Power and Energy seem more related to Physics than Electronics and indeed they probably are. However, the power equations is necessary to calculate the correct choice of fuse, the current required by a certain output device and what rating resistor to use in a given situation. The power equation influences circuit design and construction and the energy concept is necessary to understand power.

The label on my kettle tells me the Voltage and Power but I need to work out the current

**Definition:** Power = Energy transferred / time taken and is measured in Watts. 1 Watt is 1 Joule of energy transferred every second.

However, this is not very useful in electronics as we don't have an easy way to measure energy transferred or the time taken. What we are comfortable measuring in an electronics lab are quantities such as voltage, current and resistance. The various definitions can be combined as follows to give a useful result:

V = E / Q and I = Q / t ... E = V x Q and t = Q/I ... P = E/t = V x Q x I/Q = V x I

So, not a definition, but a useful working equation in electronics, we have **P = I x V**

This can be rearranged to give: **V = P / I** and **I = P / V**

Here V is the potential difference across the component in question and I is the current flowing through the component.

By combining P = V x I and V = I x R we can also use **P = I ^{2}R**

By combining P = V x I and I = V/R we can also use **P = V ^{2}/R**

P = V x I

P = V

P = I

1. What is the power disipated in a 6V bulb that takes a current of 200mA?

200mA = 0.2A and therefore P = 6 x 0.2 = 1.2W

2. What current does a 60W mains light bulb take if the mains voltage is 230V?

I = P/V and so I = 60/230 = 0.24A

3. What voltage does a 120W bulb operate at if it draws a current of 10A?

V = P/I and therefore V = 120/10 = 12V

4. How much power is dissipated by a 100Ω resistor connected to a 12V power supply?

P = V^{2}/R therefore P = 144/100 = 1.44W (the standard resistors used in electronics are rated at 1/4W)

5. What is the power rating of a heater that has a resistance of 50Ω and takes a current of 2A?

P = I^{2}R and so P = 4 x 50 = 200W

Resistors get hot and, if they get too hot, they stop working so it is important to know the power rating of a resistor for use in a circuit. Basic resistors used in circuit building are either 1/8W or 1/4W and slightly bigger 1/2W ones are also common. For higher power applications power resistors are used and may be rated at 1W, 5W, 20W 100W etc depending on the manufacturer.

**Consider a 470Ω resistor used as a current limiting resistor with an LED.** The current through the LED and resistor is 20mA. What power rating is suitable for the resistor?

P = I^{2}R and I = 0.02A therefore P = 0.02^{2}x470 = 0.19W. A 1/8W resistor would not be appropriate (1/8W = 0.125W) as this is the maximum power rating and is less than the 0.19W being dissipated. However, a 1/4W resistor would be fine.

**It is important to fit a fuse with the correct rating to provide maximum protection.** Consider a domestic appliance such as a hair drier that runs off the 230v mains supply and dissipates 650W.

Given that 3A, 5A and 13A fuses are available, which would be most suitable?

I = P/V and so I = 650/230 = 2.8A.

Therefore a 3A fuse would probably be okay. However, in reality, a 5A fuse would be more reliable as the 3A fuse would be working at the upper end of its current carrying capability.

**A domestic HiFi has a maximum output voltage of 70v and is connected to 8Ω speakers.** What power must the speakers be able to handle, as a minimum?

P = V^{2}/R and so P = 70^{2}/8 = 612W ... quite a big HiFi.

In Physics and Electronics, units of Watts and Joules are fine. However, in the domestic situation, these aren't always appropriate. Often power is measured in kilowatts and time is measured in hours. When we buy electricity we are paying for energy, not power, so what is the domestic unit of energy?

E = P x t and so, in domestic terms, Energy paid for is measured in kWh (kilowatthours). 1 kWh is 1kW of power used for 1 hour - like boiling a 1kW kettle continuously for an hour. So, converting back to standard units, 1kWh in Joules = Power in Watts x time in seconds = 1000 x 60 x 60 = 3.6MJ. One unit of domestic electricity is 3.6MJ of energy!

© Paul Nicholls

September 2015

Electronics Resources by Paul Nicholls is licensed under a Creative Commons Attribution 4.0 International License.