In a star connected 3 phase resistive load, the phase current is 20 A. If the same resistances are connected in delta and same 3 phase Voltage is applied, what is the phase current ?

concept:​

Relationship between line and phase voltages and currents in star connected and delta connected networks:

Application:

Given, If the same resistances are connected in delta and the same 3 phase Voltage is applied,

Hence,

R_(Delta) = R_(Star)

The Circuit can be drawn as,

Power in star and delta is given by,

For Star:

P_(Star) = 3V_phI_ph

P(Star) = \(3× \frac{V_L}{√3}×\frac{V_{ph}}{R}=3× \frac{V_L}{√3}×\frac{V_{L}}{√3R}=\frac{V_L^2}{R}\) ..... (1)

For Delta:

P_(Delta) = 3VphIph

P(Delta) = \(3× V_L× \frac{V_{Ph}}{R}=3× V_L× \frac{V_{L}}{R}=\frac{3V_L^2}{R}\) ..... (2)

Hence, if we reconnect the circuit from star to delta without transformation of resistance, then power in delta connected is three times of power in star connected.

∴ P(Delta) = 3 × P(Star)

Given,

I_L = I_ph = 20 A (For star connection)

Since,

P(Delta) = 3 × P(Star)

(I_Δ)²R = 3 × (20)²R

(I_Δ)² = 3 × (20)²

∴ \(I_Δ = \sqrt{3\times20^2}=20\sqrt3\)

I_Δ = 34.64 A

Mistake Points

If we reconnect the circuit from star to delta with the transformation of resistance, then power in delta connected is the same as the power in star connected.

Proof:

For Star:

P(Star) = 3VphIph

P(Star) = \(3× \frac{V_L}{√3}×\frac{V_{ph}}{R}=3× \frac{V_L}{√3}×\frac{V_{L}}{√3R}=\frac{V_L^2}{R}\) ..... (1)

For Delta:

P(Delta) = 3VphIph

P(Delta) = \(3× V_L× \frac{V_{Ph}}{3R}=3× V_L× \frac{V_{L}}{3R}=\frac{V_L^2}{R}\) ..... (2)

Hence, P(Delta) = P(Star)