# What is the power ratio between power in Star and Delta circuit?

KaranSorout

• Member Since Jul-2006 | Jul 13th, 2006

The power ratio between Star to Delta is 1:3

Explanation:

Let’s suppose Vs be the supply voltage per phase.

So the line voltage of the supply will be Ö3Vs.

Now assume any type of load; for simplicity I’m assuming it a only resistive load.

And let it be ‘R’ per phase.

Calculation for per phase power; PD= I2R

Where I à load current (per phase)

And,

I = Ö3Vs/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

Pd = (Ö3Vs/R )2R = 3Vs2/R  watts per phase.

For 3 phases:

P3D = 3Pd = 3*3Vs2/R = 9Vs2/R watts.

PS = I2R = (Vs/R)2R = Vs2/R watts

For 3 phases: P3S = 3PS = 3 Vs2/R watts

Conclusion:

P3S / P3D  =  3Vs2/R / 9Vs2/R = 1/3

It may be better concluded by drowing figures. If any variations, can connect to me through karan.sorout@gmail.com

#### SASI KUMAR.A

• Jul 6th, 2006

The Power Ratio between STAR and DELTA isÂ  1:3

#### KaranSorout Profile Answers by KaranSorout

• Jul 13th, 2006

The power ratio between Star to Delta is 1:3

Explanation:

Let?s suppose Vs be the supply voltage per phase.

So the line voltage of the supply will be ?3Vs.

Now assume any type of load; for simplicity I?m assuming it a only resistive load.

And let it be ?R? per phase.

Calculation for per phase power; PD= I2R

Where I ? load current (per phase)

And,

I = ?3Vs/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

Pd = (?3Vs/R )2R = 3Vs2/R  watts per phase.

For 3 phases:

P3D = 3Pd = 3*3Vs2/R = 9Vs2/R watts.

PS = I2R = (Vs/R)2R = Vs2/R watts

For 3 phases: P3S = 3PS = 3 Vs2/R watts

Conclusion:

P3S / P3D  =  3Vs2/R / 9Vs2/R = 1/3

It may be better concluded by drowing figures. If any variations, can connect to me through karan.sorout@gmail.com

#### akhileshkesavanunnithan Profile Answers by akhileshkesavanunnithan

• Jul 17th, 2006

#### sibani

• Jul 31st, 2006

1:2

#### sonu14g2003 Profile Answers by sonu14g2003

• Feb 25th, 2007

its actually 1:3 and the explaination is given above

#### andyarok Profile Answers by andyarok

• Apr 11th, 2008

Wow guys what happened? Never give wrong info guys. Total Power remains the same.

Irrespective of the type of  connection  Power  remains same.  Imagine  that  u loose  1/3  of the  power  for  every  Y-delta  T.F. Dont Joke pa.

If u need explanation then will reply

#### kanchan13 Profile Answers by kanchan13 Questions by kanchan13

• Apr 22nd, 2008

power is same

its only the line currents and voltages which differ in star and delta circuits.

#### vishwas_deshmukh Profile Answers by vishwas_deshmukh

• May 6th, 2008

it is type of communication in which we use differete frequencies .communiacte  via same conductor which is use for transmission .

#### munish1983 Profile Answers by munish1983 Questions by munish1983

• Sep 30th, 2008

Power is same in both star and delta connections which is √3 VL.ILcosΦ. So power ratio is 1:1

#### scientist007 Profile Answers by scientist007 Questions by scientist007

• Feb 17th, 2009

WhateverÂ be the connection whether it is delta or star, total power remains same. So their will be 1:1
Thanks

#### heeru Profile Answers by heeru

• Mar 19th, 2009

The power ratio between the power in Star & Delta is 1/3.

#### jeet29 Profile Answers by jeet29

• Jul 6th, 2009

Power for a circuit will always be same despite of it is connected in star or delta ...because star or delta
are the only connection types which connect your equipments with the power
sources.  The only things which have a difference or the ratio is voltage
and current in star or delta connected system.

In Star,
line voltage=1.732*phase voltage
line current=phase current

For Delta,
line voltage=phase voltage
line current=1.732*phase current

#### GeeK_Percy Profile Answers by GeeK_Percy

• Oct 3rd, 2009

It is 1

Suppose v= line voltage applied
i=line current

For star connected resistances
power= 3*(v/1.732)*i=1.732*v*i

For delta
power=3*v*(i/1.732)=1.732*v*i

Both are same

#### full61 Profile Answers by full61 Questions by full61

• Feb 11th, 2010

Power is irrespective of connections whether it is star or delta, power is same in both connection and the ratio of power is 1:1.

#### saginala13 Profile Answers by saginala13

• Jun 19th, 2010

WE KNOW THREE PHASE POWER P=ROOT3VICOSPI in any circuit, whether it is sta or delta.... and one we have to remeber is for an type of load current i or P.F indvidualy change but the product icos(pi) wil not change.

#### saginala13 Profile Answers by saginala13

• Jun 19th, 2010

1:1 because for three phase power P=√3vicosΦ for both Star and Delta

#### VEERENDRAKUMAR.CH Profile Answers by VEERENDRAKUMAR.CH

• Jan 28th, 2011

the power in both the connections are equal  because

in star line currents=phase currents
line voltages=1.732*phase voltages
so power(P)=3VI=1.732*vline*iline
in delta line voltages=phase voltages
line currents=1.732*phase currents
so power(P)=3VI=1.732*vline*iline

#### jaxrobinson Profile Answers by jaxrobinson

• Jun 24th, 2011

Yes this formula of P=I^2R is right for calculating the exact power of both the circuits.

#### Mudigal Prakash

• Aug 4th, 2011

Power remains same and is 1:1, only Line and Phase currents and Voltages change. Because Power = 3*Phase Voltage *Phase current* Power factor. For star connection : phase Voltage = Line voltage / root 3, Phase current = line current, whereas in Delta connection phase voltage = line voltage, phase current = line current/root 3.

#### shruti

• Nov 24th, 2014

power ratio remains the same irrespective of the connection.
if power ratio would change why would we use a star delta starter or a star delta or delta star transformer.
so guys, power ratio is 1 for star and delta circuits...

#### AK ROUT

• Mar 31st, 2015

1:3 .do not confuse with as in case of a transformer where power remains same.Here due to change in load configuration power changes.

#### abanirout Profile Answers by abanirout

• Mar 31st, 2015

1:3. donot confuse with transformer.load configuration changes hence as described above @ksrout

#### abanirout Profile Answers by abanirout

• Mar 31st, 2015

no

• Apr 26th, 2015

There are 2 perspectives of seeing this problem

1.In case of a transformer, where the power going from primary side to secondary side is same. Therefore in its case power remains same.Note that transformer does not consume power, it just transforms it with a few losses.

2.If a star connected load(say of R in each phase) is connected across a Â 3 phase supply, Power=(âˆš3)(V line)(I line) then I line=(V lineÃ·((âˆš3)(R))), hence P=(V line^2Ã·R)

Now when same load is connected as delta, P=(âˆš3)(V line)(I line). Here I line=âˆš3(V lineÃ·R) hence power=3(V line^2Ã·R)=3P.Therefore power consumed in delta is 3times that in star.

• Sep 5th, 2015

STAR CONNECTION
Vph=Vline/1.732
Iph=Iline
POWER =3*Vph*Iph*PF (OR) 1.732*Vline*Iline*PF
Line Current = Phase Current

DELTA CONNECTION
Vph=Vline
Iph=Iline/1.732
POWER =3*Vph*Iph*PF (OR) 1.732*Vline*Iline*PF
Line Voltage = Phase Voltage

We can connect in any way of the connection, the Power remains same. There is difference in Voltage and Current

#### Girish R Pillai

• Sep 6th, 2015

Sorry, This answer is wrong because the question is for the same load.

#### Girish R Pillai

• Oct 16th, 2015

Its 1:3 if the supply source and load are same. But the equation for power , P = âˆš3. VL x IL x cos phi is the same, values changing.

#### Aritra Rakshit

• Mar 23rd, 2016

Load is not same. Because if you connect same impedance in star and again in delta. Equivalent impedance will change per line.

For Delta:
z=z1||(z2+z3)

For Star:
z=z1+z2

so, ultimately power ratio remains unity.

#### Pankaj sawale

• Jan 3rd, 2017

The explanation must be 9Vs^2/R/27Vs^2/R and then it become 1/3

#### Anil Suwalka

• May 30th, 2017

1:1

#### Pragya

• Jul 22nd, 2017

1:1

#### sandeep kumar

• Sep 15th, 2017

How much rating motor we use for star and delta connection?

#### Anbumozhi R

• Dec 3rd, 2017

Power is Same. Whether it is star to Delta conversion or Delta to star conversion. Power is 1:1

#### RUP KAMAL KUTUM

• Feb 2nd, 2018

Well first you assumed voltage per phase as Vs and Line voltage as sqrt(3) x Vs: Which is for star connection. Then, you assumed the same thing for Delta Connection(where line and phase voltages are the same) which is what went wrong here. Power consumed always remains the same irrespective of Star or Delta connection. There is a book I am reading that does the same mistake of 1st saying that Star and Delta load always consumes the same power then showing an example saying P-delta is 3 times P-star, which is wrong,