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Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems. |
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#1
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Heating effect of AC vs DC
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? |
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Heating effect of AC vs DC
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? -- The e-mail address in our reply-to line is reversed in an attempt to minimize spam. Our true address is of the form . |
#3
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Heating effect of AC vs DC
What is the 'other' formula for power? P = I*V
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#4
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Heating effect of AC vs DC
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 |
#5
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Heating effect of AC vs DC
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%. |
#6
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Heating effect of AC vs DC
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 . |
#7
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Heating effect of AC vs DC
"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. |
#8
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Heating effect of AC vs DC
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 |
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