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As we know the main function of transformer is to transfer power from one ckt to another ..so ratio will same on both side of transformer..it only change the ratio of voltage and current .
The power ratio between Star to Delta is1:3
The inpot power = The output power + the power losses
the power ratio in the circuit if the Star 1 the delta will be3 and the calculation answer above by another college and for me i remamber only1/3
Power ratio is1,
As for star:
Pt=P1+P2+P3 (P1 for phase1 ,P2 for phase2 and P3 for Phase3)
P=P1=P2=P3=Vph*Iph*cos theta
Pt=3*P
As in star connection Vline=(3)^1/2*Vph
so
Pt=3*(Vline/(3)^1/2)*Iph*cos theta
Pt-star=(3)^1/2*Vline*ILine*cos theta as ILine=Iph
Similarly for delta connection
Pt=P1+P2+P3 (P1 for phase1 ,P2 for phase2 and P3 for Phase3)
P=P1=P2=P3=Vph*Iph*cos theta
Pt=3*P
As in delta connection ILine=(3)^1/2*Iph
Pt=3*Vline*(Iph/(3)^1/2)*cos theta as Vline=Vph
Pt-delta=(3)^1/2*Vline*ILine*cos theta as Vline=Vph
As
Pt-star =Pt-delta
so
Pt-star/Pt-delta=Pt-delta/Pt-star=1
The power ratio is1:3 when referring this to a wye-delta motor starting.
normally it is1:3 ratio
The power ratio between Star to Delta is1:3
Explanation:
Lets 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 Im assuming it a only resistive load.
And let it be R per phase.
For Delta connected load:
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.
For3 phases:
P3D =3Pd =3*3Vs2/R =9Vs2/R watts.
Now for Star connected load:
PS = I2R = (Vs/R)2R = Vs2/R watts
For3 phases: P3S =3PS =3 Vs2/R watts
Conclusion:
P3S / P3D = 3Vs2/R /9Vs2/R =1/3
power is always constant in Xmer
