Recent content by Tom McCormack

Thanks for your help :smile:

Thanks for the help So in that case would you agree with me on R70 as R20x[(1+0.004x50)] R20 = From table 1 193 so 193 x 1.2 = 231.6 microhms meter therefore volt drop = 0.0002316A per meter As my answer for 3. Thanks again

Thanks for the help So in that case would you agree with me on R70 as R20x[(1+0.004x50)] R20 = From table 1 193 so 193 x 1.2 = 231.6 microhms meter therefore volt drop = 0.0002316A per meter As my answer for 3. Thanks again Reference...

Thanks again here is the questions i have.

I am not too sure i was working from the example in the learning material. what are your thoughts? Thanks Tom

Hi guys forum looks great nice to meet you all Tom x

Hi there, i have the same question and also found this a little confusing as Neil Hayes. So i have calculated my R70 as R20x[(1+0.004x50)] R20 = From table 1 193 + 1200 ( 193 is the resistance/meter of conductor 1200 is the resistance/meter of steel armour. Both given in Table 1) so 1393 x 1.2...