# Q: Derive the relationship between line current and phase current for delta connected 3 load when supplied from 3-phase balanced supply.

**delta Connection**

As shown in the figure given below, there are only three possible voltages.

- V
_{RY}– between R and Y - V
_{YB}– between Y and B - V
_{BR}– between B and R.

Hence , these are phase as well as line voltages.

There are six possible currents , IR, IY, IB are known as phase current I_{ph} .

And their combinations ( Ir – Ib ) ,( Ib – Iy ) and ( Iy – Ir ) are known as line currents. Let, the phase sequence be R Y B.

Again, all the three phase currents have same magnitude but 120^{0} phase difference.

**Relation between I**_{ph} and I_{L}

_{ph}and

Any one of the three line currents can be calculated, relation remaining same for other two phasor diagram for phase voltages is shown below. Diagram is drawn taking IR as reference.

therefor: Line current ( Ir — Ib) will be the sum of ( Ir) and (-Ib ).

The resultant (Ir —Ib) will be 30^{o} behind Ir. Hence, is 30^{0} behind Iph. The magnitude of ( Ir – Ib) can be calculated as,

Hence, magnitude of line current is under root 3 times phase current.

**Net voltage in the closed loop of a balanced delta connection**

Consider, a delta connection constructed from three impedances of equal value ‘Z’ as shown in figure.

The three phase voltages which are also the line voltages are V_{RY} , V_{YB} and V_{BR }For a balanced system,

V_{RY} = V_{m} sin 0

V_{YB} = V_{m} sin ( 0 – 120^{0} )

V_{BR} = V_{m} sin ( 0 – 240^{0} )

At any instant of time, the net voltage in the closed loop should bc equal to the sum of above three phase voltages.

Net voltage in closed delta.

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