant that determines how 'stiff' a material.
\n\nstress = Force(N)⁄Area(m^2)
\nstress = F⁄A
\n\nstrain(no units) = extension(m)⁄original length(m)
\nstrain(no units) = x(m)⁄L(m)
\n\nYoung's Modulus = stress⁄strain
\n\nYoung's Modulus = Force(N)⁄Area(m^2) ÷ extension(m)⁄original length(m)
\n\nYoung's Modulus = F⁄A ÷ x(m)⁄L(m)
\n\nYoung's Modulus = F⁄A × L(m)⁄x(m)
\n\nYoung's Modulus(F/m^2) = FL(Fm)⁄Ax(m^3)
\n\nYoung's Modulus(Pa) = FL(Fm)⁄Ax(m^3)
\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "7 1newtonsfirstandsecondlawsofmotion", "revisioncard": "Newton's first Law: An object will stay in a constant state of inertia(constant velocity/stationary) unless a Resultant force is acted on it
\nIf an objects velocity changes it means a result force is actong on it therefore there is an acceleration or deceleration
\n\nNewton's third Law: When two objects interact they exert equal and opposite forces on each other
\n\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "7 2linearmomentum", "revisioncard": "Momentum is vector quantity meaning it both has magintude and direction
\nmomentum = mass × velocity
\n\np(kgms-1) = m(kg) × v(ms-1)
\n\nIn a closed system, when two objects collide they transfer momentum and kinetic energy between each other, the total momentum is always the same meaning the total momentum is conserved.
\n\nThis means that when the objects collide the momentum before the collision is always the same as the momentum after the collision.
\n\nTo investigate momentum a linear air track would be used to reduce the effect of friction. The initial and final velocity will be determined and the mass of the objects would be determined. Which can allow prediction of the mass or velocity of the second object.
\n\nAn elastic collision is where both objects move as a result of the momentum.
\nAn inelastic collision is where one object is stopped when the two objects collide.
\nperfectly elastic: The total momentum is always conserved, the total energy is always conserved and the total kinetic energy is conserved.
\ninelastic: The total momentum is always conserved, the total energy is always conserved but the total kinetic energy is not conserved.
\n\nAmperes is one of the seven base si units
\n\ncurrent = Δcharge(C)⁄Δtime(s)
\nI = ΔQ⁄Δt
\n\nElectric Charge is Coulombs(C)
\n\nC = AS
\n\nelementary charge = 1.60×10−19 Coulombs
\n\nA proton has a relative charge of +1e and an electron has relative charge of -1e
\n\nExample:
\nThe current in a lamp is 6.2 A. Calculate the number of electrons passing through one point in the lamp in 2.0 minutes.
\n\nSolution:
\n\nAn electron is one unit of charge, so we are finding how many one electron units of charge fit in the overall charge
\n\nC = As
\n\nC = 6.2 × 120
\n\nC = 744 C
\n\nnumber of electrons = 744⁄1.60×10−19
\n\nThe number of electrons is about 4700000000000 000 000 000.
\n\nWhen an electron is lost or gain there is a -1e or +1e
\nWhen two electrons are lost it is -2e or +2e
\nThe net charge of an object can only be a multiply of 1.60×10−19 becaues the elementary charge is the smallest possible unit. Meaning the net charge is quantized.
\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "8 4meandriftvelocity", "revisioncard": "The number density is the number of free electrons per cubic metre of material. The higher number density, the greater the number of free electrons per m3
\n\n| Material | \nType | \nNumber Density n/m-3 at (300K) | \n
|---|---|---|
| Copper | \nconductor | \n8.5 × 1028 | \n
| Zinc | \nconductor | \n6.6 × 1028 | \n
| Germanium | \nsemiconductor | \n2.0 × 1018 | \n
| Silicon | \nsemiconductor | \n8.7 × 1015 | \n
There is an additional equation for electric current
\nI = A nev
\n\nI = ΔQ⁄Δt
\nThe number of electrons in a given volume V of the condutor is nV where n is the elementary charge. This gives:
\nI = neV⁄Δt
\nWhen there is an electric current in the conductor,acertain valume of\n charge carriers passesagiven point each second. This valume depends an\n the cross-sectional area of the conductor and the mean drift velocityvof\n the charge carriers.
\nV⁄Δt = Av
\nSubstituting this into the previous equation for electric current gives:
\nI = neV⁄Δt = neAv
\nI = Anev
" }, { "subject": "AS Level Physics", "revisioncardtitle": "9 2potentialdifferenceandelectromotiveforce", "revisioncard": "Potential difference is the measure of the amount of energy transferred by charge carriers(electrons)
\n\nLamps will transfer energy as thermal energy and light energy by the charge carriers(electrons)
\n\nThe transfer of a large amount of energy can be dangerous meaning a large potential difference can be dangerous
\n\n1V = 1 JC−1
\n\nVolt = Energy transferred⁄Charge
\n\nV = W⁄Q
\n\n1000 V means that 1000 J of energy is transferred every 1 Coulombs of charge
\n\nPotential difference is described when a charge carrier(electron) loses energy transferring it to Heat/Light energy
\n\nelectron -> lamp -> heat/light Energy
\n\nElectromotive Force is described when a charge carrier(electron) gains energy from a source
\n\nBattery: chemical energy -> cell/battery -> electron
\n\nSolarPanel: light energy -> lithium plates -> electron
\n\nElectromotive force = Energy transferred⁄Charge
\n\nE = W⁄Q
\n\nEnergy transferred can be calculated by determinig the electromotive force and charge.
\n\nW = VQ
\n\nW = EQ
\n\nIf the current and time is given it is:
\n\nW = V × As
\n\nW = E × As
\n\n\n\n\n" }, { "subject": "AS Level Physics", "revisioncardtitle": "9 7resistanceandresistivity", "revisioncard": "The resistance R of a wire is directly proportional to its length L.
\nR = k × L
\n\nWhen the cross-sectional area of the wire increases, the opposite happens - the resistance drops.
\nFor any given p.d, doubling the cross-sectional area will double the current in the wire, halving the resistance.
\n\nR = k × 1⁄A
\n\nR = k ×L⁄A
\n\nR = ρL⁄A
\n\nρ = RA⁄L
" } ] }