Star vs Delta Connection: Impedance and Power Analysis

This question explores the behavior of three equal impedances when connected first in a star configuration and then in a delta configuration across the same balanced 3-phase supply. We need to understand how this change affects the phase current, line current, and overall power consumption.

Understanding Star and Delta Connections

In a 3-phase system:

• Star (Y) Connection: Has a common neutral point. The line voltage ($V_L$) is $\sqrt{3}$ times the phase voltage ($V_{ph}$), and the line current ($I_L$) is equal to the phase current ($I_{ph}$).
  • $V_{ph} = \frac{V_L}{\sqrt{3}}$
  • $I_L = I_{ph}$
• Delta ($\Delta$) Connection: Forms a closed loop. The line voltage ($V_L$) is equal to the phase voltage ($V_{ph}$), and the line current ($I_L$) is $\sqrt{3}$ times the phase current ($I_{ph}$).
  • $V_{ph} = V_L$
  • $I_L = \sqrt{3} \times I_{ph}$

Analyzing the Impedance Changes

Let the impedance per phase be denoted by $Z$. Since the three impedances are equal:

Star Connection Calculations

When the three equal impedances ($Z$) are connected in star:

• Phase Voltage: $V_{ph(star)} = \frac{V_L}{\sqrt{3}}$
• Phase Current: $I_{ph(star)} = \frac{V_{ph(star)}}{Z} = \frac{V_L}{\sqrt{3}Z}$
• Line Current: $I_{L(star)} = I_{ph(star)} = \frac{V_L}{\sqrt{3}Z}$

Delta Connection Calculations

When the same three equal impedances ($Z$) are connected in delta across the same supply:

• Phase Voltage: $V_{ph(delta)} = V_L$
• Phase Current: $I_{ph(delta)} = \frac{V_{ph(delta)}}{Z} = \frac{V_L}{Z}$
• Line Current: $I_{L(delta)} = \sqrt{3} \times I_{ph(delta)} = \sqrt{3} \times \frac{V_L}{Z}$

Comparing Currents and Power

Phase Current Comparison

Comparing the phase currents:

Ratio: $\frac{I_{ph(delta)}}{I_{ph(star)}} = \frac{V_L/Z}{V_L/(\sqrt{3}Z)} = \sqrt{3}$

The phase current in the delta connection is $\sqrt{3}$ times the phase current in the star connection. This means the phase current increases, but not necessarily tripled or doubled as stated in options 1 and 2.

Line Current Comparison

Comparing the line currents:

Ratio: $\frac{I_{L(delta)}}{I_{L(star)}} = \frac{\sqrt{3} V_L/Z}{V_L/(\sqrt{3}Z)} = \sqrt{3} \times \sqrt{3} = 3$

The line current in the delta connection is three times the line current in the star connection. Option 3 is therefore incorrect.

Power Consumption Comparison

The power consumed by an impedance $Z$ with resistance $R$ and current $I$ is $P = 3 \times I_{ph}^2 \times R$. Alternatively, using line current, $P = 3 \times I_L^2 \times |Z| \times \cos(\phi)$, where $\cos(\phi)$ is the power factor.

Let's compare using phase currents:

• Power in Star: $P_{star} = 3 \times (I_{ph(star)})^2 \times R = 3 \times \left(\frac{V_L}{\sqrt{3}Z}\right)^2 \times R$
• Power in Delta: $P_{delta} = 3 \times (I_{ph(delta)})^2 \times R = 3 \times \left(\frac{V_L}{Z}\right)^2 \times R$

Comparing the power:

Ratio: $\frac{P_{delta}}{P_{star}} = \frac{3 \times (V_L/Z)^2 \times R}{3 \times (V_L/(\sqrt{3}Z))^2 \times R} = \frac{(V_L/Z)^2}{(V_L/(\sqrt{3}Z))^2} = \frac{V_L^2/Z^2}{V_L^2/(3Z^2)} = 3$

The power consumed in the delta connection is three times the power consumed in the star connection.

Conclusion

Based on the analysis, when three equal impedances are switched from a star connection to a delta connection across the same balanced 3-phase supply:

• Phase current increases by a factor of $\sqrt{3}$.
• Line current triples (increases by a factor of 3).
• Power consumed increases three-fold (increases by a factor of 3).

Therefore, the statement 'Power consumed will increase three-fold' is correct.