Amps to watts
Amps to watts
I have a load of 3.73 amps @ 12volts If I have a power supply rated @ 12v 30 watts will this run The load that I have? From what I was told the load will be over a little bit but I would like to get your guys input on this
 Chris Smith
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Re: Amps to watts
3.73 times 12 volts is 44.76 watts, so no in theory and in practice.

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Re: Amps to watts
DON'T DO IT! House fires are very common this time of year precisely because of this kind of mismatching. An inescapable product of the use of electricity is heat. Sometimes light and smoke and noise too. From my understanding a Watts rating usually refers to AC power values. The equation to find AC watts is more complex than the equation for DC watts. Calculating DC watts (DC power, DC load) is (rule of thumb) Amps times Volts. The values for the same given AC supply Voltage and AC load, which is what I'll assume you you have, are far more complex and, from my experience, the complex variables add AT LEAST another 10 to 15% to the resultant Wattage. Transformer (AC to AC power supply) ratings:  Watts are given for a purpose. Even taking the simplest DC calculation. 3.73A x 12V = 44.76 Watts. (AC calcs would give a higher load wattage rating) your load and power supply are not even closely matched. The problem is that a well built power supply will get bogged down trying to supply the load and may try to sustain the load long enough that eventually BOTH the power supply AND the load burn up. The chances are that a poorly built power supply will burn up immediately but it could it still do lots of damage to the load and the power supply's environs. If you are referring to Christmas ornaments, the device (load) may work slowly or be quite dim while the power supply is an accident waiting to happen. A 12 VAC 80 Watt power supply (a garden lighting standard I think) would be the ticket. Better to have power in reserve than too little to start with.<p>[ December 02, 2003: Message edited by: perfectbite ]<p>[ December 02, 2003: Message edited by: perfectbite ]</p>
Re: Amps to watts
my p71
Perchance I am reading between the lines for info, but if this could be relevant to an engineering breadboard mock up of the fade out dimmer circuitry and that is the load imparted by TWO of the lamps, just use one lamp for TESTING as the resulting decay time would be equally imparted, as the timing circuitry is not load sensitive. That way you could adjust and see your desired delay using the power supply you had. Then you could even go out and use the vehicles battery for testing at full 2 lamp loading.
73's de Edd
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Perchance I am reading between the lines for info, but if this could be relevant to an engineering breadboard mock up of the fade out dimmer circuitry and that is the load imparted by TWO of the lamps, just use one lamp for TESTING as the resulting decay time would be equally imparted, as the timing circuitry is not load sensitive. That way you could adjust and see your desired delay using the power supply you had. Then you could even go out and use the vehicles battery for testing at full 2 lamp loading.
73's de Edd
[email protected] .........(Interstellar~~~~Warp~~~Speed)
[email protected]........(FirewalledSpam*Cookies*Crumbs)
Re: Amps to watts
Thanks guys for the help I was told times it by ten and that would be close but not close enough <p>
Edd this kind of has to do with the dome light circuit I happened to find at the very last second a domlight circuit that would do as I want it but the problem is the circuit for the light that i was going to buy only can handle 30 watts so my domelight app for my truck is way overbord on the watts that this product can carry. <p>So I might have to build your circuit
Edd this kind of has to do with the dome light circuit I happened to find at the very last second a domlight circuit that would do as I want it but the problem is the circuit for the light that i was going to buy only can handle 30 watts so my domelight app for my truck is way overbord on the watts that this product can carry. <p>So I might have to build your circuit

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Re: Amps to watts
Or you could set your car on fire.
Re: Amps to watts
<blockquote><font size="1" face="Verdana, Helvetica, sansserif">quote:</font><hr>Originally posted by wayne:
P=EI<hr></blockquote><p>O.k Wayne you lost me<p>explain further
<p>
Myp71
P=EI<hr></blockquote><p>O.k Wayne you lost me<p>explain further
<p>
Myp71

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Re: Amps to watts
P=EI is ONLY for DC
Re: Amps to watts
<blockquote><font size="1" face="Verdana, Helvetica, sansserif">quote:</font><hr>Originally posted by perfectbite:
P=EI is ONLY for DC<hr></blockquote>
Not true. It works exactly the same for resistive loads (light bulbs, heaters, etc.). If you take phase angle into account, it also works for reactive loads (motors, etc.).
P=EI is ONLY for DC<hr></blockquote>
Not true. It works exactly the same for resistive loads (light bulbs, heaters, etc.). If you take phase angle into account, it also works for reactive loads (motors, etc.).

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Re: Amps to watts
Ron H, You are correct. The basic equation for both DC and AC loads is P=EI but because of the phase angle factor for AC P=EI then becomes P=E x I x something else. I recall using 6 terms plus a divisor for a 220 VAC circuit one time to find the Amps (we had an 'engineer' who would chisel the nameplates off motors so he could refer to the numbers when he was on the phone! The boss told him he could do it!!!!) Sometimes all you knew was that the in place motor was supplied with whatever VAC and had such and such a frame. You had to go to the manufacturer for the driven device's speed, rotation etc. So I know whereof I speak. DC is MUCH simpler. Volts x Amps = Watts.
Re: Amps to watts
P(power) = V(voltage) * I(current) * cos (theta)
E = V and is voltage according to an old tradition (I learned it in the Navy almost 20 years ago). theta is the angle between the voltage and current and is known as the phase angle. In DC the phase angle is zero, also if the load is purely resistive the phase angle is zero. So, in those cases P=EI=VI. The I and V are rms values which in the case of a sinusoidal wave is the peak value Vp/sqrt(2) or Ip/sqrt(2).
this is real power as in what is available to do work or make heat.
Q(reactive power) = V*I*sin(theta) is the power contained in the fields of the inductances and capacitances of the system.
The sum of P and Q is S(apparent power) and is shown as
S = P + Q
S = V*I*cos(theta) + V*I*sin(theta)
S = V*I
I cursorally checked a book as I wrote that so there is undoubtedly mistakes present. Sue me.
Rick
E = V and is voltage according to an old tradition (I learned it in the Navy almost 20 years ago). theta is the angle between the voltage and current and is known as the phase angle. In DC the phase angle is zero, also if the load is purely resistive the phase angle is zero. So, in those cases P=EI=VI. The I and V are rms values which in the case of a sinusoidal wave is the peak value Vp/sqrt(2) or Ip/sqrt(2).
this is real power as in what is available to do work or make heat.
Q(reactive power) = V*I*sin(theta) is the power contained in the fields of the inductances and capacitances of the system.
The sum of P and Q is S(apparent power) and is shown as
S = P + Q
S = V*I*cos(theta) + V*I*sin(theta)
S = V*I
I cursorally checked a book as I wrote that so there is undoubtedly mistakes present. Sue me.
Rick
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