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Appendix 4, section 6.1 equation 6

Former Community Member
Former Community Member

 

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Hi I am working through equation 6 for an ambient temperature of 50 degrees for 1mm 90degC thermoplastic cable (Table 4E2A)

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I get a correction factor of 0.95 for a 2A load. 

Do i divide the mV/A/m by this factor ? 

The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

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    The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

    Isn't the overall factor increasing from 0.82 (for just the high ambient temperature) to 0.95 (now accounting for the reduced conductor temperature, due to the reduced load, as well) - which seems to be going in the right direction to me.

       - Andy.

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  • 0

    The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

    Isn't the overall factor increasing from 0.82 (for just the high ambient temperature) to 0.95 (now accounting for the reduced conductor temperature, due to the reduced load, as well) - which seems to be going in the right direction to me.

       - Andy.

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