A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 leading is

Understanding Synchronous Motor Current at Rated Load

This problem involves understanding how the line current drawn by a three-phase synchronous motor changes when the power factor is altered, while the load and line voltage remain constant. A key principle is that the real power consumed by the motor at a specific load is primarily determined by the mechanical load and efficiency, not the power factor directly, although power factor impacts the current drawn.

Input Power Calculation Formula

The apparent power ($S$) drawn by a balanced three-phase system is given by the formula:

`$S=\sqrt{3}\times V_L\times I_L$ `

Where:

- `$V_L$ ` is the Line Voltage
- `$I_L$ ` is the Line Current

The real power ($P$), which represents the actual power consumed to do work, is calculated as:

`$P=\sqrt{3}\times V_L\times I_L\times \cos \varphi$ `

Where:

- `$\cos \varphi$ ` is the Power Factor (PF)

Analyzing the Motor Conditions

We are given two scenarios for the synchronous motor:

- Scenario 1: Rated load, unity power factor (PF = 1), Line current ($I_1$) = 200 A.
- Scenario 2: Same rated load, power factor (PF) = 0.5 leading, Line voltage constant. We need to find the new line current ($I_2$).

The problem states the motor operates "at rated load" and considers the "same line voltage and load". This implies that the real power input required by the motor remains constant in both scenarios (assuming efficiency is constant or the load refers to the mechanical output power required).

Step-by-Step Calculation

1. **Calculate Real Power ($P$) from Scenario 1:** Using the formula `$P=\sqrt{3}\times V_L\times I_1\times P{F}_1$ $P=\sqrt{3}\times V_L\times 200\text{A}\times 1$ So, the real power input is $P=200\sqrt{3}{V}_L$ ` Watts.
2. Use Real Power for Scenario 2: Since the real power ($P$) and line voltage ($V_L$) are the same in Scenario 2, we can write: `$P=\sqrt{3}\times V_L\times I_2\times P{F}_2$ Substitute the known values: $P{F}_2=0.5$ $200\sqrt{3}{V}_L=\sqrt{3}\times V_L\times I_2\times 0.5$ `
3. **Solve for the new Line Current ($I_2$):** We can cancel out the common terms `$\sqrt{3}{V}_L$ from both sides: $200=I_2\times 0.5$ ` Rearrange the equation to solve for $I_2$: `$I_2=\frac{200}{0.5}$ $I_2=400\text{A}$ `

Conclusion

When the synchronous motor operates at a power factor of 0.5 leading under the same load conditions and line voltage, the line current drawn increases to 400 A. This happens because to deliver the same real power (work), a lower power factor requires a higher current, and the reactive component of the current increases significantly when operating at a leading power factor.