# AS Core Physics 1

**9** Equations for uniformly accelerated motion:

- Where $s$ = displacement ($\mathrm{m}$), $u$ = initial velocity ($\mathrm{m}\text{}{\mathrm{s}}^{-1}$), $v$ = final velocity ($\mathrm{m}\text{}{\mathrm{s}}^{-1}$), $t$ = time ($\mathrm{s}$) and $a$ = acceleration ($\mathrm{m}\text{}{\mathrm{s}}^{-2}$):

$s=$$\frac{(u+v)t}{2}$

$v=u+at$

$s=ut+\frac{1}{2}a{t}^{2}$

${v}^{2}={u}^{2}+2as$ - If an object is uniformly accelerating, average velocity = $\frac{1}{2}(u+v)$

**10, 11** Displacement/velocity/acceleration - time graphs:

- On a
**displacement-time**graph,

- the gradient at a point is the velocity at that time (the**instantaneous velocity**)

- to get the average velocity, divide the overall displacement (i.e. the final displacement minus the initial displacement) by the time of the entire graph - On a
**velocity-time**graph,

- the gradient is equal to the acceleration

- the area under the graph is the displacement - On an
**acceleration-time**graph,

- the area underneath is the**change in velocity**. If the area is 0 square units, the object is moving with constant average velocity - Remember that displacement, velocity and acceleration are vector quantities (see 12), so they can be positive or negative. So usually you would give a decelerating object a negative acceleration

## How the above are derived:

- To find the units of the gradient of a graph, divide the $y$ quantity by the $x$ quantity. For example, the gradient of a velocity-time graph is $\mathrm{m}\text{}{\mathrm{s}}^{-1}$ $\xf7$ $\mathrm{s}$ $={\mathrm{m}\text{}\mathrm{s}}^{-1}\times {\mathrm{s}}^{-1}=$ $\mathrm{m}\text{}{\mathrm{s}}^{-2}$, acceleration
- To find the units of the area underneath a graph, multiply the $y$ quantity by the $x$ quantity. For example, the area under a velocity-time graph is $\mathrm{m}\text{}{\mathrm{s}}^{-1}$ $\times $ $\mathrm{s}$ $=$ $\mathrm{m}\text{}{\mathrm{s}}^{0}$ $=$ $\mathrm{m}$, displacement

## Techniques:

- To find the gradient of a curved graph, find the tangent at the point you are measuring at and use $m=$ $\frac{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\text{}\mathrm{i}\mathrm{n}\text{}y}{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\text{}\mathrm{i}\mathrm{n}\text{}x}$
- If you're finding the area underneath a curved graph, split it into small chunks (boxes and triangles, or trapezia). Calculate the area of each chunk and then sum these areas to find the approximate distance travelled

**12, 14** Scalar and vector quantities:

- A vector has a magnitude and direction
- A scalar quantity has a magnitude only
- Vectors are drawn with arrows above them and are sometimes printed in
*bold italics* - They can also be drawn as an arrow. The direction of the arrow indicates the direction of the vector, and the length is proportional to its magnitude
- Add vectors to find the
**resultant**(overall magnitude and direction): draw them tip-to-tail and use a ruler/trigonometry to calculate the resultant (distance from the tail of the first to the tip of the last)

## Examples of scalar quantities:

- Mass
- Electric charge
- Energy
- Temperature
- Time
- Length
- Speed
- Work

## Examples of vector quantities:

- Force
- Displacement
- Momentum
- Weight
- Velocity
- Acceleration

**13** Resolving vectors:

**Resolving**a vector is the process of taking its overall magnitude and direction and using it to find its magnitude in two directions at right angles to each other. For example, if you are travelling at $10\text{}\mathrm{m}\text{}{\mathrm{s}}^{-1}$ north-west, your velocity in the west direction would be about $7\text{}\mathrm{m}\text{}{\mathrm{s}}^{-1}$ ($10\times \mathrm{cos}45$°)- Use $\mathrm{sin}$ and $\mathrm{cos}$ to calculate the two component forces from the resultant - the hypotenuse
- This can also be done by drawing a right-angled triangle with the hypotenuse as the overall vector (the one you are resolving) and measuring the sides with a ruler to find the components
- The two components can now be separately used for calculations

**14** Finding a resultant vector:

- This is the opposite of resolving - here you'll have the two components and will need to find the resultant (hypotenuse)
- Use Pythagoras' theorem to calculate the magnitude (magnitude = $\sqrt{{a}^{2}+{b}^{2}}$)
- Use $\mathrm{tan}$ to calculate the direction (sketch the triangle first)
- If you are asked to do this for two components which are not at right angles to each other, draw them tip-to-tail, fill in the third line to create a triangle, and use a protractor and ruler to find the resultant

**15** Projectiles:

- The vertical motion and horizontal motion of a projectile are independent of each other
- Assuming air resistance is negligible, a projectile will continue at constant horizontal velocity
- Vertically, a projectile will obey the equations of motion for uniform acceleration (see 9)
- Find the time for the projectile to reach its highest point by using the vertical component of the initial velocity (acceleration $=$ $-9.81\text{}\mathrm{m}\text{}{\mathrm{s}}^{-2}$)
- To get the total travel time, multiply this by 2 (because the projectile will rise to its maximum height and then fall back down again at the same rate)
- By multiplying the horizontal component of the initial velocity by the total travel time, the horizontal distance travelled can be calculated

**16** Free-body force diagrams:

- Free-body force diagrams show all of the forces acting on a body with arrows
- The diagrams should not show the forces that the body exerts on other bodies
- You can use free-body force diagrams to help calculate if an object is in equilibrium
- A dot can be used to represent the body because it is the centre of gravity
- If possible, you should make the lengths of the arrows proportional to the size of the force they represent. For example, if a body is accelerating in free fall, the $mg$ arrow should be longer than the atmospheric drag arrow
- Examples of forces on objects:

- $F$ (friction)

- $mg$ (gravity)

- $R$ (reaction force due to Newton's third law) - If you know the numeric value of forces, you should write those next to the arrow

**17, 20** Newton's laws:

**Newton's first law**states that an object continues at a constant velocity/remains at rest, provided that there is no external force acting on it (a force is needed to change velocity)**Newton's second law**states that acceleration is produced when force acts on a mass, and this acceleration is directly proportional to the magnitude of the force**Newton's third law**states that forces always occur in equal, opposite pairs. The pairs are also always of the same type (e.g. both kinetic). In other words, if object A exerts a force on object B, object B will exert an equal and opposite force on object A

## $\mathrm{\Sigma}F=ma$:

- force
**$\mathrm{\Sigma}F$**($\mathrm{N}$) $=$ mass**$m$**($\mathrm{k}\mathrm{g}$) $\times $ acceleration**$a$**($\mathrm{m}\text{}{\mathrm{s}}^{-2}$) - $\mathrm{\Sigma}$ is the symbol for 'the sum of'. So $\mathrm{\Sigma}F$ is the sum of all forces. Because force is a vector quantity, $\mathrm{\Sigma}F$ represents the resultant (overall) force
- This equation uses Newton's second law because it shows that $F$ and $a$ are directly proportional
- If we substitute $a=0$ into the equation, we get $\mathrm{\Sigma}F=0$, showing that there is no overall force if an object is not accelerating - i.e. Newton's first law

## Terminal velocity:

**Terminal velocity**is reached when the weight of a falling object is equal to the upwards force (e.g. drag). Since there is no overall force, the acceleration will be 0 (Newton's first law)

**18** Gravity:

- Close to the surface of Earth, $g=$ $9.81\text{}\mathrm{N}\text{}{\mathrm{k}\mathrm{g}}^{-1}$
- This means that any object in free fall has an acceleration of about $9.81\text{}\mathrm{m}\text{}{\mathrm{s}}^{-2}$
- weight
**$mg$**($\mathrm{N}$) $=$ mass**$m$**($\mathrm{k}\mathrm{g}$) $\times $ gravity**$g$**($\mathrm{N}\text{}{\mathrm{k}\mathrm{g}}^{-1}$) - gravitational field strength
**$g$**($\mathrm{N}\text{}{\mathrm{k}\mathrm{g}}^{-1}$) $=$ force**$F$**($\mathrm{N}$) $\xf7$ mass**$m$**($\mathrm{k}\mathrm{g}$)

**19** Determining the acceleration of a freely falling object (CP 1):

- Light gates can be used to accurately measure the time for an object to fall a specific distance
- Set up two light gates and measure the distance between them with a metre rule
- Drop a long thin object through the top light gate and with a data logger, record the time it takes to pass through them
- This should be repeated three times (or more if one result appears to be anomalous) and then with different distances
- Because we know the displacement, starting velocity ($0\text{}{\mathrm{m}\mathrm{s}}^{-1}$), and time, an equation can be selected to calculate the acceleration; $s=ut+\frac{1}{2}a{t}^{2}$
- This equation can be rearranged to $y=mx+c$ form, from which we can find that on a graph of displacement against time squared, the acceleration due to gravity is the gradient multiplied by 2
- This answer should be close to the actual value, about $9.81\text{}{\mathrm{m}\mathrm{s}}^{-2}$

**21** Momentum:

- momentum
**$p$**($\mathrm{k}\mathrm{g}\text{}\mathrm{m}\text{}{\mathrm{s}}^{-1}$) $=$ mass**$m$**($\mathrm{k}\mathrm{g}$) $\times $ velocity**$v$**($\mathrm{m}\text{}{\mathrm{s}}^{-1}$)

**22** Conservation of linear momentum:

- Momentum is always conserved, this is the law of conservation of linear momentum. So total momentum after a collision $=$ total momentum before the collision
- Because $p=mv$, this can be expressed as ${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$ where

- $m$ is the mass ($\mathrm{k}\mathrm{g}$)

- $u$ is the initial velocity ($\mathrm{m}\text{}{\mathrm{s}}^{-1}$)

- $v$ is the final velocity ($\mathrm{m}\text{}{\mathrm{s}}^{-1}$)

## Momentum and Newton's laws:

- Newton's second law can be expressed in terms of momentum; the rate of change of momentum is directly proportional to the resultant force acting on the object
- Due to Newton's third law, the momenta of both objects are equal. For example, when firing a gun, the bullet's momentum and the gun's momentum (due to recoil) are equal and opposite

**23** Moments:

- If you have a plank of wood lying on a pivot, the further away from the pivot you place a mass, the more it will rotate

- This ability to rotate can be calculated and is called a**moment**

- The total clockwise moment about a point is the sum of all moments which cause a clockwise rotation about that point

- The total counter-clockwise moment is the opposite of this - moment about a point
**$M$**($\mathrm{N}\text{}\mathrm{m}$) = force**$F$**($\mathrm{N}$) $\times $ perpendicular distance (i.e. distance from the force to the point)**$x$**($\mathrm{m}$) - The perpendicular distance, $x$, is the distance connecting the pivot point and line of action of the force
- At equilibrium, the clockwise moment is equal to the counter-clockwise moment and all forces acting on the object must have a resultant force of zero

**24** Equilibrium:

- An object is in equilibrium if its forces are balanced
- Due to Newton's first law, an object in equilibrium will not be accelerating
- The vectors of the forces acting on a body can be drawn tip-to-tail. If they connect to create a closed shape, the body is in equilibrium

## Centre of mass/gravity:

- The centre of mass is a location on a body where if a force is applied, the effect is the same as if the force was applied equally to the whole body
- It is not necessarily in the centre of an object - if it's heavier on one side, the centre of mass will be closer to that side
- At equilibrium, the moments around the centre of mass will sum to zero. This can be used to find the centre of mass of a body
- An object will fall over if, when drawn rotated at the angle you are testing for, a vertical line drawn straight down from the centre of mass
**doesn't**pass through the edge that would usually be in contact with the surface it's on

**25** Work:

- Work is done when a force moves something, transferring energy
- For example, if you lift something up, you do work against gravity, and kinetic energy is transferred to gravitational potential energy
- Work is defined as: work
**$\mathrm{\Delta}W$**($\mathrm{J}$) $=$ force**$F$**($\mathrm{N}$) $\times $ displacement (in the direction of the force)**$\mathrm{\Delta}s$**($\mathrm{m}$) - To calculate the work done when the force is acting at an angle to the distance, calculate the component of the force in the direction of motion and multiply that by the displacement

**26** Kinetic energy:

- The kinetic energy of a moving object can be calculated with ${E}_{\mathrm{k}}=\frac{1}{2}m{v}^{2}$, where

- ${E}_{\mathrm{k}}$ is the kinetic energy ($\mathrm{J}$)

- $m$ is its mass ($\mathrm{k}\mathrm{g}$)

- $v$ is its speed ($\mathrm{m}\text{}{\mathrm{s}}^{-1}$)

**27, 28** Gravitational potential energy:

- GPE is the energy something gains when lifted up (i.e. the work done to lift it). This is because energy is conserved, and this GPE will be converted into kinetic energy when the object falls
- change in gravitational potential energy
**$\mathrm{\Delta}{E}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$**($\mathrm{J}$) $=$ mass**$m$**($\mathrm{k}\mathrm{g}$) $\times $ gravitational field strength**$g$**($\mathrm{N}\text{}{\mathrm{k}\mathrm{g}}^{-1}$) $\times $ change in height**$\mathrm{\Delta}h$**($\mathrm{m}$) - For example, if an object of mass $10\text{}\mathrm{k}\mathrm{g}$ is lifted by $10\text{}\mathrm{m}$, it gains $10\text{}\mathrm{k}\mathrm{g}$ $\times $ $9.81\text{}\mathrm{N}\text{}{\mathrm{k}\mathrm{g}}^{-1}$ $\times $ $10\text{}\mathrm{m}$ $=$ $981\text{}\mathrm{J}$

**29** Power:

- Power is the energy transferred or work done per second
- power
**$P$**($\mathrm{W}$) $=$ energy transferred**$E$**($\mathrm{J}$) $\xf7$ time**$t$**($\mathrm{s}$) - power
**$P$**($\mathrm{W}$) $=$ work done**$W$**($\mathrm{J}$) $\xf7$ time**$t$**($\mathrm{s}$)

**30** Efficiency:

- Although energy is never created or destroyed, it can be converted into non-useful energy (e.g. heat from a lightbulb)
- efficiency $=$ useful energy output $\xf7$ total energy input
- efficiency $=$ useful power output $\xf7$ total power input
- Multiply these by 100 to get the
**percentage efficiency**

**31, 32, 34, 35** Current:

- Current is measured in amperes ($\mathrm{A}$) with an ammeter (connect in series)
- it is the rate of flow of charged particles through a component in a circuit
- There is never a 'build-up' of current in a circuit

## Charge:

- Charge is measured in coulombs ($\mathrm{C}$)
- $1\text{}\mathrm{C}$ $=$ the charge in 1 second with $1\text{}\mathrm{A}$ of current
- $1\text{}\mathrm{A}$ $=$ $1\text{}\mathrm{C}\text{}{\mathrm{s}}^{-1}$
- $I=$ $\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}t}$ where

- $I$ is the current ($\mathrm{A}$)

- $t$ is the time ($\mathrm{s}$)

- $Q$ is the charge passing during that time ($\mathrm{C}$)

## Potential difference:

- pd is measured in volts ($\mathrm{V}$)
- It is the work done per unit of charge moved
- $V=$ $\frac{W}{Q}$ where

- $V$ is the potential difference ($\mathrm{V}$)

- $W$ is the work done ($\mathrm{J}$)

- $Q$ is the charge that the work is being done on ($\mathrm{C}$)

## Conservation of charge:

- Charge and current are conserved in a circuit
- In a series circuit, the current through each component is the same, but the potential difference across high resistance components will be higher because more work needs to be done to move the charge carriers through them
- At a junction in a parallel circuit, the sum of the current flowing in equals the sum of the current flowing out

**33** Electrical resistance:

- Charge flows more easily if there is less resistance in a circuit
- Resistance is measured in ohms ($\mathrm{\Omega}$), and $1\text{}\mathrm{\Omega}$ $=$ $1\text{}\mathrm{V}\text{}{\mathrm{A}}^{-1}$, because $R=$ $\frac{V}{I}$
- There will be a resistance of $1\text{}\mathrm{\Omega}$ if $1\text{}\mathrm{V}$ makes a current of $1\text{}\mathrm{A}$ flow through it

## Ohmic devices:

- If the resistance of a material is constant over a range of pds (under constant physical conditions), it is considered ohmic; so the pd is directly proportional to current ($V\propto I$)
- Since there will be error when measuring current and pd in a conductor, a device can be considered ohmic within the limits of experimental uncertainty if $I$ against $V$ can be plotted on a graph with a straight line passing within all of the error bars/boxes

**36** Combining resistors:

- In a series circuit, ${V}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={V}_{1}+{V}_{2}+{V}_{3}+...$, therefore $I{R}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=I{R}_{1}+I{R}_{2}+I{R}_{3}+...$ and ${R}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={R}_{1}+{R}_{2}+{R}_{3}+...$
- In a parallel circuit, the same can be done to get $\frac{1}{{R}_{total}}$ $=$ $\frac{1}{{R}_{1}}$ $+$ $\frac{1}{{R}_{2}}$ $+$ $\frac{1}{{R}_{3}}$

## Deriving for a parallel circuit with emf $\epsilon $:

- ${Q}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={Q}_{1}+{Q}_{2}+{Q}_{3}+...$
- $Q=It$ so ${I}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={I}_{1}+{I}_{2}+{I}_{3}+...$
- $V=IR\Rightarrow \epsilon =IR$ so $\frac{\epsilon}{{R}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}$ $=$ $\frac{\epsilon}{{R}_{1}}$ $+$ $\frac{\epsilon}{{R}_{2}}$ $+$ $\frac{\epsilon}{{R}_{3}}$ $+...$
- The $\epsilon $s cancel to give $\frac{1}{{R}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}$ $=$ $\frac{1}{{R}_{1}}$ $+$ $\frac{1}{{R}_{2}}$ $+$ $\frac{1}{{R}_{3}}$ $+...$

**37** Power:

- power
**$P$**($\mathrm{W}$) $=$ energy transferred**$\mathrm{\Delta}W$**($\mathrm{J}$) $\xf7$ time taken**$\mathrm{\Delta}t$**($\mathrm{s}$) (this is in the mechanics section of the formulae sheet but it works for electricity too) - power
**$P$**($\mathrm{W}$) $=$ potential difference**$V$**($\mathrm{V}$) $\times $ current**$I$**($\mathrm{A}$) - work done
**$W$**($\mathrm{J}$) $=$ potential difference**$V$**($\mathrm{V}$) $\times $ current**$I$**($\mathrm{A}$) $\times $ time**$t$**($\mathrm{s}$) - The equations in this section can be combined to form

- $\mathrm{\Delta}W=$ $\frac{V}{t}$

- $P={I}^{2}R$

- $P=$ $\frac{{V}^{2}}{R}$

**38** Current-potential difference graphs:

- With an ohmic device, the current-potential difference graph will be a straight line with a positive gradient
- Filament bulbs have a similar relationship (current increases with potential difference), however the graph would not be a straight line, with rate of change decreasing as potential difference increases:
- With an NTC thermistor, resistance decreases as temperature increases, with resistance decreasing fastest at lower temperatures (it is an exponential function):
- Diodes do not conduct until a certain potential is reached and only allow charge through in one direction:

**39** Resistivity:

**Electrical resistivity**is a measure of how well a material resists, regardless of size- Resistance is
**directly proportional**to the length of a conductor and**inversely proportional**to cross-sectional area - These properties can be combined to find
**resistivity**:

resistance**$R$**($\mathrm{\Omega}$) $=$ resistivity**$\rho $**($\mathrm{\Omega}\text{}\mathrm{m}$) $\times $ length**$l$**($\mathrm{m}$) $\xf7$ cross-sectional area**$A$**(${\mathrm{m}}^{2}$) $=$ $\frac{\rho l}{A}$ - A good conductor has low resistivity
- Insulators have very few free electrons compared to conductors which have many
- When a semiconductor is heated, electrons are freed making it a better conductor. Similarly, LDRs (light-dependent resistors) free electrons when lit

## Comparing resistivities:

- Resistivities can range by many orders of magnitude, so a logarithmic scale is therefore often used when displaying them on a graph
- When the temperature of a resistor is increased, the lattice ions vibrate more. This reduces the kinetic energy of the delocalised electrons and therefore reduces the drift velocity
- In different resistors, charge carrier density varies widely and therefore resistivities for different materials can be very different

**40** Determining the resistivity of a material (CP 2):

- Find the diameter of the wire with a micrometer in several places and find the mean. Convert to metres and find the cross-sectional area with $A=\pi {r}^{2}$
- Attach the wire to a ruler and use crocodile clips to create connections
- Attach a power supply, an ammeter (in series) and a voltmeter (in parallel)
- Adjust the length and record the length, current and pd
- Plot a graph of resistance against length and find the gradient. The line should be straight and pass through the origin
- Multiply the gradient by the cross-sectional area in ${\mathrm{m}}^{2}$ to get the resistivity
- Resistivity is only for a specific temperature, and current in a resistive wire increases temperature, so aim to keep the current low or the wire at a constant temperature

**41** $I=nqvA$:

- Charged particles have a
**drift velocity**, their average velocity in the direction of flow (it is quite low as they won't travel in a straight line) - current
**$I$**($\mathrm{A}$) $=$ number of charge carriers per ${\mathrm{m}}^{3}$**$n$**(count/${\mathrm{m}}^{3}$) $\times $ electron charge (from data sheet)**$q$**($\mathrm{C}$) $\times $ average drift velocity**$v$**($\mathrm{m}\text{}{\mathrm{s}}^{-1}$) $\times $ cross-sectional area**$A$**(${\mathrm{m}}^{2}$) - An electron has the charge $1.60\times {10}^{-19}$ $\mathrm{C}$ and this constant is represented by the symbol $e$, or in this equation, $q$

**43, 44** Potential divider circuits:

- A
**potential divider**circuit produces an output potential difference which is a fraction of the input potential difference - A potential divider is shown below:
- When there are two resistors in a potential divider circuit, the output resistance is calculated with ${V}_{\mathrm{o}\mathrm{u}\mathrm{t}}=$ $\frac{{R}_{2}}{{R}_{1}+{R}_{2}}$ $\times {V}_{\mathrm{i}\mathrm{n}}$
- You need to memorise this equation. The numerator of the fraction must be the resistance of the resistor which you're taking the output potential difference from (so ${R}_{2}$ in the circuit diagram above)
- So the output voltage will be greatest if you have a high resistance resistor between your $+$ and $-$ terminals for the output
- An LDR has high resistance in the dark and low with lots of light. An NTC thermistor has high resistance at low temperatures. Therefore, an LDR/thermistor could be used in place of ${R}_{1}$

**45** Internal resistance and emf:

- emf,
**electromotive force**($\epsilon $), is a measure of energy (not force) transferred by a power supply for each coulomb of charge

- The emf of the power supply is equal to the sum of potential difference across all other components in a series circuit

- In a parallel circuit, the pd across each component is equal to the emf of the power source - All power sources have internal resistance due to the materials they are constructed from
- It is indicated on a circuit diagram with a cell/battery marked with $\epsilon $ (or the numeric value for emf) in series with a resistor labelled with a lower-case $r$ (or the numerical value for internal resistance), and a dotted line box drawn around the two
- The external load's resistance is marked with an upper-case $R$
- ${V}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={V}_{1}+{V}_{2}$, therefore $\epsilon =V+Ir$ and $\epsilon =IR+Ir$
**Terminal pd**is the potential difference across the terminals. At zero current flowing, this is the same as the emf- So terminal pd is represented by the $V$ in the above equations
- $Ir$ is sometimes referred to as
**lost volts**

**46** Finding the emf and internal resistance of a cell (CP 3):

- The above equation ($\epsilon =V+Ir$) can be rearranged to get $V=\epsilon -Ir$
- By comparing this to the equation of a straight line graph ($y=mx+c$), the $y$-intercept of a pd-current graph is the emf and the negative of the gradient is the internal resistance (the graph should have a negative gradient and not pass through the origin)
- Use this graph to calculate the internal resistance and emf of a cell. Use a range of resistors or a variable resistor. Connect the voltmeter in parallel with the load resistor and ammeter in series, as shown in the below circuit diagram

**47** Resistance and temperature:

- Some conductors have a resistance which linearly increases with temperature,
**PTC**(positive temperature coefficient) - Many are also NTC (inverse relationship)

## Cells:

- A cell is shown below. The longest vertical line is the positive terminal
- An easy way to remember this is to draw a P and N like this:

## Modelling current:

- Conductors have charged particles (
*free electrons*) which can move. An applied voltage causes electrons (negative) to move towards the positive terminal, creating a current - Because current is a measure of the rate of flow of these particles, it depends on their travel speed

## Modelling resistance:

- Materials are made from vibrating ions, and an increase in temperature (energy) makes them vibrate more vigorously
- Electrons moving through the material are slowed due to collisions with the vibrating ions, and there are more collisions at higher temperatures. This is why most conductors are PTC
- Additionally, more electrons are released with a rise in temperature. In some materials,
**semiconductors**, this outweighs the resistance increase due to vibrating atoms, leading to a negative temperature coefficient, resistance decreasing at higher temperatures - NTC thermistors are usually made from a semiconductor. This is why resistance decreases as their temperature increases
- Semiconductors are useful in electronics - they can be combined to form diodes and logic gates

**48** LDRs:

- Some semiconductors release electrons when they absorb photons, this property is used in light-dependent resistors (LDRs):

*Resistance-light intensity graph for an LDR* - So LDRs work in the same way as NTC thermistors, except that the energy comes from light instead of heat energy