Loop impedance calculation

[helio007]

Hi all ,

The measured value of loop impedance for a circuit is 2.4Ω. If the temperature at the time of the test was 20°C and the cable is 70°C (factor 1.2) what is the corrected value. Ze = 0.4Ω:

the answer that I got is 2.8 but according to my calculations it would be 2.88 ohms follow up my work ;
2.4x1.2/1000 it feels like im missing something
I've also tried
2.4+0.4×1.2/1000
please guys don't want anyone criticising me just shed some light and I'll do the rest . Tia

 

[westward10]

Don't use Ze in your calculations, remove it and then add back on.
Why are you dividing by 1000.

 

[helio007]

Don't use Ze in your calculations, remove it and then add back on.
Why are you dividing by 1000.

Must be mixing up with voltage drop ib×mv/a/m×l/1000 thank you much appreciated, I get it now . Enjoy the rest of your Sunday.

 

[westward10]

Apply your correction factor to the internal impedance (Zs-Ze) then add Ze back on.

 

[helio007]

Apply your correction factor to the internal impedance (Zs-Ze) then add Ze back on.

(2.4-0.4)×1.2+0.4 = 2.8ohms