# Calculating watt power question

Discussion in 'General Electronics Chat' started by magnetman12003, Jun 7, 2010.

1. ### magnetman12003 Thread Starter Active Member

May 10, 2008
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I have a question that I need an answer for. This relates to WATT POWER calculations.

I have a power source that puts out 305-315 volts Dc after The AC circuit that feeds it is rectified using a full wave bridge rectifier.

After applying a resistor load to the 305 DC output the voltage across the resistor drops to 250 volts DC.

Figuring out the circuit WATT POWER do I use the 305 DC output voltage or the 250 volt DC drop in my WATT POWER calculations?

Say my current is .0105 amps. Using the 305 volts output DC my Watt value is 3.2025

Using the 250 volt DC drop my Watt value is 2.625

Which is correct to figure Watt power?

Tom

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Use the actual voltage with the load connected - 250V

3. ### Mike33 AAC Fanatic!

Feb 4, 2005
349
25
I would use the rated, initial (no load 305v) voltage, personally. Just to be cautious.

The voltage is dropping under load because the power supply has its own 'resistance' that is working in conjunction with your load resistor...source resistance, also called source impedance. Picture a voltage divider, that is what is going on with the impedance and your load resistor. So, the work is actually being done even tho you don't see it. Being conservative in calculations is a good way to provide a safety margin.
I'm sure a guru will come in and explain it better ;o)

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I think the OP was asking what the actual power in the load would be with the particular load connected. So with the load connected the power in the load would be determined by the actual power supply terminal voltage at that load condition.

They may well be interested in the choice of load resistor rating, but that didn't appear to be what the OP was asking. I agree that if they want to know what a cautious choice of load power rating might be, then basing their calculations on the supply open circuit voltage would be a sensible approach.

5. ### Mike33 AAC Fanatic!

Feb 4, 2005
349
25
Oh, you're right on, TNK. Like many, I am still learning much even tho I have that "senior member" ranking thing next to my name ;o) A resistive load will of course use power at its individual V*I irrespective of any other circuit voltages. Since the OP has the current and voltage based on measurement, no problems there.

I was thinking along the lines of keeping in mind the power supply ratings and such...using a conservative approach so as to not overstress it. It is using more current than what is being measured at the load due to its own impedance is what I was referring to.