3-Phase Star Config.

[RM11541]

Hey Guys,

Just been looking at the C&G 2365 L2 Textbook and don't understand the answer to one of the questions as given below:

A 400/230V star-connected system has three equal loads of 11.5 Ω per phase. The line and phase currents will be, respectively:

a) 34 A and 34 A
b) 20 A and 20 A
c) 34 A and 20 A
d) 20 A and 34 A

The answer is given as (b).

I could wait to cover this material on the course but I'm reading ahead a bit. As I understand it [IL=IP] and [VP=VL / √3] in a star configuration.

It looks like the phase (and line) current has been calculated using the phase voltage (i.e. IP=230/11.5=20A). Presumably the line voltage could be used to determine line current? If so, how would the calculation be done? Obviously not as straightforward as IL=VL/R.

Any explanations welcome.....

Cheers,

Raymond.

 

[Simonslimline]

In star configuration IL = IP and VP = VL / 1.732

In Delta configuration VL = VP and IP = IL / 1.732

:stooge_curly:

 

[RM11541]

Cheers....The basic arithmetic (i.e. square-root three) I'm reasonably OK with. The concept of three-phase circuits where it gets a bit wooly. Still I'm sure it will all become clearer with time as the course goes on.

 

[telectrix]

if you look at the star config. you see that the line current of each phase is equal to the phase current. as the phase voltage is 230V ( L-N ).then Ip = Il = 230/11.5 = 20A.

 

[RM11541]

So following on from this, in a star-config we know IP=IL=20A. And if VL=400V, then the "combined" phase loads across L1-L2 would be 400/20=20Ω ?

 

[telectrix]

there's no value assumed for that. the load is generally given as per phase . each phase has it's own individual breaker/fuse. it's worth noting that with a balanced load ( equal load per phase ) then there is zero current in the neutral.

 

[RM11541]

Cool....Cheers for the info.