Tin Can
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rtbcomp
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V = I x R
In the triangle cover the unit you want to calculate and the formula remains visble.
The example you show is the relationship between volts, amps and ohms (resistance)
For Watts (Power) W = I x V.
The two formulae can be combined to give
W = I^2 x R (I squared R) for example.
In the triangle cover the unit you want to calculate and the formula remains visble.
The example you show is the relationship between volts, amps and ohms (resistance)
For Watts (Power) W = I x V.
The two formulae can be combined to give
W = I^2 x R (I squared R) for example.
Tin Can
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Thanks for the replies, eventually found it under Ohms Law as per tigler http://www.sengpielaudio.com/calculator-ohmslaw.htm
"Sorted" - Mark
"Sorted" - Mark
funny its one of the few equations i remember 
Tin Can
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Thanks for the comments, i was really looking to 'fill' in the 'triangle' in the correct sequence
Rtbcomp's reply was too 'scientific' for my grey matter "The two formulae can be combined to give - W = I^2 x R (I squared R) for example" and made me go cross eyed.
Tiglers comment was more on my wave length and so i searched ohms law and was directed to the web site i linked to and i now have a 'pretty picture' of a triangle, correctly labeled up sitting on the wall next to me
Thanks Guys
Rtbcomp's reply was too 'scientific' for my grey matter "The two formulae can be combined to give - W = I^2 x R (I squared R) for example" and made me go cross eyed.
Tiglers comment was more on my wave length and so i searched ohms law and was directed to the web site i linked to and i now have a 'pretty picture' of a triangle, correctly labeled up sitting on the wall next to me
Thanks Guys
saw rtb's comments and thought thats it .
Luke
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The triangle works by covering up the value you are trying to find, then reading off the 'formula' that is shown.
So,
To find V, cover V, you're left with I x R
To Find I, cover I, you're left with V / R
To find R, cover R, you're left with V / I
The other thing to remember is, the formulas only work if you use the correct units - Amps, Watts, Volts, Ohms, not milliamps, Kilowatts etc.
(Ok, strictly speaking, they still work, you just get the answer in the units you've used, rather than the units you might expect as standard)
So,
To find V, cover V, you're left with I x R
To Find I, cover I, you're left with V / R
To find R, cover R, you're left with V / I
The other thing to remember is, the formulas only work if you use the correct units - Amps, Watts, Volts, Ohms, not milliamps, Kilowatts etc.
(Ok, strictly speaking, they still work, you just get the answer in the units you've used, rather than the units you might expect as standard)
rtbcomp
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It works for AC into a resistance, resistive circuits have a power factor of 1 IIRC. 1 volt AC rms into 1 ohm gives 1 amp rms and 1 watt of power.
That is correct. It's induction motors , or other inductive loads that spoil the party !!
rtbcomp
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Hardly academic, just think of all the heaters in a home, incandescent bulbs, kettles, immersion heaters, cookers &c &c.
Hello rtbcomp,
in the context of this forum where I have seen past queries about Wattage where only the voltage and full load currents are quoted are concerning inductive devices, and particularly where welding sets are concerned the load is variable. Often (K)VA is quoted, but the light and heating devices are usually rated in Watts anyway?
My point was basically to give a more accurate reply.
Alec
in the context of this forum where I have seen past queries about Wattage where only the voltage and full load currents are quoted are concerning inductive devices, and particularly where welding sets are concerned the load is variable. Often (K)VA is quoted, but the light and heating devices are usually rated in Watts anyway?
My point was basically to give a more accurate reply.
Alec
jugs
Dont wish to offend & annoy, but...
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