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subject: Interpreting Specifications for Audio Electronics III [print this page]

In this article, we will examine the basic units for technical

specifications. Defining and calculating will be covered. We'll

also cover power dissipation in resistance. Finally, all three

articles will be tied together by solving problem exercises for

review.

1. Multiple & Sub Multiple Units.

Using the basic units, volts, ohms, & amperes are practical in

most power circuits. There are many variations of these units, so

we'll focus on what you need to know for reading specifications.

In the following chart, I will give you the prefixes, followed by

symbols used for them, relation to the basic unit, and finally

examples used in specifications.

Prefix- Symbol- Unit- Example

mega - M - 1,000,000 or 1x10 to 6th - 5,000,000 ohms = 5Mohms

kilo - k - 1,000 or 1x10 to 3rd - 18kV = 18,000V

milli - m - 0.001 or 1x10 to -3rd - 48mA = 0.048A

micro - u - 0.000001 or 1x10 to -6th - 15uA = 0.000048A

Before we move on, let's do a practice exercise on converting

units to help your understanding. Change the following to basic

units in powers of ten.: 6mA, 5kohms, and 3 uA. The solutions

would be: 6 x 10 to -3A, 5 x 10 to 3rd ohms, and 3 x 10 to -6A.

In decimal form, they would be 0.003A, 5,000 ohms, and 0.000003A.

2. Calculating Power in a Circuit and Power Dissipation in

Resistance.

The unit for electrical power is the watt (W). 1 watt of power

equals the work done in 1 second by 1 volt of potential difference

moving 1 ampere. For our purposes, time will be omitted since you

will not find it used in specs. Very often.

Power in watts equals volts x amperes or P = V x I. If a

toaster 5A from the 120 volt power line. How much power is used? P

= V x I or 120V x 5A = 600W (watts).

If you know what the power and voltage is, you can calculate

what the current is.

P = V x I so I = P/V. Using the previous problem, I = 600W/120V =

5A. Proving there is a direct relationship between units.

Power dissipation: Heat is produced when current flows in a

resistance. The reason is because friction between the free moving

electrons (from basic electricity) and atoms obstructs the path of

electron flow. This concept lead to using fuses for circuit

protection.

This concludes the series of articles on reading specifications

for audio circuits. Future articles will include: understanding

specs. for microphones, and how to choose a speaker to fit yourneeds. God bless, Bradley Angell.

Below are some practice problems. Answers can be found on the

articles page of our website.

1. With 10V across 5ohms [R] , the current is _____A.

2. When 10V produces 2.5A, R = _____ohms.

3. With 8 A through a 2ohm resistor = ______V.

4. A current of 2A and a voltage of 25V, P = _____W.

5. Will a 1A fuse work for the above problem? Y/N

Interpreting Specifications for Audio Electronics III

By: Bradley Angell

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