Difference between revisions of "Electrical Power"
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| − | | style="height:20px; width: | + | | 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: | + | | style="height:20px; width:300px; text-align:center;" |A [[Electrical Kettle|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]]. |
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| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
I = 625mA = 0.625A | I = 625mA = 0.625A | ||
V = 96V | V = 96V | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
I = 9.2A | I = 9.2A | ||
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V = 230V | V = 230V | ||
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| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>P = IV</math> | <math>P = IV</math> | ||
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<math>P = 60W</math> | <math>P = 60W</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>P = IV</math> | <math>P = IV</math> | ||
Revision as of 20:37, 2 March 2019
Contents
Key Stage 4
Meaning
Electrical power is the rate of electrical energy transfer in an component.
About Electrical Power
- The SI Units of electrical power are Watts.
- Electrical power is the work done by an electrical current per unit time.
Equations
Power, Work Done and Time
NB: You must remember this equation.
Power = (Electrical Work Done)/(time)
\(P=\frac{W}{t}\)
Where\[P\] = Electrical Power.
\(W\) = Electrical Energy Transferred or Work Done by an electrical current.
\(t\) = The time over which energy is transferred.
Power, Current and Potential Difference
NB: You must remember this equation.
Power = (Current) x (Potential Difference)
\(P=IV\)
Where\[P\] = Electrical Power.
\(I\) = Electrical Current through a component.
\(V\) = Potential Difference across the component.
Power, Current and Resistance
NB: You must remember this equation.
Power = (Current)2 x (Resistance)
\(P=I^2R\)
Where\[P\] = Electrical Power.
\(I\) = Electrical Current through a component.
\(R\) = The resistance of the component.
Power Potential Difference and Resistance
NB: You must remember this equation.
Power = (Current) x (Potential Difference)
\(P=\frac{V^2}{R}\)
Where\[P\] = Electrical Power.
\(V\) = Potential Difference across the component.
\(R\) = The resistance of the component.
Example Calculations
| A light bulb is supplied a current of 625mA. The potential difference across the terminals on the light bulb is 96 volts. Calculate the power of the bulb correct to two significant figures. | A kettle is plugged into the mains supply which operates at 230V. It receives a current of 9.2Amps Calculate the power of the kettle correct to two significant figures. |
| 1. State the known quantities
I = 625mA = 0.625A V = 96V |
1. State the known quantities
I = 9.2A V = 230V |
| 2. Substitute the numbers into the equation and solve.
\(P = IV\) \(P = I \times V\) \(P = 0.625 \times 96\) \(P = 60W\) |
2. Substitute the numbers into the equation and solve.
\(P = IV\) \(P = I \times V\) \(P = 9.2 \times 230\) \(P = 2116W\) \(P \approx 2100W\) |