need load calculation for power supply

[raj90012000]

i have a power supply of 5v 10A dc output
i want to check its amps by multimeter so can any body tell me how can i calculate load for it and how much watts it consumes

[triode101]

First, you want to get a supply of power resistors from a nearby surplus and/or used electronics store.

5volts at 10 amps is 50 watts, so you will need one 50 watt resistor, or five 10-watt resistors, or fifty 1-watt resistors.

5 volts / 10 amps = 1/2 ohms. So your load needs to be 0.5 ohms to draw the maximum power from your PS.

To get a 1/2 ohm load, you could use one 0.5-ohm 50-watt resistor across the output. This would draw all the power your PS can deliver. To verify that it is delivering that amount of power, measure the voltage across the resistor. If it is 5 volts, your PS is good. If it is less than 5 volts, your PS is weaker than the specs say it is.

But you don't have to use only one resistor. For example, you could use two 1-ohm resistors in parallel. In that case, each resistor only needs to be rated for 25 watts.

Four 1-ohm resistors can be wired series/parallel to yield still a 1-ohm unit with 4 times the power handling ability.

Now wire two such groups in parallel and you have eight resistors giving you 1/2 ohm. This way, you only need 7 watt resistors.

Taking that a step further, with 32 1-ohm resistors, four banks of eight like above in series/parallel, your total resistance is still 1/2 ohm, but each resistor now only needs to handle about 2 watts.

You don't need to use resistors of the same value. To get the result of putting two resistors in parallel use RT = (R1 * R2) / (R1 + R2)

To get several resistors in parallel use 1 / (1 / R1 + 1 / R2 + 1 / R3 + 1 / R4 . . .)

You don't have to do this geometrically that is. Any combination of series + parallel that gets the total resistance to 1/2 ohms will work as long as each resistor is able to handle the power across that resistor.

To get the power across one resistor use W = I^2 R. For example, four 20-ohm resistors in parallel will give you 5 ohms total. If you put that across your output, it will draw 1 amp at 5 volts, putting 5 volts across each resistor. That means each resistor needs to handle 1^2 * 5 = 5 watts. The total output at that point would be 5 * 4 = 20 watts, about 40 percent of your PS capability.

To measure power use at a lower level than max power, use resistors of different values. For example, put a 5-ohm resistor in series with a 3-ohm resistor for a total of 8 ohms. 5 volts into 8 ohms should draw 5/8 Amps. 5 v * 5/8 Amps = 25/8 watts, meaning your resistors need to distribute about 4 watts. So use a 2-watt-5 ohm resistor in series with a 2-watt 3-ohm resistor. Now if you measure the voltage across the 3-ohm resistor you can use ohm's law to find the true current through it and thus compute the actual power coming out of the PS.

The max power equation, though, again, is that you need a 1/2-ohm load that can handle 50 watts.

You get this from Ohm's law and the power laws:

E = I * R (Volts = Amps * Ohms)
P = E * I (Watts = Volts * Amps)
P = I^2 * R (Watts = Amps (squared) times Ohms)

Another way to say it is, put any physically large resistor of a value greater than about 10 ohms across the output after measuring its resistance and then measure the voltage across it. That will tell you the current though it and the power coming out of the PS into it.

If the supply is not regulated, the voltage will probably be greater than 5 volts if you put a light load on it.