Im doing automotive engineering and im struggling to get a grip on Ohms law. Was wondering if anyone here understood it that could give me a hand with figuring out a way to make it stick in my brain.
desertcolt07,
Mar 29, 10:05pm
v=ir
desertcolt07,
Mar 29, 10:06pm
voltage = current x resistance
crazy126,
Mar 29, 10:07pm
the issue im having is working out the ohms and amps when all i have is volts and watts.
tonyrockyhorror,
Mar 29, 10:10pm
Divide Watts by volts (W = V x A). That gives amps. then V = I x R transposes to R=V / I.
You have calculated the resistance.
crazy126,
Mar 29, 10:11pm
thanks ill have a crack and see how i go. Im doing well with all the practical stuff but its just ohms law im struggling with.
tonyrockyhorror,
Mar 29, 10:19pm
Don't forget the units. Watt Volts Amp Ohms
They might throw a curveball and say it was 100mW which you have to factor in by dividing by 1000 (milli) or the units of the calculation will not be what you expected.
For example: if you have a 50W load on 12V then you have 50W / 12V = 4.17A. 12V = 4.17A x R R = 12V / 4.17A R = 2.88 Ohms
Also: W = I² x R R = V² x I W = V² / R (that would be the most direct route to resistance, if you can remember it)
tonyrockyhorror,
Mar 29, 10:21pm
Do enough of them and it becomes second nature.
crazy126,
Mar 29, 10:23pm
lol thanks for that that actually helps heaps.
i will get there by the end of the year lmao
tonyrockyhorror,
Mar 29, 10:29pm
I can't remember that stuff, especially the V² etc. only the basic (V = I x R) & (W = V x I) and then I derive the rest by substitution and transposition i.e. replace V with (I x R) in the W = V x I and you get W = I x R x I which is W = I² x R.
crazy126,
Mar 29, 10:37pm
lol i can remember the triangles and can spout them off to anyone that wants to know its just applying them to some calculations that i struggle with. mind u im not a big theory based learner
crazy126,
Mar 29, 11:18pm
yeah i know the triangles really well its just getting the math to work that i struggle with
cj_fool,
Mar 30, 12:06am
You only need to know two equations, V = I x R and P = V x I
Just remember those and learn to rearrange them and you will be fine
nightboss,
Mar 30, 4:07am
Once you learn "formula triangles" you will easily solve any equation where you need to rearrange formula.
& on top of all this remember all this applies to "resistive" circuits only throw in a few things like inductors/coils/sloenoids and it's a whole new ball game basically rememberthe triangle & I2 X R =w
tonyrockyhorror,
Mar 30, 7:11am
I've never even heard of them and never needed them.
nightboss,
Mar 30, 7:23am
If you don't need them, good for you. However the OP said they were having trouble with formulae. So I suggested a method I know has helped others unlock what can be a difficult concept. We all visualise thing slightly differently, I find the above method suitable for those that need to see the workings to understand them.
tonyrockyhorror,
Mar 30, 7:34am
I'm just saying it seems to me to be a long winded and non-intuitive method to formula transposition, like the way some people calculate GST with 23 and 7 or whatever it is for 15% - there's no obvious correlation between what you're doing and the figures you're using and that increases the chance of getting it wrong.
One of the first things you learn is that if you do something to both sides of the formula it gives the same result. And that something divided by itself is always 1.
V = I x R Divide both sides by I V / I = I x R / I = I / I x R = R Thus R = V / I
loonee-dial-111,
Mar 30, 7:44am
Most of the time I^2 R ("I squared R") is referred to for heat losses in cables or equipment. I.e. if you increase current through a cable, the heat(power) produced increases exponentially. V^2 /R is the other power related one - i.e. if you increase the voltage, the heat or wattage increases exponentially. I find the practical aspects help me remember it.
loonee-dial-111,
Mar 30, 7:44am
Most of the time I^2 R ("I squared R") is referred to for heat losses in cables or equipment. I.e. if you increase current through a cable, the heat(power) produced increases exponentially. More heat = failure, so it is a reminder of how say if a cable has a current rating, and you exceed it by a small amount, it makes a big difference. V^2 /R is the other power related one - i.e. if you increase the voltage, the heat or wattage increases exponentially. I find the practical aspects help me remember it.
fordcrzy,
Mar 30, 7:48am
TWINKLE TWINKLE LITTLE STAR. POWER EQUALS i SQUARED R
nightboss,
Mar 30, 7:58am
As long as the OP finds a method that works and is easily remembered. Good to see so many of us are keen to chip in with what works for us.
skin1235,
Mar 30, 8:10am
a long, long, long time ago a tutor stressed us to remember the two golden rules
I Want Valerie and Is Valerie Ready
I = amps(current) V= volts( potential ) R = resistance(ohms)
I= amps W= watts(power or work done) V = volts
those 4 units are about it for automotive work, batt V, amp draw, wattage and resistance the triangle works if you know where to put the value, placed from left, top, then right, cover the one you need and the others will show you how to get it - if alongside multiply, ifabove divide the top by the bottom
perhaps someone more recent or maybe a sparky could confirm
whqqsh,
Mar 30, 8:15am
I only know of Boyles Law (steam/gases pressure) Benfords Law (mathematics & statistics) Coles Law (the ratio of mayo to cabbage)
sandypheet,
Mar 30, 8:19am
Ohms law states that the current flowing through a conductor is directionally proportional the the voltage across the ends,providing the temperature remains constant.Now that was drummed into me when I was an apprentice 40 years ago,still dont realy know,but I have never forgotten it.Shit,it may be wrong.
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