# Converting and Using Units in Equations

Discussion in 'Homework Help' started by WyMan, Apr 24, 2014.

1. ### WyMan Thread Starter New Member

Apr 24, 2014
2
0
Hello,

I'm new here and new to the world of electricity in general. I'm currently going through training to work on electronics (Avionics systems) within Aircraft. I'm loving it but I have a lot of difficulty figuring out proper units in equations. I can do the math but making sure the decimal is in the right spot and the unit is correct is where I have difficulty. I was wondering if anyone had any tips and tricks to how they learned and applied it.

An example would be:

How much power does 20mA through a 5kΩ resistor produce?

P= I^2 XR
P= 20^2 X5
P= 2000
P= 2.0 W (I know this is the answer but I have no idea how to get here from 2000)

Thanks!

Wy

2. ### amilton542 Active Member

Nov 13, 2010
495
64
How does 2000 W amount to 2 W ?

3. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
The metric prefixes are, essentially, multipliers that we add to a value in order to denote something is X orders of magnitude bigger/smaller.

See here.

In other words, the 20 mA means 20 milli-Amperes, or, according to the table, 20 * .001 Amps, or, 0.02A.

Similarly, the kilo- prefix means multiply by 1000, so 5k Ohms becomes 5000 Ohms.

I=0.02A
R=5000 Ohms
P= I^2*R =(0.02A)^2*(5000 Ohms)=2 Watts

See the eBook if you need more clarification.

WyMan likes this.
4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515

Which Kingston would that be? The calypso one?

Engineering uses a convention about units.

They don't like decimal points, since using them leads to too many mistakes.

Furthermore an engineer is someone who can count up to 1000.

So what engineers do is to work in multiples of 1000 or 1/1000.

So using current as an example.

1 milliamp is 1/1000 amps

1 amp is 1 amp

1 kiloamp is 1000 amps

The same prefixes are used for all units.

We count current from 1/1000 (= .001) amps to 0.999 amps as milliamps

current from 1 - 999 amps as amps

current from 1000 to 999,000 amps as kiloamps.

This comes in handy in electronics as follows

Circuit currents are usually in the milliamp range, circuit resistances are usually in the kilo-ohm range and voltages in the volts range so Ohm's law becomes

milliamps x kilo-ohms = volts

WyMan likes this.
5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,049
939
First, in the equation P = (I^2) R, what are the units for I (current) and R (resistance)?

Make sure the values you use are in the proper units. You have made a mistake in both values. Resistance is in Ωs. You use a value of 5Ω, but the specified value is in kΩ. So instead of 5, it should be 5000. Same for mA.

Once your values are in the correct units, the decimal place will take care of itself.

P.S., wow, you can get a lot of good replies in the time it takes to type one on a smartphone.

WyMan likes this.
6. ### WyMan Thread Starter New Member

Apr 24, 2014
2
0
Thanks for the help. I really appreciate it. My degree is in History and English I'm a lot out of my element here but enjoying it as I learn something new.

7. ### WBahn Moderator

Mar 31, 2012
18,290
4,958
The best way is to recognize that 20 is not a current and 5 is not a resistance. Just as it is nonsensical to say that a building is 50 tall since "50" is not a height. It might be 50 feet, or 50 meters, or even 50 stories. A measurement has two parts that are both needed for the measurement to have meaning -- the numerical part and the units part. So use them religiously and you will make far fewer mistakes and you will catch most of the ones you do make (and you WILL make them -- we all do).

So you have 20 mA flowing through a 5 kΩ resistor.

You know that

P = (I^2)(R)

P = (20 mA)^2 (5 kΩ)
P = 400 (mA)^2 * 5 kΩ
P = 2000 (mA)(mA)(kΩ)
P = 2000 (mmk)(A^2)(Ω)
m = milli = 10^-3
k = kilo = 10^3

so (m)(k) = 1, giving us

P = 2000 (m)(A^2)(Ω)

Now, 1 (A^2)(Ω) = 1W. This is the same as saying that 1A through a 1Ω resistor produces 1W of heat. If you don't remember this, you can get it mathematically as follows:

1 V = 1 J/C (1 joule per coulomb)
1 Ω = 1 V/A (1 volt per ampere)
1 A = 1 C/s (1 coulomb per second)
1 W = 1 J/s (1 joule per second)

Thus,

1 (A^2)(Ω) = 1 (A^2)(V/A) = 1 VA = 1 (J/C)(C/s) = 1 J/s = 1 W

Either way you come at it, you now have

P = 2000 mW = 2 W

8. ### vk6zgo Active Member

Jul 21, 2012
677
85
You are trying to cancel out the powers of 10 for mA & kΩ,but you can't do that,as I needs to be squared prior to multiplying it by R.

9. ### daviddeakin Active Member

Aug 6, 2009
207
27
In the days of slide rules it was very handy to take short cuts, like dividing volts by kilohms and knowing your answer would be in milliamps, not amps.

We are now spoilt by pocket calculators, so we don't really need to take such short cuts. If in doubt, always enter your values into the equation in their honest-to-goodness base units (ohms, volts, amps etc), and you know the answer will be in no-foolin' real McCoy units too. If the calculator gives you and answer like 2x10$^{-6}$ volts you know you have 2uV.

If you do a lot of calculations you will become familiar with the shortcuts by yourself, but I tend to avoid them as I don't really trust myself not to make a silly mistake!

10. ### KeepItSimpleStupid Well-Known Member

Mar 4, 2014
1,362
230
Way back when I learned the CGS (centimeters grams seconds) system and MKS (meters-kilograms-seconds) system.

Some deal. Amps, Volts Watts or mA mV and Watts,

and you can see from

R=1V/1A = 1 W or 1 mV/1mA =? 1 * 0.001/1 * 0.001

The 0.001's cancel and you still get 1 ohm.

So, it's a little shortcut.

With engineering notation, you can easily use what you were given,

20 mA = 20e-3 and 5K = 5E3

so your P=(I^2)*R becomes P = ((20e-3)^2) A * 5e3 Ohms which is 20 mA * 5K in engineering notation.

When doing calculations always carry the units,

I should do it in LATEX.

Anyway P=(I^2)(R) and P=( 20 mA)^2 * 5K ohms is acceptable to me,

Units are (amps * amps)/ohms, but ohms - volts/amps

so you get

amps * amps * volts / amps ; the resulting units are amps*volts and V*I is power.

So, it's important that the answer is "dimensionally correct". Just like to get area of a 6 inch x 1 foot section, you have to convert units.

Area = 6 inch * 1 foot; so the area is 6 inch-feet which is correct, but not what we are accustomed too, The answer we expect would be sqin or sqft.

The numbers get multiplied and the units get multiplied.

We would best solve this problem with:

Area = 6 inch * (1 foot/12 inches) * 1 foot

Now we get the dimensionally correct square foot unit. The 1ft/12in is a conversion.

For now with electrical measurements, stick with Amps, Volts, Ohms and Watts.

Once you know that resistance in ohms is the ratio of voltage in Volts to the current in Amps and that Power in Watts is equal to Voltage in Volts * current in Amps you can derive all of the other eqns.

e.g. R=V/I and P=VI to get P=(V^2)/R etc.

Last edited: Apr 30, 2014