I would like to test the current carrying capability of a connector that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10 amps at
300V) and just want to verify performance in this application. For reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at 208VAC?

Surfer wrote:

I would like to test the current carrying capability of a connector that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10 amps at
300V) and just want to verify performance in this application. For reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at 208VAC?



RMS?

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The e-mail address in our reply-to line is reversed in an attempt to
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What is the 'other' formula for power? P = I*V

Surfer wrote:
I would like to test the current carrying capability of a connector that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10 amps at
300V) and just want to verify performance in this application. For reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at 208VAC?

Hi Surfer...

Good plan, hope you share how this is done and we'll all be
zillionaires

Sorry, couldn't resist, meant no disrespect.

Dunno where your formula for power came from; but think you really
meant P (watts) = I * E

Your example above - 10 amps at 3 volts would be 30 watts; and
10 amps at 208 volts would be 624 watts. There's no free lunch

Take care.

Ken

To test a connector set, note both parts of the connection, the
connector set must be tested in a normal circuit application. Measure
the voltage drop across the connector and the current in the circuit.
Then you can calculate the power on the connector set. Make sure to
use true RMS values when measuring AC current and voltage.

Only then can you apply the proper voltage on the connector at the
expected current to test the heating of the connector.

As with all connectors, they may also fail due to voltage arcing or
excessive current surges causing arc marks which increase the effective
resistance of the connector. This is why all real electrical
maintenance programs on critical equipment use a thermal imager or
temperature device to periodically check the operating temperature of
connections and other devices to see if there has been an increase due
to impending failure.

As for your actual test. Forget about any resistive load except for
the resistance of the connector itself. You will need to make sure
that the wiring resistance is at least 10 times less than the connector
resistance in order to test it by applying a constant current source of
10 amps. Otherwise the wiring will seriously skew the results by more
than 10%.

Ken Weitzel wrote:



Surfer wrote:

I would like to test the current carrying capability of a connector
that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10
amps at
300V) and just want to verify performance in this application. For
reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC
power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at
208VAC?


Hi Surfer...

Good plan, hope you share how this is done and we'll all be
zillionaires

Sorry, couldn't resist, meant no disrespect.

Dunno where your formula for power came from; but think you really
meant P (watts) = I * E

Your example above - 10 amps at 3 volts would be 30 watts; and
10 amps at 208 volts would be 624 watts. There's no free lunch

but that's a bit irrelevant to the question of how much the connector
will heat, isn't it?

Take care.

Ken



--
The e-mail address in our reply-to line is reversed in an attempt to
minimize spam. Our true address is of the form .

"Surfer" writes:

I would like to test the current carrying capability of a connector that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10 amps at
300V) and just want to verify performance in this application. For reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at 208VAC?


10 amps over the connector will heat it up exactly the same whether
you run 3V over it or 208V. The only relevant voltage is the voltage
drop across the connector, which is dependant on the resistance of the
connector and current run through it.

The R in your formula above is the R across the connector, not in the
load.

The only relevance of the 300V voltage spec is for isolation.

To compare AC with DC you need to use the RMS current of the AC.

On Tue, 08 Nov 2005 04:22:14 GMT, Ken Weitzel
wrote:



Surfer wrote:
I would like to test the current carrying capability of a connector that's
used in a 208VAC 3-phase circuit. I have the manufacturer's spec (10 amps at
300V) and just want to verify performance in this application. For reasons
that would take too long to explain, I want to test this connector out of
circuit. It was suggested to just use a low voltage supply, like a PC power
supply and set up a resistive load that would deliver 10 amps.

Since the formula for power, P=I^2*R doesn't include a value for voltage,
would 10 amps at 3VDC provide the same heating effect as 10 amps at 208VAC?

Hi Surfer...

Good plan, hope you share how this is done and we'll all be
zillionaires

Sorry, couldn't resist, meant no disrespect.

Dunno where your formula for power came from; but think you really
meant P (watts) = I * E

The formula is correct, but there seems to be the assumption that the
resistance would be the same, and of course it isn't.

Tom


Your example above - 10 amps at 3 volts would be 30 watts; and
10 amps at 208 volts would be 624 watts. There's no free lunch

Take care.

Ken