Could someone please explain how the value of sin is calculated when working out voltage drop for cables above 16mm? Do we use the values provided in BS 7671 appendix 4 such as resistance reactance and impedance of a conductor?
Thank you
To answer my own question a - there are some limiting Ze values here: Earth fault loop impedance revision of ENA Engineering Recommendation P23 e.g. 90% of 100A supplies will have a supply impedance ≤ (0.25+j0.23)Ω
- Andy.
To answer my own question a - there are some limiting Ze values here: Earth fault loop impedance revision of ENA Engineering Recommendation P23 e.g. 90% of 100A supplies will have a supply impedance ≤ (0.25+j0.23)Ω
- Andy.
which if I remember my O-level trig correctly means we can equally write it as:
0.3397 = cos(42.614º)*0.25 + sin(42.614º)*0.23
(perhaps slightly perverse in the immediate context, but might compare with the appendix 4 approach)
- Andy.
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