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Author Topic: volts times amps = watts  (Read 12850 times)
dodgy rog
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« on: March 13, 2008, 04:46:53 PM »

hi could someone please help me, i would like to no more about the simple equations, what are they ,i seem to remember some kind of triangle equation, all input much appreciated.
i would like to do the math to calculate the difference in running say a DC TV and a AC TV, same for ac lowenergy bulbs against DC lowenergy bulbs?

i have my 24V 700ah forklift battery after that its a clean slate, fresh piece of paper,
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« Reply #1 on: March 13, 2008, 05:00:46 PM »

I think I know what you are after Rog, if not someone else will understand I'm sure.

You are right Voltage (V) x Current (A) = Power (W)

if you arrange them in a triangle with W on top and V & A on the base all you need to do is cover the one you wish to find woih your finger.


V  A

So V = W/A  A= W/V and W = V x A.

Is that what you were looking for?

My AC theory is sadly lacking but the principles are the same.

100W lamp / 240V = 0.4 Amps
11W lamp / 240V = 0.05 Amps

You can remember Ohms law withe the same triangle.

I  R  Where R is resistance on Ohms. Enjoy

Mark Time

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St Ives

« Reply #2 on: March 13, 2008, 05:11:00 PM »

Hi Rog,
         something I can help with at last  Smiley

want to find out how much power you need:-

P(watts)=V(volts) x A(amps)
Want to know how many amps are going to be drawn:-

A=P/V  (power divided by voltage)

Excluding all other things it is as simple as that (ducks and hides  Wink )

So.. with your 700 Ah battery at 24V you have 16800w (16.8kw)

You could run a 1kw heater for 16.8 hours and you would end up with a completely dead and knackered battery. It would draw 41.67 amps per hour from your battery so you have 41.67 x 16.8=700

This also assumes you have no losses elsewhere (from cable, battery etc).

I'll go and hide and let the real doctors diagnose you now  angel


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« Reply #3 on: March 13, 2008, 05:35:16 PM »

In practical terms, a bulb of the same wattage consumes the same amount of current, whether it be ac or dc - but it is worth bearing in mind the route it takes to get there from the battery, and the losses involved Wink
For argument's sake - taking the same state of battery charge, and assuming that you're running it down to the same level each time
if you power it directly from the battery - no losses - 100 hours
inverter (85% efficient) - 85 hours
switch-mode voltage converter  - 93 hours
- this is NOT precise, but gives a fairly good comparison Wink
The other practical point to bear in mind is cable size - a 12 or 24v system requires far heavier cables/fuses/switches than mains voltage (10 to 20 times heftier!) due to the amperage! Cool
Another point worth mentioning is that you should NEVER equate the rated amp/hrs of a battery to what you'd ever dream of taking out of it........even the very best batteries have their lives considerably shortened by deep discharge!  Cool

Unpaid volunteer administrator and moderator (not employed by Navitron) - Views expressed are my own - curmudgeonly babyboomer! -
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« Reply #4 on: March 13, 2008, 05:44:10 PM »

Yes VA is Watts for DC....but for AC only if the voltage is a true sine wave and the load is resistive like a heater.

For modified sine wave or a motor (inductive) the average cheap multimeter will give slightly inaccurate results for V and A on the AC ranges.

For AC the power  is VA*Cos(p)   where p is the phase shift angle between V and A seen on an oscilloscope and they are RMS values.
                                                     Good quality professional meters will read true RMS values.

This is often simplified to VA*pf   where pf is the power factor( Cos(p ) ), which is 1.0 for a resistive load.

« Last Edit: March 13, 2008, 05:46:39 PM by Paulh_Boats » Logged

30 tube thermal,
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LED lighting in every room
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« Reply #5 on: March 22, 2008, 06:59:39 PM »

RMS values. Now this is where it gets interesting, and painful for my poor old brain cell.

In AC the voltage can be described as having 2 values because a sine wave produces a 'peak voltage' and an RMS voltage. The peak is to the very top of the sine wave where as the RMS is the root mean square value of that voltage which is found by dividing the peak by 1.414. The RMS voltage is the value that produces the vaule that is normally used in everyday parlance.

If you are looking for wattage you will be using the RMS voltage and looking at a restive load, like a fire, but if you start to use motors then as PB states you will bring a 'power factor' into it. Then it really gets interesting, inductive or capacitive.

Oh hell, I need another beer!! Tongue

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