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 piano falls down a flight of stairs. In this fall 5.4kJ of energy is transferred from the gravitational potential energy store into the kinetic energy store of the piano. It take 0.80 seconds to reach the bottom of the stairs. Calculate the average power of this transfer correct to two significant figures. | A 'weight lifter' lifts a 2000N weight a distance of 1.2 metre from its original position. They do this in 0.70 seconds. Calculate the power output of the 'weight lifter' correct to two significant figures. |
| 1. State the known quantities
W = 5.4kJ = 5400J t = 0.80s |
1. State the known quantities
F = 2000N d = 1.2m t = 0.70s |
| 2. Substitute the numbers into the equation and solve.
\(P = \frac{W}{t}\) \(P = \frac{5400}{0.80}\) \(P = 6750W\) \(P \approx 6800W\) |
2. Substitute the numbers into the equation and solve.
\(W = \vec F \vec d\) \(W = \vec F \times \vec d\) \(W = 2000 \times 1.2\) \(W = 2400J\) \(P = \frac{W}{t}\) \(P = \frac{2400}{0.7}\) \(P = 3428.57W\) \(P \approx 3400W\) |