ohms law clarification

Good day,can someone please give me more clarity on the below.I don't understand this part that I got from the internet.   Definition of 1 Ohm The unit by which electrical resistance is measured. One ohm is equal to the current of one ampere which will flow when a voltage of one volt is applied I understand how current,resistance and volt operate,I just don't understand this specific sentence from the net. Thanks in advance,You may roar at me if I'm imprudent.Do it once,do it right - This means no short cuts. You never get paid twice for having to do the job twice because it wasn't done right the first time.

If you have a 1 ohm resistor, placing 1V across it will result in a 1A current flowing thru the resistor. If you have a 1 ohm resistor, a current of 1A thru it will cause a voltage drop of 1V across it. The current flowing thru a resistance if 10 Ohm connected across a 12V battery: from V=IR, I=V/R = 12V/1Ohm = 1.2Amps.

I'm coming there slowly. So just an example. if you have a small fan in your car that needs a voltage of 12volts and current of 10amps to power it,and the distance from the power supply (12v car battery) to the location where the fan will be mounted is 3feet,I need a 10amp fuse and a 16ga wire rite? Now what I understand is that my fan is getting the correct volts and the correct current to operate correctly. R= V/I R=1.2ohms. Correct me if I'm wrong. So there's a resister of 1.2 ohms built in this fan that controls the current flowing from the battery to the fan. which means,if there was no resister in this fan,what would have happened to fan?was the fan going to burn out as it is already receiving its correct amount of current flowing? When are resisters used except on leds which are miliamps .I hope I'm not confusing you. What will happen if I used a 20amp fuse and a 14ga wire for the distance of 3feet for a fan that only needed 10amps?I hope you notice where I'm getting lost in ohms law.Do it once,do it right - This means no short cuts. You never get paid twice for having to do the job twice because it wasn't done right the first time.

Another thing that is confusing me. Example. if you have five of 5meter led strips with a current of 2amps each which gives me 10amps together,running from a 12volt dc power supply with the current of 10 amps ,plucked in to a 220volt ac wall. So if I add another 5meter with the current of 2 amps which make it 12 amps on a 10amp supply,the leds will try to draw 12amps out of a 10amp power supply and the power supply will burn rite? Why does the 10amp power supply not burn out 1meter of led strip as the strips need to be five 5meters to be 10amps.I think my question relates to the below. Can you use components with lower amp ratings on power supplies with higher amp ratings?where as I already know if I use components with higher amp ratings will kill the supply with the lower amp rating.Do it once,do it right - This means no short cuts. You never get paid twice for having to do the job twice because it wasn't done right the first time.

Ah, I'm totally with you. This stuff (Ohm's Law) is so simple, but like so many simple (in retrospect) things can be , tricky as all heck to initially understand. So let's see if I can help the penny drop... Starting with your 2nd last reply... Forgive my pedanticness, but IMO we should cover important associated issues. I hope that's not too much. All I can suggest is consider each aspect or issue on its own before moving on. That'd be easy if we were in a classroom where I could gauge your understanding. (parenthesised text) is usually a clarification whilst [bracketed text] - especially if [bold] is extra FYI - maybe to help clarify by analogy or another example, or for your future use (ie, prevention of common mistakes).    For your 10A fan you would not use a 10A fuse. I'd probably used 15A since fuses (and many things like wiring and components) should NOT be run at their full rating. A general rule for fuses is to not normally run at more than 70% of rating. Hence 10A/.7 = 14A => 15A. If it were a 12A fan I might still use 15A fuse. That means it runs at 12/15 = 80% of rating but that may be ok. (70% is a ROT (Rule Of Thumb) and like all ROTs they are generalisations and not specific to all situations. ROTs may be a simple "first guess" design estimate. Besides, Rules are meant to be broken right? (But make sure you can justify why you broke them when the authorities come after you!! ) 16ga wire has a resistance of 4.016 Ohms per 1000' => 3 x 4.016/1000 = 0.012 Ohms (12mR). Ok, so I've started using "R" instead of Ohms. That is common on webpages and in electronic documentation since (1) the Ω symbol can is not a normally accessible character and (2) the Ω symbol does not always show correctly on all systems. [ In fact Ω often shows as "W" which is real confusing because W = Watts is a legitimate electrical value and you think the author is an idiot by specifying a 1.2W resistor for the fan! ]. Needless to say, we - or I - would rather not write Ohms or Volts each time; we use single letter abbreviations wherever possible. Using "R" has its practicality too. Because decimal-points are easy to lose or hard to see, instead of printing a resistor or circuit value as 1.2Ω or 1.2R, it will be marked as 1R2. That is especially common on SMDs (surface mount devices/components). Back to the 16ga. So 12mR over your 3' run means a 10A x 12mA = 120mV = 0.12V voltage drop across it (from V=IR). Hence there is a 0.12V voltage drop across your 3' 16ga wire @ 10A, or 2 x .12 = 0.24V is your GND is the same length. Hence you need a 12.12V or 12.24V battery. Lucky batteries are really 12.6-12.7V eh, or ~14V when being charged? But forget the latter pedanticness re 12.24V or 14V batteries etc. One bit at a time. KIS = Keep It Simple and stick with simple 12V etc. I should say that I am not used to working with wire gauges. (I use a different method to size wires - namely similar to what I did above.... I decide the max voltage drop I want and thereby work out the maximum wire resistance and hence determine the minimum wire diameter (gauge).) I see from powerstream's American Wire Gauge Table that 16ga is suitable for 22A in chassis or 3.7A in transmission wiring. So I might therefor use 10ga (55A chassis or 15A transmission) so 10A is fine irrespective of whether it's hanging in the air or the only wire without other wire or chassis heat conduction. You can read other threads about fusing - ie, how it is to protect the wiring (ie, the distribution, not the load/fan) and hence is placed as close to the power source (battery) as possible and the fuse must be rated no higher than the LOWEST rated thing its protecting. Back to the fan and its "1.2R resistance". The fan does not have a built in 1.2R resistor, The fan itself has a 1.2R resistance. IE - its wiring is effectively 1.2R. [ Not that I want to go down this path (yet!), but a fan is actually an inductor - ie, a coil - and hence has an impedance. Impedance is merely (merely? ha!) "AC resistance", but that's for another day. Being an inductor (motor) means it will typically have a high inrush current. Hence another reason for using a fuse larger than its normal running current of 10A (tho "slow blow" fuses could also be used). [But also if you realise that in practice, if it was 1.2R and hence 1.2A @ 12V, it will "normally" be 14.4V/1.2R = 11.75A at 14.4V - the normal maximum long-term voltage of 12V automotive systems. (Most modern 12V vehicles regulate to 14.2V but 14.4V is the normal maximum above which most lead-acid batteries will gas excessively and hence be damaged if excessive.) ] So you can't "remove" the fan's 1.2R resistance. But if you could (eg, by removing windings), then yes, its resistance (impedance) will drop, hence it takes more current and can thus overheat and melt or flame if it can't handle that current or heat etc. EG - if you halve its resistance, its current will double with the same voltage applied (I=V/R). So instead of 10A it will take 20A. In that case our 15A fuse with 15A or 25A wire will blow. So to will a 20A fuse with 25A wire (thou it will take a bit longer to blow, but fuse blowing (or "clearing" aka rupture) times are also for another day! Don't worry about WHAT it is in a load or component that limits its current etc. (That happens later as you learn more.) Just accept that it is whatever it is - eg, 1.2R, or 10A @ 12V, or 20mA @ 2.2V (LEDs) etc. You (simply??) need to supply whatever voltage it requires and it will look after itself. Later you'll understand that Ohms Law applies to each individual component, and each individual series "string" of components, and parallel combinations; and the circuit as a whole. Your last part (2nd last question) is another tricky. I have partly covered it with my 15A fuse & 25A wire and how you simply provide the correct voltage (the load "regulates" itself). It doesn't matter how big or fat the supply (including wiring) is as long as it is the correct voltage. You probably know the latter but don't realise it. You have a battery that can supply well over 100A (eg, to the starter motor) yet it does not blow 10A fans or 12W (1A) bulbs or 20mA LEDs. Many fear that adding more or bigger batteries or alternators can blow loads but no - they can't provided they are the same voltage. For that I use the good old electricity-water analogy.... Voltage is water pressure. Water pressure increases with height/depth. Current is the quantity of water (whether thru the pipe or available in the tank or dam). Resistance is the hose or pipe and end-device (bucket, nozzle etc). Picture a huge dam full of water. You have a pipe thru the wall (anywhere below the water level) into the side of a glass or bucket. The glass has the same water level as the dam. You lower the glass - it fills. You lift the glass and it fills (empties into) the dam. (Imagine that - your insignificant glass can push water into that huge dam!) It's a case of pressure - not quantity. It doesn't matter how fat the pipe is, it's all about the level of the glass or bucket. Likewise electrical loads. As long as they are rated for the same voltage - ie, the same pressure (the water surface height) they'll be fine. Hence why your MegaWatt (mega-Amps) power station doesn't not blow your 300mA phone charger or your 12V 60AH battery (that can supply say 200A @ 12V) doesn't blow your 250mA (3W) dash bulbs. The above may not be understandable per se, but I suggest accept it is an observable fact or knowledge. It's when you "reverse engineer" that that it become understandable. EG - why doesn't a 12V 3W bulb blow? (P=VI so I=P/V = 3W/12V = 250mA.) A 3W bulb is 12V/250mA = 48 Ohms (48R). So lets blow it by putting in 500mA. V=IR so V = 500mA x 48R = 24V. Si it'll then have 24V across it, or, in other words, we need 24V across it to get 500mA thru it. But we only have a 12V battery so we can't get 24V and hence 500mA thru it. So how much can we get? Well, we answered that above: I=V/R = 12/48 = 250mA & .25A x 12V = 3W for which it is rated for. We can only increase the current thru a load by decreasing its resistance OR increasing its applied voltage. Hence why automotive systems must not exceed a certain voltage (usually taken to be 16V but some loads may tolerate less). And loads must handle their current & power dissipation at their rated voltage. A 3W bulb will handle its 3W or heat and light. If we halve its resistance to make a 6W bulb, we may to increase its thermal capability to handle the extra heat (tungsten bulbs are typically 95% heat and 5% light) - so we double its ability to remove heat (or use different materials). I hope the above helps tho I know I have written a lot and digressed somewhat, and no doubt I can rewrite it much better. But I did cover a various related or practical consideration. Sometimes they confuse more but sometimes they help understanding, or prevent confusion further down the track. PS - maybe I should have emphasised that Ohm's Law V=IR means that the voltage drop across a resistance equals is resistance times the current thru it.

The last question you posted has been answered in my last reply. BUT I hope you mean you are plugging your 12V LEDs into a 12V supply - NOT the AC wall socket!! (I refuse to fly so I can't attend your funeral.) And note that 12V LED strips are 12V - NOT automotive 12V nor for 12V batteries (which are ~12.6V or more fully charged). They need a "12V regulated" supply. Each 12V string would be connected in parallel across the 12V supply. Each string takes its Amperage so if 2A each, the 3 strings would take a total of 6A. FYI - 220VAC @ 10A can supply up to ~180A @ 12V; probably 150A in practice. So with a big enough 220VAC to 12V DC converter you could supply up to 75 12V 2A strings of LEDs.

After 1 day of reading your answers makes it very clear for me ,thank you I understand now why my led blew up few years back as I connected it straight to 12V and it could only take 2v. I also understand that no matter how big and fat the wire is,as long as the voltage is the same. So the only way I will damage a component is when the components voltage is less then the 12V supply. So what causes wires to burn? If a 0gauge wire ,and a 250amp fuse is right for a 3000W car amp. what will happen if I use a 12gauge wire and a 50amp fuse to the same amp and the same voltage 12V? I assume the 50amp fuse will burn out,and if there was no fuse in the circuit the wire would have burn?or will the wire not burn,but the amp will not put out its full power due to more resistance and lack of current passing through. I think this will be my last question as you already made everything very clear to meDo it once,do it right - This means no short cuts. You never get paid twice for having to do the job twice because it wasn't done right the first time.

Heat melts things and causes fire etc or burns component conductors. The heat generated by a component is its resistance times it current-squared. So a 1 Ohm wire or load needs to dissipate twice the heat of a 2 Ohm load/wire at the same voltage. FYI - at the same current (ie twice the voltage of the 2 Ohm) the 1 Ohm generates FOUR times the heat, hence why overvoltage can kill rather quickly. [ That's from V=I*R & P=V*I, hence P=V*V/R or P=I*I*R ] So a thinner wire has n-times the resistance hence n-times the heat to get rid of (assuming it passes the same current). Since it is thinner, it also has less copper thermal conductivity, and less surface area to convect or conduct or radiate heat. A fuse is essentially a sacrificial wire. Instead of melting your (say) 250A wire, you melt your safely rupturable fuse instead.

Good day Old spark,just want to thank you for your patience with me.you have a very good way of explaining and making things clear to someone.thumbs up Do it once,do it right - This means no short cuts. You never get paid twice for having to do the job twice because it wasn't done right the first time.