Electrical Power

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

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 20kΩ resistor has a current of 31mA. Calculate the power of the resistor 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 = 31mA = 31x10-3A R = 20kΩ = 20x103
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. \(P = I^2R\) \(P = (31 \times 10^-3)^2 \times 20 \times 10^3\) \(P = 19.22W\) \(P \approx 19W\)

Finding Power from Potential Difference and Resistance

A potential difference is applied across a 98kΩ resistor. Calculate the power	Question
1. State the known quantities	1. State the known quantities
2. Substitute the numbers into the equation and solve.	2. Substitute the numbers into the equation and solve.