The power ratio between Star to Delta is 1:3

Explanation:

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

So the line voltage of the supply will be Ö3V_s.

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

And let it be ‘R’ per phase.

For Delta connected load:

Calculation for per phase power; P_D= I²R

Where I à load current (per phase)

And,

I = Ö3V_s/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

            P_d = (Ö3V_s/R )²R = 3V_s²/R  watts per phase.

For 3 phases:

            P_3D = 3P_d = 3*3V_s²/R = 9V_s²/R watts.

Now for Star connected load:

P_S = I²R = (V_s/R)²R = V_s²/R watts

For 3 phases: P_3S = 3P_S = 3 V_s²/R watts

Conclusion:

P_3S / P_{3D  =}  3V_s²/R / 9V_s²/R = 1/3

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

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

power is same

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