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Calculating load requirements

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SavannahLion:
How does one calculate the load requirements of an unknown circuit? I know I'm drawing 5 volts but how does one figure the watts or amperage of an unknown circuit?

MonMotha:
Power is related to voltage and current: P = V * I.  You need to know 2 of the 3 to compute the other.  If you have a resistive load and you know the resistance R, you additionally know (via Ohm's law) that V = I * R and various substitutions let you come up with P = V^2 / R or P = I^2 * R, but then you need to know R.

Note that many loads are NOT resistive.  many loads are "constant power" - their current increases as the voltage decreases so that P stays roughly constant.  Some loads behave somewhere in between, and some esoteric loads behave really weirdly (like fluorescent tubes).

If you just have a "black box" circuit, you have no idea how voltage will relate to current.  It's possible to build a circuit that does just about anything.

If you have a multimeter, you can measure the current directly.  You have to put the meter "in-line" with the circuit.  That is, you unhook one supply lead, and hook the meter up between them on the current (amperage) setting.  For AC loads, there also exist "clip-on" type meters, though their accuracy varies, and this doesn't work on DC, which it sounds like you've got.

SavannahLion:

--- Quote from: MonMotha on January 17, 2011, 12:23:47 am ---Power is related to voltage and current: P = V * I.  You need to know 2 of the 3 to compute the other.  If you have a resistive load and you know the resistance R, you additionally know (via Ohm's law) that V = I * R and various substitutions let you come up with P = V^2 / R or P = I^2 * R, but then you need to know R.

--- End quote ---

Yep, that's my problem alright. Not enough info to compute.


--- Quote ---Note that many loads are NOT resistive.  many loads are "constant power" - their current increases as the voltage decreases so that P stays roughly constant.  Some loads behave somewhere in between, and some esoteric loads behave really weirdly (like fluorescent tubes).

If you just have a "black box" circuit, you have no idea how voltage will relate to current.  It's possible to build a circuit that does just about anything.

--- End quote ---

It's not technically black box I guess. It's a circuit I built (will build. If my fiance, kids and animals leave me alone long enough, I might actually finish assembling the breadboard).


--- Quote ---If you have a multimeter, you can measure the current directly.  You have to put the meter "in-line" with the circuit.  That is, you unhook one supply lead, and hook the meter up between them on the current (amperage) setting.  For AC loads, there also exist "clip-on" type meters, though their accuracy varies, and this doesn't work on DC, which it sounds like you've got.

--- End quote ---

As luck would have it, Equus actually has the manual for my DMM online so I was able to look up which of the appropriate switches I should use. Naturally, it's a function I never used until now. Go figure.

I haven't read the instructions entirely yet so this may sound rather obvious.

When I usually build bread board circuits, I usually use a battery to test and power the circuit. In this case, the final circuit is going to be inserted into a PC where the 5v line is specified to be a max 10mA (though one source cites as high as 500mA). Is it acceptable to get a reading using the battery or do I want to drop the circuit into it's final environment then get a reading? I would imagine the reading is going to be the same regardless of the source or am I missing something fundamental here?


fixed bad markup

danny_galaga:

Keep adding to it until something goes pop. Then you know you should have had one less than that  ;D

SavannahLion:

--- Quote from: danny_galaga on January 17, 2011, 02:49:48 am ---
Keep adding to it until something goes pop. Then you know you should have had one less than that  ;D

--- End quote ---

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