A three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf leading. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line a to the neutral, 15 kW from line b to the neutral, and 9 kW from line c to the neutral. Determine the current in neutral line.

Name: A three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf leading. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line a to the neutral, 15 kW from line b to the neutral, and 9 kW from line c to the neutral. Determine the current in neutral line.
Uploaded: 2023-06-17T17:47:52-08:00
Duration: 3 min 33 s
Channel: Sri K

          A three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf leading. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line a to the neutral, 15 kW from line b to the neutral, and 9 kW from line c to the neutral. Determine the current in neutral line.

A three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf leading. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line a to the neutral, 15 kW from line b to the neutral, and 9 kW from line c to the neutral. Determine the current in neutral line.

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Fundamentals of Electric Circuits

Charles K. Alexander, Matthew N.O.… 3rd Edition

Chapter 12

Instant Answer

Solved by Expert Sri K

06/17/2023

Step 1

The motor load is given as 260 kVA at 0.85 power factor leading. The line current \( I_{\text{line}} \) for a balanced three-phase load can be calculated using the formula: \[ I_{\text{line}} = \frac{S}{\sqrt{3} \cdot V_{\text{line}}} \] where \( S \) is the Show more…

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