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Electrical Power

4,946 bytes added, 09:36, 30 November 2021
→‎Meaning
==Key Stage 4==
===Meaning===
'''Electrical power''' is the rate of [[Electrical Energy Transfer|electrical energy transfer]] in an a [[Electrical Component|component]].
===About Electrical Power===
<math>P=\frac{W}{t}</math>
Where:
<math>P</math> = '''Electrical Power'''.
<math>P=IV</math>
Where:
<math>P</math> = '''Electrical Power'''.
<math>P=I^2R</math>
Where:
<math>P</math> = '''Electrical Power'''.
<math>P=\frac{V^2}{R}</math>
Where:
<math>P</math> = '''Electrical Power'''.
===Example Calculations===
====Finding Power from Work Done and Time====
{| class="wikitable"
| style="height:20px; width:200px300px; text-align:center;" |A piano falls down a flight of stairs. In this fall 5.4kJ of An [[energy]] is [[Energy Transfer|transferred]] from the [[Gravitational Potential Energy Store|gravitational potential energy store]] into the [[Kinetic Energy Store|kinetic energy storeLED]] uses 842J over a period of the piano. It take 0.80 120 seconds to reach the bottom of the stairs. Calculate the average '''power''' of this transfer the [[LED]] correct to two [[Significant Figures|significant figures]].| style="height:20px; width:200px300px; text-align:center;" |A '[[weight]] lifter' lifts a 2000N [[weight]] a distance of toaster uses 180kJ while toasting some bread for 1.2 [[metreminute]] from its original position. They do this in 0.70 and 27 [[second]]s. Calculate the [['''power]] output ''' of the '[[weight]] lifter' toaster correct to two 2 [[Significant Figures|significant figures]].
|-
| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities'''
W = 5.4kJ = 5400J842J
t = 0.80s120s| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities'''
F W = 2000N180kJ = 180x10<sup>3</sup>J
d t = 11min27s = 87s|-| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' <math>P=\frac{W}{t}</math> <math>P=\frac{842}{120}</math> <math>P=7.01616W</math> <math>P\approx 7.0W</math>| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' <math>P=\frac{W}{t}</math> <math>P=\frac{180 \times 10^3}{87}</math> <math>P=2068.2m9655W</math>
t <math>P\approx 2100W</math>|} = 0===Finding Power from Potential Difference and Current===={| class="wikitable"| style="height:20px; width:300px; text-align:center;" |A light bulb is supplied a current of 625mA. The [[Potential Difference|potential difference]] across the terminals on the [[Electrical Bulb|light bulb]] is 96 [[volt]]s. Calculate the [[power]] of the [[Electrical Bulb|bulb]] correct to two [[Significant Figures|significant figures]].| style="height:20px; width:300px; text-align:center;" |A kettle is plugged into the [[Mains Electricity|mains supply]] which operates at 230V. It receives a current of 9.2Amps Calculate the power of the kettle correct to two [[Significant Figures|significant figures]].70s
|-
| style="height:20px; width:200px300px; text-align:left;" |'''21. [[Substitute (Maths)|Substitute]] the numbers into State the known quantities in [[equation]] and [[Solve (Maths)|solveSI Unit]].s'''
<math>P I = \frac{W}{t}</math>625mA = 0.625A
<math>P V = \frac{5400}{096V| style="height:20px; width:300px; text-align:left;" |'''1.80}</math>State the known quantities in [[SI Unit]]s'''
<math>P I = 6750W</math>9.2A
<math>P \approx 6800W</math>V = 230V|-| style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
<math>W P = \vec F \vec dIV</math>
<math>W P = \vec F I \times \vec dV</math>
<math>W P = 2000 0.625 \times 1.296</math>
<math>W P = 2400J60W</math>| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
<math>P = \frac{W}{t}IV</math>
<math>P = I \frac{2400}{0.7}times V</math>
<math>P = 34289.57W2 \times 230</math>
<math>P \approx 3400W= 2116W</math>
<math>P \approx 2100W</math>
|}
 
====Finding Power from Current and Resistance====
{| class="wikitable"
| style="height:20px; width:300px; text-align:center;" |A [[Electrical Bulb|bulb]] uses a [[Electrical Current|current]] of 1.4A and has a [[Electrical Resistance|resistance]] of 9.1Ω. Calculate the '''power''' of the [[Electrical Bulb|bulb]] correct to two [[Significant Figures|significant figures]].
| style="height:20px; width:300px; text-align:center;" |A [[wire]] has a resistance of 15mΩ. Calculate the '''power''' dissipated by the [[wire]] when a [[Electrical Current|current]] of 3.5A is in the [[wire]] correct to two [[Significant Figures|significant figures]].
|-
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 
I = 1.4A
 
R = 9.1Ω
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 
I = 3.5A
 
R = 15mΩ = 15x10<sup>-3</sup>Ω
|-
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>P = I^2R</math>
 
<math>P = 1.4^2 \times 9.1</math>
 
<math>P = 17.836W</math>
 
<math>P \approx 18W</math>
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>P = I^2R</math>
 
<math>P = (3.5)^2 \times 15 \times 10^-3</math>
 
<math>P = 0.18375W</math>
 
<math>P \approx 0.18W</math>
|}
 
====Finding Power from Potential Difference and Resistance====
{| class="wikitable"
| style="height:20px; width:300px; text-align:center;" |A [[Potential Difference|potential difference]] of 6.8V is applied across a [[Electrical Bulb|bulb]] with a [[Electrical Resistance|resistance]] of 13Ω. Calculate the '''power''' of the [[Electrical Bulb]] correct to two [[Significant Figures|significant figures]].
| style="height:20px; width:300px; text-align:center;" |A 20kΩ [[Electrical Resistor|resistor]] has a [[Potential Difference|potential difference]] of 56V. Calculate the '''power''' dissipated by the [[resistor]] correct to two [[Significant Figures|significant figures]].
|-
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 
V = 6.8V
 
R = 13Ω
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 
V = 56V
 
R = 20kΩ = 20x10<sup>3</sup>Ω
|-
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>P=\frac{V^2}{R}</math>
 
<math>P=\frac{6.8^2}{13}</math>
 
<math>P=3.5569W</math>
 
<math>P\approx 3.6W</math>
 
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>P=\frac{V^2}{R}</math>
 
<math>P=\frac{56^2}{20 \times 10^3}</math>
 
<math>P=0.1568W</math>
 
<math>P\approx0.16W</math>
 
|}
 
===References===
====Edexcel====
 
:[https://www.amazon.co.uk/gp/product/1292120223/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120223&linkCode=as2&tag=nrjc-21&linkId=068ecf40278c32406a7f1c6e66751417 ''Electrical power, pages 155, 174, GCSE Physics, Pearson Edexcel '']
:[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Electrical power, pages 395, 408, GCSE Combined Science, Pearson Edexcel '']
 
====OCR====
:[https://www.amazon.co.uk/gp/product/0198359837/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359837&linkCode=as2&tag=nrjc-21&linkId=3c4229e8b023b2b60768e7ea2307cc6f ''Electrical power, pages 114-115, Gateway GCSE Physics, Oxford, OCR '']
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