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This text is meant to accompany class discussions. It is not everything there is to know about the basics of electricity. It is meant as a  prep for class. More detailed notes and examples are given in the class notes, presentations, and demonstrations (click here.)

Click for the questions that go with this reading
Watts versus Volt•Amps or V•A


Watts are Volts times Amps when the you are talking about direct currents, and a few a.c. devices. Watts equals Volt•Amps for battery powered devices, light bulbs, heaters and old electronics -pre 1996. Technology and alternating circuits have changed the way power is measured. There is something called "REAL" power and "apparent" power. Apparent power is used by electrical engineers who have a much deeper understanding of what is happening electrically when electrical power is supplied or consumed. Suffice to say when talking about modern electronics, Watts does not equal Volt•Amps. Watts measures the real power. But for most electronic devices built after 1996, (0.60)•Apparent power = Real Power. In other words (0.60)(Volt•Amps) = Watts.

The device above is called a UPS, Uninterrupted Power Supply. It is battery that turns on when the power goes out. When connecting the UPS to a computer you need to look at the VA. If you bought a 500 VA uninterrupted power supply then you can hook up (0.60)550VA = 330 Watts worth of modern electronics.

For everything we do in the introductory physics class, a Watt equals a Volt•Amp ...always. (However, it's good to know how world outside of this course deals with Watts. That's we this topic is brought up.)


Battery Life


Recall that

Therefore by rearranging the equation

Charge = Current•Time

In terms of units this is

Coulombs = A•s

The number of Coulombs stored in a battery is equal to the Amps that can be drawn out of the battery times the length of time it takes to draw the Amps out. This gives the number of Coulomb's of charge stored in the battery. Battery manufacturers have given consumers a similar way to measure the amount of charge in a battery. On the package it lists the amount of charge in units called A•h or mAh. Recall that the "m" symbol is used to indicate milli units, x 10-3. These units are pronounced AmpHours or milliAmpHours. to the right is an image of a package of rechargable batteries. The storage capacity of each battery is prominantly displayed on the package. It is 1000 mAh which converts to 1 Ah.



A pack of batteries is labeled such that is says each battery has a charge of 2300 mAh. If these rechargeable batteries are used in the remote for a Wii video game controller that draws 0.23 Amps, then how much time will it take before the batteries go "dead?"

Q = 2300 mAh = 2300x10-3 Ah = 2.3 Ah.
You will need to convert the current units to the same scale. Either convert both to Amps or milliAmps. In this case the Amps was chosen.
I = 0.23 A
t = ?

Since the charge is now given in mAh, (milliAmpHours) the time will also have units of hours.

Notice how the units of Amps divides out leaving the units of hours for time.

Note: Adding batteries to a device by placing them in a row, end-to-end, does not increase battery life.

Adding batteries side does increase battery life. Nearly all appliances do not add batteries like this. Even though they look side by side, they are still connected end to end.

All of the battery charging questions will use I=Q/t. An easier way to solve these battery charging problems is by using dimensional analysis. This could have been done for the example problem by taking AmpHours and dividing it by Amps.

Refer to the notes for an example problem (Click here)

How appliances get their energy

The stuff that plugs into your walls does not need a specific voltage. They need a difference in voltages. Half of your house is wired with +120 V and the other half of your house is wired with -120 V. They are all compared to the neutral line of 0 Volts. This means that each plug has potential difference of 120 V between the small and large slots. Potential difference is the difference in voltages between two locations.(This was discussed earlier too.)

The cost of electricity

The electric company charges home owners for the work , electrical work, they do to supply your energy needs. Recall that work is he change in energy. Instead of charging your a rate like

and reporting the energy you use in 1/2 million units, the electric company charges you in units of

The actual rate of $0.10 will vary depending on where you live and the demand on the electric company versus the available supply. In a way the kW•h unit is a better unit. It allows you to do some quick dimensional analysis calculations. These dimensional analysis calculations are different than conversion type calculations. instead of keeping units and just changing them, you will be eliminating units. See the example below.

Suppose electricity costs $0.12/(kW•h). A hair dryer is plugged into the wall. The hair dryer uses 1875 W on the high heat and high air flow setting. How long would it take to use up 5 cents worth of electricity?


A 20 cu.ft. refrigerator uses 1200 W. It runs for about 7 hours a day. Later this refrigerator is replaced uses with a newer model that runs for 5 hours a day and uses 600 W. If electricity costs $0.08/kWh, then how much time does it take for each refrigerator to use up $0.25 worth of electricity?



by Tony Wayne ...(If you are a teacher, please feel free to use these resources in your teaching.)