Showing posts with label Electronics. Show all posts
Showing posts with label Electronics. Show all posts

How to Learn Basic Algebra Fast and Free for Beginners — Rules, Equations, Solutions, Examples, Formulas, Problems and Answers

Latest update: April 13, 2024. Note: copyright info at end of page, there is history there. I never should have removed this tutorial.

This lesson is for people who think they can't learn algebra. Yes, you can.

Alternate titles for this page would be...
The Quick Basic Math Rules for Algebra
In Algebra How Do You Solve for X in Equations
In Algebra How Do You Simplify Equations
How to Learn Algebra Fast - How to Do Equations - Complete Beginner Algebra Lessons - Plus Problems with Solutions

Here is your online, complete, free, beginner algebra and equations tutorial. Easily and quickly learn algebra on your own. It is recommended that one does not attempt to do the entire tutorial in a single session; bookmark and return as desired.

If you already know arithmetic (including fractions and decimals), then you already know algebra. You just don't know you know yet. If you understand the answers to the following statements, proceed with this page; otherwise, it is probably not a good idea.
4 + 5 = 9
17 - 13 = 4
5 times 7 = 35
70 divided by 35 = 2
80 divided by 25 = 3.2

The Basics

Example #1
Algebra is nothing more than merely substituting letters for numbers. As an example:

       4 + 2 = 6

So, if we say the letter A is temporarily equal to 4, i.e.:

      A = 4

And the letter B is temporarily equal to 2, i.e.:

      B = 2

Then A plus B must equal 6, i.e.:

      A + B = 6

Example #2

      A = 5

      B = 3

So,

      A + B = ?

Well, if we replace the letter A with 5, then the question becomes:

      5 + B = ?

And then when we replace the letter B with 3, we have:

      5 + 3 = ?

Problem solved.

A side note: Algebra likes to use the letter X in place of the question mark. So the correct way to have stated the above question would have been to say:

      A = 5

      B = 3

      X = A + B

What is X? The answer is:

      X = 8

You have just learned the basic concept of algebra.

Example #3: Subtraction

      A = 19

      B = 14

      X = A - B

What is X?

We plug in the numbers and we get:

      X = 19 - 14

      X = 5

Multiplication and Division
Of course multiplication and division in algebra are just the same as in arithmetic.

Multiplication Example
(The asterisk sign (“*”) is used to replace the word “multiply.”)

      A = 20

      B = 5

      X = A * B

What is X? We plug in the numbers and we get:

      X = 20 * 5

      X = 100

Division Example
(The “/” sign is used to replace the word “divide.”)

      A = 20

      B = 5

      X = A/B

What is X? We plug in the numbers and we get:

      X = 20/5

      X = 4

Let's Mix Things Up
You now know all the arithmetic functions of algebra. Algebra lets you mix and combine these functions.

For example:

      A=1

      B=2

      C=3

      D=4

      X=A+B+C+D

      X=10

Let’s include subtraction:

      X=A + B + C - D

      X=(1+2+3) - 4, or X = 6 - 4, which is 2, or

      X = 6 - 4 = 2

Yes, there can be more than one equal sign in an equation. Instead of saying,

      A=42
      B=42
      C=42
      D=42

You can say,

      A=42
      A=B=C=D

Or just say,

      A=B=C=D=42

Side note. You have been solving equations since the first paragraph.

Some Random Example NASA Formulas

In Algebra How Do You Solve for V? Basic / beginner algebra volume formulas.

About the NASA Formula Examples
Note the "d²" in the volume formula for the cylinder. Yes, the upper "2" means the variable "d" is squared or itself times itself or "d" to the second power.

Note the "a³" in the volume formula for the cube. Likewise, the upper "3" means the variable "a" is cubed or itself times itself times itself or "a" to the third power.

Notice how some of the variables in the formulas are directly adjacent to each other. This is the standard used to indicate the variables are multiplied.

Examples
The rectangular prism formula or equation, V = a b h, means volume is equal to "a" times "b" times "h".
The top half of the volume for the sphere formula or equation, "πd³", means pi times d after d has been cubed. If d was equal to 5, then d³ would equal 125, making the equation π times 125 or 125π.
Yes, the horizontal slash in the sphere and cylinder formulas means divide by the lower number, 6 and 4 respectively.
As mentioned, "π" is the well-known symbol for pi. The approximate value of pi is 3.14159; this approximation serves most everyday purposes just fine.
More Multiplication Practice

Example #1

      A=1
      B=2
      C=3
      D=4

      X=A+B*C-D

What is X?

Simplify and solve.

When you see an equation has multiplication and division mixed into it, the rule is to do the multiplication and division first, then do the +’s and -‘s.

So the equation above really means,

      X = A + (B*C) - D or

      X = 1 + (2*3) - 4 or

      X = 1 + (6) - 4

      X = 3

The “(“ and “)” are used to indicate what parts of the equation to do first.

It should be noted X=A and A=X are mathematically equivalent.

Just Like the Pros
What you have been and are doing is just simplifying, i.e., breaking down the equation one piece at a time; just like the mathematicians do it. The mathematicians are no more able to look at an equation and instantly come up with the answer any better than the rest of us can. In other words, they can’t grasp the whole equation either. They just solve and proceed from line to line, trusting they solved the previous lines correctly.

Example #2
Here is another one:

      A=1, B=2, C=3, D=4, E=5, F=6

      ((D * B) + (F - 7)) + A) * C = X.

What is X?

This time there is more than one set of parentheses. When that happens, the rule is to do the innermost ones first. So let’s start solving this equation by breaking it down.

The (D*B) and the (F-7) are the innermost parts of the equation.

Let’s start with the (D*B).

      D * B = 4 * 2 = 8,

so we simplify the equation to,

      (8 + (F-7) + A) * C = X

Next is the (F-7).

      F - 7 = 6 - 7

This results in a number one less than zero, so we say negative one or -1.

(Another example would be 15-20. This results in a number 5 less than zero, so we say negative 5 or -5.)

The equation now looks like,

      (8 + (-1) + A) * C = X

Let’s get the A and C taken care of; the equation is now,

      (8 + (-1) + 1) * 3 = X

Next we add up the numbers inside the parenthesis.

-1 plus 1 equals zero of course.

Or you could have said: -1 plus 8 equals 7. The 8 is called a positive number, just as the -1 is called a negative number. Adding a positive number to a negative number is really just subtracting the negative number from the positive number. In other words:

      8 + (-1) = 8 - 1 = 7 or 1 + (-1) = 1 -1 = 0

Either way, our equation now looks like,

      (8 - 1 + 1) * 3 = X, which is

      (8) * 3 = 24 = X, or

      8 * 3 = 24 = X, or

      X = 24

Simplified a step at a time and solved.

If you didn’t know negative numbers before, now you do. For the sake of completeness, the next section is about what else one should know about negative numbers.

More About Negative Numbers

Numbers plus negative numbers result in lesser numbers. Keep in mind -10 is a lesser number than -5, etc.

Numbers minus negative numbers result in larger numbers. For example, whereas 9-5 = 4, but 9-(-5) = 14. In other words, minus minus results in a positive increase aka a lesser lesser or a larger larger. Minus a minus is exactly the same as plus a plus, e.g. -(-25)=25.

This is a good time to mention that in mathematics, two negatives equal a positive when applied to minus a minus subtraction, or any multiplication, or any division.

For multiplication:
Negative numbers times positive numbers equal negative numbers, e.g. -5 * 4 = -20.
Negative numbers times negative numbers equal positive numbers, e.g. -5 * -4 = 20.
You already knew positive numbers times positive numbers equal positive numbers.
For division, the same rules apply:
Negative numbers divided by positive numbers (or vice versa) equal negative numbers, e.g. -5/4 = -1.25 and 5/-4 = -1.25.
Negative numbers divided by negative numbers equal positive numbers, e.g. -5/-4 = 1.25.
You already knew positive numbers divided by positive numbers equal positive numbers.

More Example NASA Formulas

In Algebra How Do You Solve for V? Learning and doing volume formulas.

Using Spreadsheets
Spreadsheet software or applications will happily do the arithmetic and sort out the negatives versus the positives for you once you have replaced all the variables. It even knows to do the innermost before the outermost, etc. As an example, suppose you have simplified an equation to the following mess:

      X=((5-3)* 52)-21+((6+7)/(34-12))

If your spreadsheet software is MS Excel or you are using cloud Google Drive, you can exclude the X and just copy/paste the following into a single cell:

      =((5-3)* 52)-21+((6+7)/(34-12))

The spreadsheet will immediately solve the equation and give back the answer of 83.5bunchmoredigits. If you have the software or Google Drive access, go ahead and try it.

If you are experienced at spreadsheet calculations, you can, of course, do equations with the variables still in place; substituting the variables with cell locations or range names.

Another Division Example
Might as well keep it simple and use the previous variables.

      A = 5
      B = 34
      C = 21

      X=((A-3)* 52)-C+((6+7)/(B-12))

We replace the variables with the assigned numbers and we are right back where we started from:

      X=((5-3)* 52)-21+((6+7)/(34-12))

The arithmetic then gives us:

      X = 83.59090909...

Dividing by Zero
This is a good time to mention you cannot divide by zero.

For example:

      If A=1
      If A=2
      If A=3

      X = 5 + 10/(3-A)

Now if A=1, then

      X=5+10/(3-1)=5+10/2=5+5=10

Now if A=2, then

      X=5+10/(3-2)=5+10/1=5+10=15

If, however, we attempt to declare the variable A as A=3, the following occurs:

      X=5+10/(3-3)=5+10/0. (invalid)

At this point the equation becomes invalid. There is no answer to the question, “What is 10 divided by 0?”. An equation immediately becomes invalid when a divide-by-zero scenario occurs. Software applications are designed to recognize this when it happens. Plugging whatever-divided-by-zero into a spreadsheet used to give interesting results, before applications were modified to detect this.

What You've Learned
The basic concept of algebra is just plugging the numbers into the variables, and then doing the arithmetic. One merely keeps simplifying the equation until it is solved. You now have a full understanding of that concept. Yes, you have been using variables since the first paragraph.

Final Example
Here is the last example. It is presented in a different format. The question, however, remains the same. What is X? You already know everything needed to solve this equation.

      A=1, B=2, C=3, D=4, E=5
      T=-1, U=-2, V=-3

      (6X/8)+(2T+4)=((CD/2)-AD)+V

It should be noted 6X means the same as 6*X; and AD means the same as A*D. Other examples would be: 3A=3*A=A*3, 5Y=5*Y=Y*5, -2C=-2*C=C*-2, etc.

We plug the numbers into the variables, and the equation now is:

      (6X/8)+((2*-1)+4)=((3*4)/2)-(1*4)+-3

Some simplifying arithmetic gives us:

      (6X/8)+-2+4=(12/2)-4+-3

More arithmetic then gives us:

      6X/8 +2=6-4+-3

More arithmetic gives us:

      6X/8+2=-1

We can’t solve X as the equation is currently stated; so we will have to move things around and do more arithmetic.

Important Note
Whenever you change the actual value on one side of the equation, you must do the same on the other side of the equation. Example: 7=7. If you subtract 3 from the left side, then you must subtract 3 from the right side; thus 4=4. The same rule applies for addition, multiplication, and division.

Let’s subtract 2 from both sides of our equation.

      6X/8+2=-1

Then becomes:

      6X/8=-3

We have to get rid of the “divide by 8” part of the left side of the equation. So we multiply both sides of the equation by 8.

      6X/8=-3

Then becomes:

      6X=-24

We must make the X stand alone, so we divide both sides by 6.

      6X=-24

Then becomes:

      X = -4 (The Answer!)

How Do We Know If We Have the Right Answer?
To find out, we go back to the original equation and replace X with -4. We then simplify (reduce) the equation as before to its simplest form. If the simplest possible construct is valid; then, by definition, the statement “X=-4” is valid.

Here is the original equation.

      A=1, B=2, C=3, D=4, E=5
      T=-1, U=-2, V=-3

      (6X/8)+(2T+4)=((CD/2)-AD)+V

We don’t have to re-solve the parts that didn’t have the X in it to begin with, so we have:

      (6X/8)+2 = -1

We replace the X with -4, giving us:

      ((6*-4)/8)+2=-1

Simplifying gives:

      (-24/8)+2=-1

Which is:

      -3+2=-1

Which is:

      -1=-1

This construct is valid and simple enough to know X=-4 is valid.

To take it to the very end, you can multiply both sides by -1, giving us:

      1=1

What Would Have Happened, If Instead of Correctly Calculating X=-4, We Had Erroneously Calculated X=16?
The equation simplification/reduction would have proceeded smoothly to this point:

      (6X/8)+2 = -1 (as above)

When the 6X is replaced with 6*16, we get:

      (96/8)+2=-1 (false)

When further simplified says:

      12+2=-1 (false)

Which is

      14=-1 (false)

The resulting false statement by definition means the original calculation of "X=16" is a false statement.

The Adventure Continues...

There is a lot more (much, much more) to algebra, but it is really only an expansion of what you have already learned. Algebra is the basis of all other mathematics; including geometry, trigonometry, calculus, and so on. A good understanding of algebra is required to succeed at the other mathematics. Mathematics, itself, is the foundation of most other disciplines. This foundation is not just necessary for the sciences such as physics, electronics, chemistry, biology, astronomy, and so on. A mathematical foundation is necessary for many careers; including marketing, economics, architecture, and many, many others.

May all your calculations be prosperous ones!

- End of Lesson, but more to follow -

Re: Using Mobile?
Home: site intro and featured articles/resources.
View Web Version: displays Main Menu article categories (will be located below), additional site info (below and side), search function, translation function.

I first published this tutorial at another website on 10/22/2010. However, to keep the information current, I then relocated it to websitewithnoname.com. I then made the mistake of moving it elsewhere. I have now moved it back to websitewithnoname.com A long road for trying to do good. This copyrighted tutorial has helped and served people well for years. If you find this tutorial currently printed or posted anywhere else but here, please let me know in comments. I originally wrote this page because of my knowing someone who was perfectly smart enough to understand algebra but had the mental block that they couldn't.

I never should have moved this page from here. Here are some of the previous comments when it was here:

Previous Comments from long ago:

profile image
dorothy jordan 

5 years ago

Im so thankful for all the help an support,i have a big test coming up,was so worried,but now ,i feel so secure,thanks for ur help.

Kristen Howe profile image
Kristen Howe 

5 years ago from Northeast Ohio

Math was always a weak subject for me. This would be helpful and useful for anyone who's struggling with it. Voted up for useful!

Jean Bakula profile image
Jean Bakula 

7 years ago from New Jersey

Where were you when I was struggling with Algebra in 7th grade? And you even have pictures. I tried so hard, I used to stay after school for extra help. Finally the teacher agreed to pass me with a D if I would stop coming. I thought it was a good deal, although I was an A student in other subjects, I just couldn't grasp it. My son is now a teacher, and tells me she was a failure as a teacher for not trying harder!

profile image
abseli 

7 years ago

Hi there! I could have sworn I've been to this website before but after browsing through some of the post I realized it's new to me. Anyways, I'm definitely delighted I found it and I'll be bookmarking and checking back frequently! make your computer run like brand new

Marderius profile image
Marderius 

7 years ago from Alabama

Well 2/3 x 5 7/8 ..first we can change 5 7/8 into an improper fraction giving us 47/8 now we can cross multiply... we will have 2/3 x 47/8= as you can see 2 can go into itself one time and into 8 four times.... now we have 1/3 x 47/4 giving us a final answer of 47/12 or 3 11/12 ! Hope this was helpful

profile image
Carole Garfield 

7 years ago

I helping someone in fraction like 2/3 x 5 7/8 = Can someone take me through this

carlarmes profile image
carlarmes 

8 years ago from Bournemouth, England

my son found this hub useful, thank you for the content.

Angela_1973 profile image
Angela_1973 

9 years ago

Good hub on algebra, I hate it though, I speak like 5 languages but can never do math. Have a lot of respect for people in the science field

kathryn1000 profile image
kathryn1000 

9 years ago from London

That is really good.Congratulations.

*Algebra helps your brain*

To forget about troubles and pain.

So dive in and swim

While wearing a grin,

And soon you'll be back here again

Electronics Math & Ohm's Law - How to Calculate Watts, Amps, Volts, Ohms - Math Formulas, Examples, Answers

Latest update: March 6, 2024. Page URL indicates original publication date; meanwhile, times change and the updates continue.

How to Calculate: Power in Watts, EMF in Volts, Current in Amps, Resistance in Ohms.

A handy math guide for those electrical and electronic math questions.

  • How to Quickly and Easily Find Electronics Answers
  • Using Ohm's Law and Its Derivatives
  • Electronics and Electrical Math Solutions
  • Includes Complete Lessons and Examples

The following formula templates alone might immediately provide the solution.


It is guessed you are here to figure out a math answer to a particular electrical or electronics problem.

This is the place to figure out watts, amps, volts, or ohms from any of the other two by using Ohm's law and its derivatives. The math is surprisingly simple. You should have your answer in no time. Don't forget the templates and table of contents.

In most circumstances, the only math required is multiplication and division. Ohm's law and its derivatives uses some basic letters to represent watts, amps, volts, and ohms.
  • "P" is the industry standard to designate power by the unit of measurement, watts. "W" is sometimes used.
  • "I" is the industry standard to designate current by the unit of measurement, amps.
  • "E" and "V" are both used to designate electromotive-force by the unit of measurement, volts. The industry formula standard used to be "E", but now both "E" and "V" are being used interchangeably.
  • "R" is the industry standard to designate resistance by the unit of measurement, ohms.
And that's all there is to it. No degree in rocket surgery required. No need to memorize, definitions are reprinted as needed.

If your inquiry concerns a particular appliance, device, etc.; check to see if there is any sort of specifications label, metal plate, or even just a sticker. Even if it doesn't provide the outright answer, it will hopefully have enough other information to enable you to calculate the answer from the templates.  If you happen to have the manual (maybe it is still online?), then you may be lucky indeed. As an example, if it tells you it consumes 200 watts and you know your house voltage is 120 volts, then you can easily calculate how many amps it uses and/or what its internal resistance in ohms will be.


Comprehensive List of Ohm's Law Formulas and Examples


Templates and Table of Contents

Here is a list of formulas and templates. With any luck, you will find one you can use and won't have to bother selecting the related title for the included lessons and examples. This is a large file, if you do make a selection it may take a few seconds to display the appropriate section.

Calculate how many WATTS from volts, amps, ohms (Expanded Lesson/Examples).


Formula Templates:

P = EI
Volts  *  Amps  =  Watts

P = E2/R
Volts Squared  /  Ohms  =  Watts

P = I2R
Amps Squared  *  Ohms  =  Watts

Calculate how many AMPS from watts, volts, ohms (Expanded Lesson/Examples).


Formula Templates:

I = P/E
Watts  /  Volts  =  Amps

I = E/R
Volts  /  Ohms  =  Amps

I = √(P/R)
Square Root of  ( Watts  /  Ohms )  =  Amps

Calculate how many VOLTS from amps, watts, ohms (Expanded Lesson/Examples).


Formula Templates:

E = P/I
Watts  /  Amps  =  Volts

E = IR
Amps  *  Ohms  =  Volts

E = √(PR)
Square Root of  ( Watts  *  Ohms )  =  Volts

Calculate how many OHMS from volts, amps, watts (Expanded Lesson/Examples).


Formula Templates:

R = E/I
Volts  /  Amps  =  Ohms

R = E2/P
Volts Squared  /  Watts  =  Ohms

R = P/I2
Watts  /  Amps Squared  =  Ohms

Lessons

There are four, independent, separate tutorials with examples on this page. Simply select the expanded how-to Lessons/Examples sub-link in the above template table of contents specifically addressing that which you you wish to find. Thanks to the laws of physics; whether it be trying to calculate how many amps, watts, ohms, or volts; Ohm's law and it's derivatives will always provide three different, possible ways for finding the answer.

What Is a VOM ( Electronics Definition ) And Some General Notes...

VOM is the acronym for Volt Ohm Milliammeter, more specifically, it is known as a multimeter or a multitester. The usual VOM can measure AC and DC voltage, current in milliamps, and resistance in ohms and megohms. For the purposes of this page, it is usually needed to find the resistance. Once the number of ohms are known, more of the templates and formulas can be used when the usual volt/amp/watt amounts aren't available.

When it comes to test instruments, skip the cheap ones. What a test instrument tells you will in turn result in your making critical decisions. As such, a quality test instrument is much more important than the usual former RadioShack novelty toy, piece of wiring, batteries, etc. And whatever you do, do not buy a kit to make your own test instrument. Buying and building kits for other things is fine, but leave the VOM manufacturing to the professionals with quality reputations (this is the voice of personal experience talking).

Do not buy a VOM until you truly know what you are doing. Cheaper meters can be extremely inaccurate when it comes to measuring certain ranges of resistance, etc. Even voltage and milliampere measurements can be suspect. Really research the subject first. Only by from reputable sources.

Here is an article from Wired Magazine that I really liked, it gets into the more esoteric and physics aspects: How Do You Define Electric Field, Voltage, and Current? The article even covers what to do if you happen to find yourself near a live, downed power line.

The Ohm's Law math lessons and examples follow or select the lessons/examples links from the above formula templates.

A rough map of world power usage.


How Many WATTS - How to Calculate or Convert Watts to and from Any Two of Either Volts, Amps, or Ohms

(P=watts, E=volts, I=amps, R=ohms)

Includes amps to watts and volts to watts.

Watts is the composite measurement of electromotive force and current, otherwise known as voltage and amperage. It is how we quantify electrical energy amounts and usage.

Three ways to figure out the electrical energy amount, measured in watts...

#1. P = EI — Watts Are Equal to Volts Times Amps

 (P=watts, E=volts, I=amps, R=ohms)

Some examples...

  • Tungsten filament light bulb. 120 volts times .8333 amps equals 100 watts. 120 * .8333 = 100
  • Microwave oven. 120 volts times 5.8333 amps equals 700 watts. 120 * 5.8333 = 700
  • Microwave oven. 120 volts times 9.1666 amps equals 1100 watts. 120 * 9.1666 = 1100
  • Some air conditioners. 240 volts times 4 amps equals 960 watts. 240 * 4 = 960
  • Car battery. 12 volts times 3 amps equals 36 watts. 12 * 3 = 36
  • Car voltage when engine is running. 14.5 volts times 3 amps equals 43.5 watts. 14.5 * 3 = 43.5
  • Car battery. 12 volts times 15 amps equals 180 watts. 12 * 15 = 180
  • Car voltage when engine is running. 14.5 volts times 15 amps equals 217.5 watts. 14.5 * 15 = 217.5
  • Most laptop batteries. 19 volts times 3.5 amps equals 66.5 watts. 19 * 3.5 = 66.5
Side note: the prefix, "milli", means one one-thousandth.
  • There are 1000 milliwatts in a watt.
  • There are 1000 millivolts in a volt.
  • There are 1000 milliamps in an amp.

 More examples...

  • A toy using a 9-volt battery consumes 250 milliamps (.25 amps). Multiplying 9 volts by 250 milliamps calculates out to 2.25 watts. 9 * .25  = 2.25
  • A 350-millivolt sub circuit uses 455 milliamps (.455 amps). Multiplying 350 millivolts by 455 milliamps indicates that part of the circuit is using 159 milliwatts (rounded) of energy. 350 * 455 = 159.25
  • A 4.5 volt LED array uses 75 milliamps. Multiplying 4.5 volts by .075 shows the LED array consumes 337.5 milliwatts. 4.5 * .075 = 337.5

#2. P = E²/R — Watts Are Equal to Volts Squared Divided by Ohms

 (P=watts, E=volts, I=amps, R=ohms)

 Some examples...

  • 110 volts squared, then divided by 65 ohms equals 186.15 watts. 110²/65 = 12100/65 = 186.15
  • 120 volts squared, then divided by 125 ohms equals 115.2 watts. 120²/125 = 14400/125 = 115.2
  • 70 volts squared, then divided by 42 ohms equals 116.67 watts.70²/42 = 4900/42 =116.67
  • 12 volts squared, then divided by 24 ohms equals 6 watts. 12²/24 = 144/24 = 6
  • 12 volts squared, then divided by 100 ohms equals 1.44 watts. 12²/100 = 144/100 = 1.44
  • 6 volts squared, then divided by 100 ohms equals 360 milliwatts. 6²/100 = 36/100 = .36
  • A motor requires 40 volts and has an internal resistance of 25 ohms. 40 volts squared, then divided by 25 ohms has a total energy usage of 64 watts. 40²/25 = 1600/25 = 64
  • There are 7.5 volts running through a component with 5 ohms resistance. Its wattage would be a total of 11.25 watts. 7.5²/5 = 56.25/5 = 11.25

#3. P = I²R — Watts Are Equal to Amps Squared Times Ohms

 (P=watts, E=volts, I=amps, R=ohms) stopping point

 Some examples...

  • 1 amps squared, multiplied by 30 ohms equals 30 watts. 1² * 30 = 1 * 30 = 30
  • 5 amps squared, multiplied by 30 ohms equals 750 watts. 5² * 30 = 25 * 30 = 750
  • 14 amps squared, multiplied by 2 ohms equals 392 watts.14² * 2 = 196 * 2 =392
  • 100 milliamps squared, multiplied by 30 ohms equals 30 milliwatts. .100² * 30 = .01 * 30 = .03
  • 334 milliamps squared, multiplied by 15 ohms equals 1.6725 watts. .334² * 15 = .1115 * 15 = 1.6725
  • 750 milliamps squared, multiplied by 5 ohms equals 2.8125 watts. .750² * 5 = .5625 * 5 = 2.8125



How Many AMPS - How to Calculate or Convert Amps to and from Any Two of Either Watts, Volts, or Ohms

(I=amps, E=volts, P=watts, R=ohms)

Includes volts to amps and watts to amps...

It's current and amperage that makes those power meters spin and flips those fuse box switches and circuit breakers on occasion. The 1500-watt space heater is a good example. Microwave ovens can be a close second. An unexpected short circuit in an appliance or house wiring is what causes buildings to burn down if the circuit breaker doesn't do its job.

Three ways to figure out current in amps...

#1. I = P/E — Amps Are Equal to Watts Divided by Volts

 (I=amps, E=volts, P=watts, R=ohms)

Some examples...

  • The aforementioned space heater. 1500 watts divided by 120 volts equals 12.5 amps current. 1500/120 = 12.5
  • The aforementioned microwave oven. 1100 watts divided by 120 volts equals 9.17 amps current. 1100/120 = 9.17
Turning both of those on at once will flip a 15-amp circuit breaker right there. A 20-amp circuit breaker wouldn't be too thrilled with it either.

More examples...

  • 2 watts divided by 6 volts equals .33333 amps current. 2/6 = .34
  • 5 watts divided by 12 volts equals .416666 amps current. 5/12 = .417
Side note: the prefix, "milli", means one one-thousandth.
  • There are 1000 millivolts in a volt.
  • There are 1000 milliamps in an amp.
  • There are 1000 milliwatts in a watt.

 More examples...

  • A 140-watt computer circuit board uses 360 volts from a step-up transformer. This is not a circuit board you want to mess with. Dividing 140 watts by 360 volts shows a current of 389 milliamps running through it. 140/360  = .389 amps (or 389 milliamps)
  • A 300-milliwatt circuit board is connected to a 3-volt power supply. Dividing 300 milliwatts by 3 volts indicates the circuit board requires a current of 100 milliamps (.1 amps). .3/3 = .1
  • A 20-watt device uses standard 120-volt house current. Dividing 20 watts by 120 volts reveals the device is using .1666 amps or 167 milliamps. 20/120 = .167

#2. I = E/R — Amps Are Equal to Volts Divided by Ohms

 (I=amps, E=volts, P=watts, R=ohms)

 Some examples...

  • 240 volts divided by 500 ohms calculates to a current of 480 milliamps. 240/500 = .480
  • 110 volts divided by 2000 ohms calculates to a current of 55 milliamps. 110/2000 = .055
  • 12 volts divided by 250 ohms calculates to a current of 48 milliamps. 12/250 = .048
  • A tiny, hobby motor needs 3 volts to operate and has an internal resistance of 40 ohms. 3 volts divided by 40 ohms indicates a usage of 75 milliamps. 3/40 = .075
  • There are 9 volts running through a controller with an internal resistance of 135 ohms. 9 divided by 135 equals a current usage of 67 milliamps. 9/135 = .066666

#3. I = √(P/R) — Amps Are Equal to the Square Root of the Quotient of Watts Divided by Ohms

 (I=amps, E=volts, P=watts, R=ohms)

Contrary to the general introduction, this third and last resort does involve the use of square roots; so break out the calculator, spreadsheet, or search engine if you haven't done so already.

Basically, all one does is divide watts by ohms; then just find the square root of the quotient to determine the amperage.

"" is the symbol for square root.

 Some examples...

  • 100 watts divided by 4 ohms gives us a quotient of 25. The square root of 25 is 5 amps.  √(100/4) = √25 = 5
  • 900 watts divided by 5 ohms gives us a quotient of 180. The square root of 180 is 13.42 amps (rounded).  √(900/5) = √180 =13.4164
  • 40 watts divided by 40 ohms gives us a quotient of 1. The square root of 1 is 1 amp.  √(40/40) = √1 =1
  • 5 watts divided by 100 ohms gives us a quotient of .05. The square root of .05 results in an answer of 224 milliamps (rounded).  √(5/100) = √(.05) =.2236. Square roots of numbers less than 1.0 are odd that way.



How Many VOLTS - How to Calculate or Convert Volts to and from Any Two of Either Watts, Amps, or Ohms

(E=volts, P=watts, I=amps, R=ohms)

Includes amps to volts and watts to volts.

Unlike with most watts and amps questions, voltage and voltage-drop questions usually have to do with circuit boards and their subcomponents. However, here are also some basics...
  • Typical US house voltage is 120 volts; though for certain appliances, voltage is boosted to 240 volts.
  • The car battery standard is 12 volts.
  • The laptop standard is most often 19 volts.
  • Standard carbon or alkaline batteries (whether sizes D, C, aa, aaa, etc.) are all 1.5 volts each. Putting them in series is simply additive. As an example, if you see a 6-volt flashlight being advertised, you know it will require four batteries.

Three ways to figure out volts...

#1. E = P/I — Volts Are Equal to Watts Divided by Amps

 (E=volts, P=watts, I=amps, R=ohms)

Some examples...

  • 500 watts divided by 5 amps equals 100 volts. 500/5 = 100
  • 12 watts divided by .1 amps equals 120 volts. 12/.1 = 120
  • 150 watts divided by 2 amps equals 75 volts. 150/2 = 75
  • A 6-watt car instrument cluster has half an amp running through it. Is the car engine running or not? Dividing the 6 watts by .5 amps gives us 12 volts. The engine is off (when the engine is running the system voltage ranges from 14 to 14.5 volts). 6/.5 = 12
  • A 600-watt starter for a small engine requires 50 amps. Dividing 600 watts by 50 amps indicates that a 12-volt battery can indeed do the job. 600/50 = 12
Side note: the prefix, "milli", means one one-thousandth.
  • There are 1000 millivolts in a volt.
  • There are 1000 milliamps in an amp.
  • There are 1000 milliwatts in a watt.

 More examples...

  • A 400-milliwatt (.4 watts) circuit board uses 80 milliamps (.080 amps). Dividing 400 milliwatts by 80 milliamps indicates it is connected to a 5-volt input. 400/80 = 5
  • A 180-milliwatt component uses 45 milliamps. Dividing 180 milliwatts by 45 milliamps equals 4 volts. 180/45 = 4

#2. E = IR — Volts Are Equal to Amps Multiplied by Ohms

 (E=volts, P=watts, I=amps, R=ohms)

 Some examples...

  • 10 amps multiplied by 12 ohms equals 120 volts. 10 * 12 = 120
  • 35 amps multiplied by 42 ohms equals 1470 volts. 35 * 42 = 1470
  • .5 amps multiplied by 6 ohms equals 3 volts. .500 * 6 = 3
  • An air conditioner requires 50 amps. The motor, pump, and other circuitry has a total resistance of 4.8 ohms (surprisingly low actually). That A/C will require 240 volts to operate. 50 * 4.8 = 240
  • There are 600 milliamps running through a circuit with a measured resistance of 5 ohms. So that would be 600 milliamps times 5 ohms, giving you 3 volts. .600 * 5 = 3

#3. E = √(PR) — Volts Are Equal to the Square Root of the Product of Watts Times Ohms

 (E=volts, P=watts, I=amps, R=ohms)

Contrary to the general introduction, this third and last resort does involve the use of square roots; so break out the calculator, spreadsheet, or search engine if you haven't done so already.

Basically, all one does is multiply watts times ohms; then just find the square root of the product to determine the voltage.

"" is the symbol for square root.

 Some examples...

  • 14 watts multiplied by 10.285 (rounded) ohms equals a product of 144. The square root of 144 is 12 volts. √(144 * 10.285) = √144 = 12
  • 300 watts multiplied by 20 ohms equals a product of 6000. The square root of 6000 is 77.46 volts (rounded).  √(300 * 20) = √6000 = 77.46
  • A 900-watt microwave oven magnetron has an internal resistance of 15 ohms. 900 watts times 15 ohms gives a product of 13,500. The square root of 13,500 is 116 volts (rounded).  √(900 * 15) = √13500 = 116.2. What with house voltages ranging from 110 to 120 volts, that will work just fine.
Side note: the prefix, "kilo", means one thousand.
  • There are 1000 volts in a kilovolt (kv).
  • There are 1000 amps in a kiloamp (ka).
  • There are 1000 watts in a kilowatt. (kw).

 An example...

  • 1,000 watts (1kw) multiplied by 10 ohms equals a product of 10,000. The square root of 10,000 is 100 volts.  √(1000 * 10) = √10000 = 100



How Many OHMS - How to Calculate or Convert Ohms to and from Any Two of Either Watts, Volts, or Amps

(R=ohms, E=volts, I=amps, P=watts)

Unlike with most watts and amps questions, resistance and ohms questions usually have to do with circuit boards and their subcomponents. However, the internal resistance of an appliance or device greatly affects how much power it uses. The classic example of this is the incandescent, tungsten filament light bulb. A single, 100-watt bulb requires almost a full amp of current at 120 volts. That can add up fairly quickly over time. Power meters love it, everyone else hates it.

Three ways to figure out resistance in ohms...

#1. R = E/I — Ohms Are Equal to Volts Divided by Amps

 (R=ohms, E=volts, I=amps, P=watts)

Some examples...

  • The aforementioned light bulb. 120 volts divided by .8333 amps equals 144 ohms resistance. 120/.8333 = 144
  • 240 volts divided by 3 amps equals 80 ohms resistance. 240/3 = 80
  • 12 volts divided by 1.50 amps equals 8 ohms resistance. 12/1.5 = 8
  • 19 volts divided by 2.3 amps equals 8.26 ohms resistance. 19/2.3 = 8.26
Side note: the prefix, "milli", means one one-thousandth.
  • There are 1000 millivolts in a volt.
  • There are 1000 milliamps in an amp.
  • There are 1000 milliwatts in a watt.

 More examples...

  • A 9-volt circuit board uses 140 milliamps (.140 amps). Dividing 9 volts by 140 milliamps indicates the board has an internal resistance of 64.29 ohms (rounded). 9/.14 = 64.29
  • A 500-millivolt component uses 120 milliamps. Dividing 500 millivolts by 120 milliamps indicates the component has a resistance of 4.17 (rounded) ohms. 500/120 = 4.17
  • A 4.5 volt LED array uses 15 milliamps. Dividing 4.5 by .015 equates to a resistance of 300 ohms. 4.5/.015 = 300

#2. R = E²/P — Ohms Are Equal to Volts Squared Divided by Watts

 (R=ohms, E=volts, I=amps, P=watts)

 Some examples...

  • 120 volts squared, then divided by 100 watts equals a resistance of 144 ohms. 120²/100 = 14400/100 = 144
  • 50 volts squared, then divided by 35 watts equals a resistance of 71.43 ohms.50²/35 = 2500/35 = 71.43
  • 6 volts squared, then divided by 4 watts indicates a resistance of 9 ohms. 6²/4 = 36/4 = 9
  • A motor requires 36 volts and uses 40 watts of power. 36 volts squared, then divided by 40 watts has a total resistance of 32.4 ohms. 36²/40 = 1296/40 = 32.4 
  • There are 1.5 volts running through a component using 2 watts. Its resistance would be 1.125 ohms.1.5²/2 = 2.25/2 = 1.125

#3. R = P/I² — Ohms Are Equal to Watts Divided by the Square of Amps

 (R=ohms, E=volts, I=amps, P=watts)

Some examples...

  • 150 watts divided by 7 amps squared. The 7 amps squared is 49, so we have 150 watts divided by 49; giving us an answer of 3.06 ohms. 150/7² = 150/49 = 3.06
  • 40 watts divided by by 20 amps squared. The 20 amps squared is 400, so we have 40 watts divided by 400, giving us an answer of .1 ohms or 100 milliohms. 40/20² = 40/400 = .1 We are pretty much looking at a 2-volt short circuit on a board that needs fixing, probably a shorted out capacitor.
  • A 500-watt refrigerator divided by 11 amps squared. 11 amps squared is 121, so we have 500 watts divided by 121, giving us an answer of 4.13 ohms (rounded).
  • A 5-watt circuit sub-board consumes 300 milliamps. So the equation is 5/.3² to give us the resistance in ohms. .3² is .09, so we have 5/.09 = 55.56 ohms (rounded) in calculated resistance.



A Final Thought...

Do be careful. The laws of physics are unforgiving.

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