QuestionJune 5, 2025

A delta-connected three -phase alternator is driving a wye-connected resistive load. The alternator output voltage is 960 V.The load resistors are 12Omega 12 Q each. What is the phase current in the alternator? 80 Amps 26.66 Amps 20 Amps 40 Amps

A delta-connected three -phase alternator is driving a wye-connected resistive load. The alternator output voltage is 960 V.The load resistors are 12Omega 12 Q each. What is the phase current in the alternator? 80 Amps 26.66 Amps 20 Amps 40 Amps
A delta-connected three -phase alternator is driving a wye-connected resistive load. The alternator output voltage is 960 V.The load resistors are 12Omega 12 Q each. What is the phase current in the alternator? 80 Amps 26.66 Amps 20 Amps 40 Amps

Solution
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Answer

46.19 Amps Explanation 1. Calculate Line Voltage of Load In a delta-wye system, the line voltage of the load is equal to the phase voltage of the alternator. Therefore, V_{line} = 960 \, V. 2. Calculate Phase Voltage of Load For a wye-connected load, the phase voltage is given by V_{phase} = \frac{V_{line}}{\sqrt{3}}. Thus, V_{phase} = \frac{960}{\sqrt{3}} \approx 554.26 \, V. 3. Calculate Phase Current in Load Using Ohm's Law, I_{phase} = \frac{V_{phase}}{R}, where R = 12 \, \Omega. So, I_{phase} = \frac{554.26}{12} \approx 46.19 \, A. 4. Determine Phase Current in Alternator In a delta connection, the phase current is related to the line current by I_{line} = \sqrt{3} \times I_{phase}. Solving for I_{phase} gives I_{phase} = \frac{I_{line}}{\sqrt{3}}. Since I_{line} = I_{phase} for the load, I_{phase} = 46.19 \, A.

Explanation

1. Calculate Line Voltage of Load<br /> In a delta-wye system, the line voltage of the load is equal to the phase voltage of the alternator. Therefore, $V_{line} = 960 \, V$.<br />2. Calculate Phase Voltage of Load<br /> For a wye-connected load, the phase voltage is given by $V_{phase} = \frac{V_{line}}{\sqrt{3}}$. Thus, $V_{phase} = \frac{960}{\sqrt{3}} \approx 554.26 \, V$.<br />3. Calculate Phase Current in Load<br /> Using Ohm's Law, $I_{phase} = \frac{V_{phase}}{R}$, where $R = 12 \, \Omega$. So, $I_{phase} = \frac{554.26}{12} \approx 46.19 \, A$.<br />4. Determine Phase Current in Alternator<br /> In a delta connection, the phase current is related to the line current by $I_{line} = \sqrt{3} \times I_{phase}$. Solving for $I_{phase}$ gives $I_{phase} = \frac{I_{line}}{\sqrt{3}}$. Since $I_{line} = I_{phase}$ for the load, $I_{phase} = 46.19 \, A$.
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