Electrical Power
Contents
Key Stage 4
Meaning
Electrical power is the rate of electrical energy transfer in a 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
Finding Power from Work Done and Time
| An LED uses 842J over a period of 120 seconds. Calculate the power of the LED correct to two significant figures. | A toaster uses 180kJ while toasting some bread for 1 minute and 27 seconds. Calculate the power of the toaster correct to 2 significant figures. |
| 1. State the known quantities
W = 842J t = 120s |
1. State the known quantities
W = 180kJ = 180x103J t = 1min27s = 87s |
| 2. Substitute the numbers into the equation and solve.
\(P=\frac{W}{t}\) \(P=\frac{842}{120}\) \(P=7.01616W\) \(P\approx 7.0W\) |
2. Substitute the numbers into the equation and solve.
\(P=\frac{W}{t}\) \(P=\frac{180 \times 10^3}{87}\) \(P=2068.9655W\) \(P\approx 2100W\) |
Finding Power from Potential Difference and Current
| 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 in SI Units
I = 625mA = 0.625A V = 96V |
1. State the known quantities in SI Units
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.
\(P = IV\) \(P = I \times V\) \(P = 9.2 \times 230\) \(P = 2116W\) \(P \approx 2100W\) |
Finding Power from Current and Resistance
| A bulb uses a current of 1.4A and has a resistance of 9.1Ω. Calculate the power of the bulb correct to two significant figures. | A wire has a resistance of 15mΩ. Calculate the power dissipated by the wire when a current of 3.5A is in the wire correct to two significant figures. |
| 1. State the known quantities in SI Units.
I = 1.4A R = 9.1Ω |
1. State the known quantities in SI Units.
I = 3.5A R = 15mΩ = 15x10-3Ω |
| 2. Substitute the numbers into the equation and solve.
\(P = I^2R\) \(P = 1.4^2 \times 9.1\) \(P = 17.836W\) \(P \approx 18W\) |
2. Substitute the numbers into the equation and solve.
\(P = I^2R\) \(P = (3.5)^2 \times 15 \times 10^-3\) \(P = 0.18375W\) \(P \approx 0.18W\) |
Finding Power from Potential Difference and Resistance
| A potential difference of 6.8V is applied across a bulb with a resistance of 13Ω. Calculate the power of the Electrical Bulb correct to two significant figures. | A 20kΩ resistor has a potential difference of 56V. Calculate the power dissipated by the resistor correct to two significant figures. |
| 1. State the known quantities in SI Units.
V = 6.8V R = 13Ω |
1. State the known quantities in SI Units.
V = 56V R = 20kΩ = 20x103Ω |
| 2. Substitute the numbers into the equation and solve.
\(P=\frac{V^2}{R}\) \(P=\frac{6.8^2}{13}\) \(P=3.5569W\) \(P\approx 3.6W\) |
2. Substitute the numbers into the equation and solve.
\(P=\frac{V^2}{R}\) \(P=\frac{56^2}{20 \times 10^3}\) \(P=0.1568W\) \(P\approx0.16W\) |
References
Edexcel
- Electrical power, pages 155, 174, GCSE Physics, Pearson Edexcel
- Electrical power, pages 395, 408, GCSE Combined Science, Pearson Edexcel