Answer / subash nishad

take line voltage is VL and Zs is the impedence of
each phase,
then in case of delta connection ,VP=VL

phase current I(delta)= VL/Zs

we know that in delta connection ,
Line current IL= root 3 times phase
current
so line current I(delta)=root3*VL/Zs
in case of star connection,Vp= VL/root3
phase current I(star)= VL/root3*Zs
weknow that in star connection,
phase current=line current
IL(star)=VL/root3*Zs

so, IL(delta)/IL(star)= root3*VL VL
---------- / ----------
Zs root3*Zs


= root3 *VL root3*Zs
--------- * ----------
Zs VL

= 3

IL(delta) =3*IL(star)

hence, delta connection will take 3 times more current
than star connection..

Answer / virendra

IF motor is started using a delta starter then VP=VL (
Phase voltage is equall to line voltage & Load current Isc
is become root3.Isc.

Now if this motor is connected in star then then VP=VL/root3
& IP= isc /roor3

It clearly shoow that in star connection motor take 1/3
times less current then starting in Delta.

Answer / pillay

What is important is, the motor is designed to operate at
what connection and what voltage.IF it is designed for the
other connection and you powewr it the other way you will
destroy the motor winding or you wont get the intended
power out put.
power is always line voltage X LINE CURRENT x PF x root 3
for 3 phase motor, and you cannot change it.

Answer / benchalli.nagaraj

in STAR the current is more,
Because of

In star: line current = Phase current
In delta: line current =/root 3 phase current

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