## Tuesday, February 22, 2011

### Why Square root of Three enters all Three phase calculations

Have you ever thought why there is always a √3 present in every 3 phase calculations, let it be a delta or star transformer.

Actually this was not derived with heavy thinking or with supercomputers; rather we have to apply the simple basics of transformers which we have studied primarily.

And keep in mind, one main fact is that the differentiation between phase voltage and line voltage. Let me explore that first.
Line voltage is nothing but the output voltage measured at transformer externally i.e. Voltage available between any 2 wires out of 3 wires. We call it as VL.
But Phase voltage is the output voltage available internally i.e. Voltage available between single phase wire and neutral. We call it as Vph.
And the same definition holds good for current available which shall be called as IL & Iph.

Our basic rule is to measure power available at secondary which is nothing but the product of Phase Voltage x Phase Current and summation of all three phases,

P             =             3 x Vph x Iph                                          …. (1)

But as you see it is very difficult to measure the phase voltages from a transformer which may involve complex connections and may be impossible rather it’s easy to measure from the lines present outside.

So here we go with a bit trigonometric work to derive the phase data from available line data.

As we know in a star connected transformer, the line voltage is nothing but the combination of two phase voltages which are out of phase by 120°. But current will be same for both phase and line. And in delta connected transformer, two currents will be out of phase by 120° and voltages will be equal.

Here we take the case of star connected transformer,
Since the line voltage is combination of two phase voltages we cannot just add together to make VL  = 2 x Vph which doesn’t make sense since there is a phase shift of 120°.

So,                                        VL  = 2 x Vph x sin(120)
VL  = 2 x Vph x √3/2
VL  = √3 x Vph
Vph  = VL  /√3                                                      …. (2)

Substituting (2) in (1),

We get                                 P = 3 x VL  /√3 x IL               since Iph  = IL

P = √3 .VL .IL

This holds good for transformers and for other loads like motors, generators etc another factor called power factor gets introduced in the equation and converted as,

P = √3 .VL .IL.cosθ