Changes

: The [[SI Unit|unit]] of '''resistance''' is the [[Ohm]] ([[Ω]]).

:<math>'''Resistance ''' = \tfrac{(Potential Difference}{Current}<)/math>:<math>R = \tfrac{V}{I}</math>Where:: R = Resistance: V = Potential Difference: I = (Current)

| style="height:20px; width:200px; text-align:center;" |'''A student [[measure]]s a [[Potential Difference|potential difference]] of 5[[V]] 5V across a [[component]] and a [[Electrical Current|current]] of 0.1[[A]]1A. Calculate the resistance.'''| style="height:20px; width:200px; text-align:center;" |'''A [[Electrical Bulb|bulb]] has a [[Electrical Current|current]] of 200[[mA]] 200mA passing through it and a [[Potential Difference|potential difference]] of 12[[V]] 12V across it. Calculate the resistance of the [[Electrical Bulb|bulb]].'''| style="height:20px; width:200px; text-align:center;" |'''Calculate the resistance of a [[buzzer]] connected in [[Series Circuit|series]] to a 6[[V]] 6V [[battery]] with an [[ammeter]] reading of 10[[mA]]10mA.'''

Force Potential Difference = 40N5V

Distance moved in the direction of the force Current = 10.2m1A

<math> W R = Fd\frac{V}{I}</math>

<math> W R = 40 \times frac{5}{0.1.2}</math>

<math> W R = 48J50 \Omega</math>

Force Potential Difference = 10,000N12V

Distance moved in the direction of the force Current = 15m200mA = 0.2A

<math> W R = Fd\frac{V}{I}</math>

<math> W R = 10,000 \times 15frac{12}{0.2}</math>

<math> W R = 15,000J60 \Omega</math>

Force Potential Difference = 40N6V Current = 10mA = 0.01A <math>R = \frac{V}{I}</math> <math>R = \frac{6}{0.01}</math> <math> R = 600 \Omega</math>|} ==Key Stage 4=====Meaning==='''Resistance''' is the [[ratio]] of [[Potential Difference|potential difference]] to [[Electrical Current|current]]. ===About Resistance===: The [[SI Unit|unit]] of '''resistance''' is the [[Ohm]] (Ω).: '''Resistance''' cannot be directly [[measure]]d. '''Resistance''' must be calculated by dividing the [[Potential Difference]] by the [[Electrical Current|Current]].: [[Electrical Conductor|Conductors]] have a low '''resistance''' and [[Electrical Insulator|insulator]]s have a high '''resistance'''.: '''Resistance''' in a [[wire]] is caused by [[electron]]s colliding with [[Positive Ion|positive ion]]s in the [[metal]] [[lattice]].{| class="wikitable"|-|[[File:ResistanceCause.png|center|500px]]|-| style="height:20px; width:200px; text-align:center;" |This [[diagram]] shows [[electron]]s moving past the [[Positive Ion|positive ions]] in a [[metal]] [[lattice]]. As they move some of them collide with the [[Positive Ion|positive ions]] in the [[lattice]] which is acts as a '''resistance''' to an [[Electrical Current|electrical current]].|}: The higher the [[temperature]] of a [[Electrical Conductor|conductor]] the greater the '''resistance''' of that [[Electrical Conductor|conductor]]. This is due to the [[metal]] [[ion]]s vibrating faster at higher [[temperature]]s which makes [[electron]]s more likely to collide with the [[ion]]s as they pass through the [[Electrical Conductor|conductor]]. ===Equation==='''Resistance''' = (Potential Difference)/(Current) <math>R = \frac{V}{I}</math> Where <math>R</math> = '''Resistance''' of a [[Electrical Component|component]]. <math>V</math> = The [[Potential Difference]] across the [[Electrical Component|component]]. <math>I</math> = The [[Electrical Current|current]] through the [[Electrical Component|component]]. ===Example Calculations=======Finding Resistance from Current and Potential Difference===={| class="wikitable"|-| style="height:20px; width:300px; text-align:center;" |'''A student [[measure]]s a [[Potential Difference|potential difference]] of 5.9V across a [[component]] and a [[Electrical Current|current]] of 0.11A. Calculate the resistance correct to two [[Significant Figures|significant figures]].'''| style="height:20px; width:300px; text-align:center;" |'''Calculate the resistance of a [[buzzer]] connected in [[Series Circuit|series]] to a 9V [[battery]] with an [[ammeter]] reading of 23mA correct to two [[Significant Figures|significant figures]].'''|-| style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' Potential Difference = 5.9V Current = 0.11A| style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' Potential Difference = 9V Current = 23mA = 23x10^-3A|-| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' <math>R = \frac{V}{I}</math> <math>R = \frac{5.9}{0.11}</math> <math> R = 53.635 \Omega</math> <math> R \approx 54\Omega</math>| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' <math>R = \frac{V}{I}</math> <math>R = \frac{9}{23\times10^{-3}}</math> <math> R = 391.304 \Omega</math> <math> R \approx 390 \Omega</math>|} ====Finding Current from Potential Difference and Resistance===={| class="wikitable"| style="height:20px; width:300px; text-align:center;" |A [[Potential Difference|potential difference]] of 9.9V is placed across an 19 Ohm [[Electrical Resistor|resistor]]. Calculate the '''current''' flowing through the [[Electrical Resistor|resistor]] correct to two [[Significant Figures|significant figures]].| style="height:20px; width:300px; text-align:center;" |A toaster has a [[Electrical Resistance|resistance]] of 27 Ohms is plugged into the [[Mains Electricity|mains]] which has a [[Potential Difference|potential difference]] of 230V. Calculate the current flowing through the toaster correct to two [[Significant Figures|significant figures]].|-| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' V = 9.9V R = 19Ω| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''V = 230V R = 27Ω|-| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>R = \frac{V}{I}</math> <math>19=\frac{9.9}{I}</math> | style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>R = \frac{V}{I}</math> <math>27=\frac{230}{R}</math>|-| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>I=\frac{9.9}{19}</math> <math>I=0.52105A</math> <math>I\approx0.52A</math> | style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>I=\frac{230}{27}</math> <math>I=8.519A</math> <math>I\approx8.5A</math>|} ====Finding Potential Difference from Current and Resistance===={| class="wikitable"| style="height:20px; width: 300px; text-align:center;" |A '''current''' of 55mA flows through a [[Electrical Component|component]] with a [[Electrical Resistance|resistance]] of 93 Ohms. Calculate the [[Potential Difference|potential difference]] across this [[Electrical Component|component]] correct to two [[Significant Figures|significant figures]].| style="height:20px; width: 300px; text-align:center;" |A 22kΩ [[Electrical Resistor|resistor]] has a '''current''' flowing through it of 6mA. Calculate the [[Potential Difference|potential difference]] across the [[Electrical Resistor|resistor]] correct to two [[Significant Figures|significant figures]].|-| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' I = 55mA = 55x10^-3A R = 93Ω| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' I = 6mA = 6x10^-3A R = 22kΩ = 22x10³Ω|-| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>R = \frac{V}{I}</math> <math>93=\frac{V}{55 \times 10^{-3}}</math>| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>R = \frac{V}{I}</math> <math>22 \times 10^3=\frac{V}{6 \times 10^{-3}}</math>|-| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''

Distance moved in the direction of the force <math>V= 0m55 \times 10^{-3} \times 93</math>

<math> W V= Fd5.115V</math>

<math> W = 20 V\times 0approx5.1V</math>

<math> W | style= 0J</math>"height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''

No '''work''' has been done because the movement is not in the direction of the [[force]]. The [[weight]] acts downwards but the movement was horizontal.<math>V= 6 \times 10^{-3} \times 22 \times 10^3</math> <math>V= 132V</math> <math>V\approx130V</math>