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Article #258

Re: Two phases to house - loss of neutral

#258
From: David
Date: Sat, 28 Nov 2009 00:28
29 lines
2376 bytes
Sylvia Else wrote:
> 
> Anyway, all a test would do is show that the answer is probably correct. 
>  It wouldn't make it any more obvious.
> 
> Sylvia.

Sylvia,

Draw yourself a vector diagram. Then with some simple trigonometry it 
*should* be more obvious.

Assume a resistive load between two of the three phases, with a load of 
1 unit current and 1 unit voltage. The load will thus be 1 unit power.

Each single phase meter will see the in phase voltage as 1 / sqrt(3) 
(240/415).

Each single phase meter will see an in phase current of 1 x cos(30). 
Remember that cos(30) = 1/2 sqrt(3).

This gives the power measured by each meter as
	V*I = 1/sqrt(3) * 1/2 sqrt(3)

The two square roots of three cancel, which, unsurprisingly leaves 1/2. 
Thus each meter records 1/2 the power in the load, and you will thus get 
billed correctly.

David
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Date: Sat, 28 Nov 2009 11:28:08 +1100
From: David <david@usenet.com>
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Subject: Re: Two phases to house - loss of neutral
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