Just Getting Started and Stuck

Yeah I'm going to be that guy...that replies to his own post...
I actually knew watts of power= current * VOLTAGE. so 8 volts and 1 amp of dc requires a resistor rated at 8 watts... is this correct? It seems way high to me.
(sorry about the first post. Did it over break from work.)

Have you read about Ohms law.

V=IR

OR I =V/R

so in your case I = 8/330 = 0.024

W = IV = 0.024 x 8 = 0.2 Watts

Or you can take short cuts like W = IV = I²R = V²/R

Suppose you had a 50 Amp, 8V supply, and you put your resistor across it, wouldn't you expect to get the same result.

Most power supplies have a fixed voltage output, and, can supply up to the specified current.

Some bench supplies will allow you to set either the voltage OR the current, but you can't set both.

Ohms law is very simple, but, will get you a long way in electronics.

Here is an analogy that might help you. Voltage is like water pressure, current is like water flow, and resistance is like a small orifice that creates back pressure when water flows through it. The power supply is like a water pump that can pump 1 amp of water flow or produce 8 volts of water pressure, but not both at the same time.

How To Calculate And Understand Resistor Values | Kitronik

My favorite: Electronics Components: Resistors - For Dummies

Power = current squared times resistance.

Power = voltage squared divided by resistance.

Power = voltage times current.

Use whichever formula you have values for.

Alright let me try again... so I have 8 volts of power and 1 amp of current. P=IE or 8*1=8watts. Am i to understandThe Power RATING of the resistor needs to be 8 watts? (My resistor is 330 ohms but is only rated up to a 1/4th watt. SO it's not usable...)
Man, these resistor things seem really weak. Maybe I'm shopping in the wrong place for them, cause I didn't see anything near 8 watts rating at microcenter. :| are 8 watt resistors common at all? Or is what I'm failing the grasp the point that 1 amp of current is rather large?
(Thanks everyone for replying by the way. I've been trying to get to the bottom of this on my own for awhile. I just wish I could find a good power source to test with... because I've tried a c battery and 1.5volts by 1.5 current is too much for my resistors as well. )

Perhaps this might help. The current of the power supply, 1 amp, is the MAXIMUM current available from it. It will supply up to 1 amp and above that other factors, usually unique to the power supply, start getting involved so forget about that here.

The resistance is the amount of "blockage" to the current. When the current gets higher this blockage has more effect on the power rating, 1/4 watt, 1/2 watt, etc., which is the total of the electricity being taken from the power supply.

Ohm's Law says the current ( I )is equal to the applied voltage (E) divided by the resistance (R). Shuffle everything around and E=I*R and R=E/I. The catch to all this is that E= applied voltage not available voltage and I = the current allowed by the resistance and the amount of voltage applied to it.

So for your situation if E= 12 volts (available and also applied here) and the resistance is 330 ohms then I=E/R, (E, 12 volts)/(R, 330 ohm) = 0.036 amps or 36 milliamps. This is the current flowing through the resistor when 12 volts is applied to the resistor. Raise the voltage and more current flows, raise the resistance and less current flows.

Power is the voltage (volts) times the current (amperes) and results in power (watts). So, 12 volts * 0.036 amps = 0.432 watts. A 1/2 watt resistor will theoretically be fine, just don't touch it. They get very hot when they are used around their maximum rating. For comfort reasons, if you are going to touch it use a 1 or 2 watt resistor. Much easier on the skin.

If you wanted full capacity current from your power supply then 12 volts/1 amp = 12 ohm resistor. Power would be 12v * 1amp =12 watts.

Good Luck, Old Salt

I see your mistake, and it took me some time to see this in my own study, Your power supply provides an electrical potential (12 V) The load, a resister in this case, determines the current drawn from the supply. The lower the resistance, the more current drawn, the higher the resistance, the less current drawn. The 1 amp rating of your supply is just the maximum current that could be drawn without burning up (or letting out the smoke as we say)

My one advice, Keep at it, keep studying. and keep coming here with your questions. There are members of this forum that know more about this subject than I could ever learn in my lifetime, and are willing to help you learn.

It's 8 volts OR 1 amp...not both at the same time! The load (resistor) determines how much current is drawn. If you put a 1k resistor, you have 8 volts, it draws 8 ma, the power is 64/1000 or 64 mw (1/4 Watt resistor is OK). If you put a 100 ohm resistor on the power supply, the current is 80 ma, the power is 64/100 or 640 mw (you need a 1 Watt resistor).

what if he adds a 20 ma LED??

Don't go there!

8 watts the least. You may increase wattage with the same 8 watts resistor having it connected as a network(parallel and series connection)

Not likely..It will increase the wattage of the circuit, but not the ability of the resistor to handle that wattage.

For example..

Adding a second identical resistor in parallel will halve the total resistance and double the circuit current, possibly destroying the power supply in the process.

The circuit wattage will double, but each resistor will still be carrying the same current with the same voltage across it, and dissipating the same wattage.

Adding another third resistor in series with those parallel two will result in an increase of total circuit resistance by 50% and a reduction in the circuit current and wattage by 33% over the original single resistor circuit. This may save the power supply and the resistors, but the entire circuit dynamics have now changed.

The whole reason for a resistor in a circuit is to control current flow through, or voltage across, some other component, putting it alone across a power supply serves no useful purpose

A good rule of thumb for resistors in parallel is that the total resistance of the parallel network will always be less than the value of the smallest resistor in the group ..ie a 1 megohm in parallel with a 1 ohm will result in a total resistance of 0.99 Ohms.

You are assuming that the rated current will actually flow through your resistor. WRONG! Your voltage (8V) is all the pressure you have to start with, so start there. 8V / 330ohms gives you .024 Amps, not 3 amps.

Voltage is electro-motive force. Your resistor will reduce that force, and if it is the only reduction then the applied voltage will "drop" across your resistor. When you place the resistor into the circuit and measure the voltage between the two ends of the resistor, you will read 8V. The resistor will allow only so much current through at 8V, in this case 0,024A which is 24mA. Now, multiply the current by the voltage and you will get the Watts. 8v x ,024A = 0.192W. A 1/4W resistor is plenty as 1/4W = 0.250W.

I too have been learning about electronics. Although I have been dabbling around the subject throughout my career as a tool maker I never had a formal course in electronics (just college level Physics) However recently, I came across a lecture series from The Geat Curses , Understanding Modern Electronics by Richard Wolfson that really got me going on the subject. The series introduces some good online circuit modeling programs.

BTW Power (Watts)= I2R ( not IR) and 12 V through 330Ω resistance will draw .036 amps current. (I=E/R) You need to study Ohm's Law.

and you need to read my post more thoroughly, because I said the actual output voltage was 8. A few times now. Yippee. And I've seen the power equation written both ways. I know I screwed up in my original post, but the thing I've really been searching for is whether or not im calculating the power rating right. (WHICH I FAILED TO DO IN ORIGINAL POST) Alright.... so
1^2 * 8 volts... = a wattage or power rating of 8 volts. IS THIS CORRECT?!?!?!? (sorry to use caps. It's just I screwed up in the original post, and since then all I wanted was verification on this.) THE VOLTAGE IS 8. THE AMPS ARE 1. I HAVE A 330 OHM RESISTOR WITH A POWER RATING OF 1/4 WATT. AM I TO UNDERSTAND THAT ANY RESISTOR I USE WILL REQUIRE A POWER RATING OF 8 WATTS?
(I honestly apologise for the caps. just wanted to get my point across and I was obviously failing...bad. I'll be more brief and to the point next time.)

Here is a question for you. Why does your meter measure 8 volts from a 12 volt power supply. Hint: the meter draws some current from the circuit, and the power supply has some internal resistance. How does your volt meter measure voltage?

That's doubtful.

Even the cheapest of analogue multimeters will have an input impedance of around 20,000 Ohms per volt (most digitals will be 1 megohm), so if it's analogue and he were to set it on the 20 volt scale, the input impedance would be 400,000 Ohms.

If the power supply is outputting 12v (and most unregulated power supplies, which is what his sounds like, would be outputting closer to 20 volts no load) the maximum current would be just 30 microamps. That's not going to cause any voltage drop across the internal resistance of the power supply.

Teaching Myself -

1) What you missed to understand / interpret is the power supply rated power! Your (PS) power supply is rated to deliver a maximum of 8 volts at 1 ampere.. or 8 Watts of maximum power! Depending on the PS design and efficiency, the PS may also be capable to supply 8 ampere @ 1 volt, which also equals its rated capacity of 8 watts or any combination thereof, as long as you do not exceed its maximum rated capacity..

2) The Load is resistor value you choose to connect to the PS. It will determine the amount of current it will pull out or draw from the PS. If the current it draws is too big, it may trip the fuse or completely burn the PS!

3) The wattage rating of your chosen resistor is directly proportional to the current it can handle and safely pass through it. Remember Mr. OHM?

hint; the smaller the R value the higher the I current for the same W!

Always remember, It is very important that you always match the capacity to any given load to maximize the power transfer!

Sorry, this link is all messed up. try www.TheGreatCourses.com

when I first started playing with electronics I had a small breadboard a pack of varied resistors and caps. I didn't do a power supply. I did all the math 1st then used a 9V battery, when i needed more power i added another battery

https://www.youtube.com/watch?v=fCoqizjnLKI&list=PLMHXYn3fM_dyM60QUx_CLj8hgvw0Gsk7N

The hostility here is too much.

<unsubscribe>

Would like to apologise to you personally gringogreg. I was rather hostile and i was wrong. It wasn't anyone else's fault that i typed way too much and wasn't clear in my original post. I'm sorry, it's just that it seemed so basic bit i couldnt make sense of it.... Everyone's posts were valuable and right from the start. It was just the one detail i overlooked it i look at the current *that the resistor will carry* not the original current *before it hits the resistor* . im sorry everyone....was just driving me mad.

Just kidding.

I have nothing to add besides the two links.

I'm all for helping, but I don't know ohms from alms.

TeachingMyself you came to the right place.

Welcome.

The mistake you are making is thinking that you will always push 1 amp through the resistor. Most (not all) power supplies you will run across will be a constant voltage, and the circuit you build will determine the current (up to the capability of the supply).

Generically (E=I*R): E (voltage) = I (current) * R (resistance). This can be rearranged algebraically: I=E/R

Specifically: I= 8 (volts) / 330 (Ohms resistance), so the current flowing through that resistor will be ~0.024... Amps

Since W (power in watts) = E (voltage) * I (current in Amps), you will have 8 * 0.0242424... ~ 0.1939 Watts. That means that a 1/4 Watt resistor will work just fine.

You can shorten the above equations but doing algebraic substitution.

I=E/R, W=E*I, therefore by substituting E/R into the W=E*I equation, you come out with W= E * E/R or W=(E*E)/R

Oh my god thank you so much. Problem solved. Thanks everyone for their help. Sorry if i usex caps....my deskjob has been slowly killing me. Thanks everyone, and ill stick adound and read thr forums. Thank you once again kilowat. I understand it now.

PLEASE keep coming back. Every day I learn something from the people on this forum, and it is a pleasure to contribute what little that I know to it.

Your welcome. And as gringogreg said: Please keep coming back.

I have worked in the mechanical/electrical/electronic/software/process control business for ~ 39 years, and I usually average 3-4 saved links from this site every week. It's an amazing site, but one word of warning: We are an ornery bunch, so you will need to grow a bit of a thick hide - and yes, it's worth it.

Being up front about what is confusing you, and what you have done to try and figure it out is a good start - keep it up!

speak for yourself! I'm sweet as hell

That's the same thing I tell my wife and daughter - they say I lie like a rug!

Yes, this is the stumbling point most beginners have. The standard jargon defining a power supply leaves many critical terms out. The complete identification for this supply is a 12V constant voltage supply with a maximum 1 ampere of current. Many a cheap power supply will provide a constant voltage output only if the input voltage is constant. Power supplies that do not require a constant input voltage are called regulated voltage supplies.

As for why this 12V supply is giving you 8V instead, there are multiple plausible reasons for this:

• The supply is failing or mislabled. New devices do fail sometimes.
• 8V*√2 is almost 12V. You might have your meter on the wrong setting of AC or DC voltage.
• Your input voltage is too low for the supply to produce the correct voltage output.
• The regulator in the power supply is getting fooled by the low power draw of just the meter for a load. This is a very atypical reason today but I had to mention it.

I would ask you why you would want to put that resistor across the output of the power supply? What is the purpose of a resistor? You are using it as a 'mini heater'? Then you should use Ohm's law to calculate a resistance that won't give more 'wattage' through the resistor than what it can dissipate as heat. If you can add a fan, i.e. blow air across it, you can lower the resistance as the moving air will carry heat from this 'mini heater' and dissipate more heat.

The normal use of a resistor is as a 'throttle device', i.e. a device to lower the voltage available to a second device such as an LED, which is current operated. You select a series resistance that allows a desired current flow through the second device. The 'voltage drop' across the resistor X the current through the resistor, gives you the watt size required so one of the devices doesn't become a 'fuse', ( a circuit opening device used to protect other more valuable things in the series (current path).

Study what the use of the devices are, then decide how to wire them together.

TeachingMyself,

You have just been exposed to a very large pitfall of this and most other forums.

It's the "yes, but".....syndrome.

Stick around. It's more fun here than anywhere else I've been and more civil than many, if the whiners are ignored.

http://www.diyaudio.com/forums/everything-else/262451-volt-ohm-amp.html