Motor Protection Circuit Breaker (MPCB) rating

calculate the Motor Protection Circuit Breaker (MPCB) rating for a 1 HP, 3-phase induction motor, we need to determine the full-load current and then select the appropriate MPCB.

Here's a step-by-step process:

### 1. Determine Full-Load Current (FLC)

The full-load current can be found using the following formula:

\[ \text{Full-load current (FLC)} = \frac{\text{Power (in HP)} \times 746}{\sqrt{3} \times \text{Voltage} \times \text{Efficiency} \times \text{Power Factor}} \]

For a 1 HP motor:

- 1 HP = 746 Watts

- Assume an efficiency (η) of 0.85 (85%)

- Assume a power factor (PF) of 0.8 (for induction motors)

- Voltage (V) for a 3-phase system is typically 400V or 415V

Let's use 400V for this calculation:

\[ \text{FLC} = \frac{1 \times 746}{\sqrt{3} \times 400 \times 0.85 \times 0.8} \]

Let's calculate this:

\[ \text{FLC} = \frac{746}{1.732 \times 400 \times 0.85 \times 0.8} \]

\[ \text{FLC} = \frac{746}{470.4} \approx 1.587 \text{ A} \]

### 2. Select MPCB Rating

MPCBs are typically rated in current ranges. For a motor with an FLC of around 1.587 A, the MPCB should be rated slightly higher to allow for starting currents and potential small overloads. Usually, a safety factor is added.

Common practice is to select an MPCB with a rating about 1.25 to 1.5 times the FLC to handle inrush currents. Here, we can use a safety factor of 1.25.

\[ \text{MPCB Rating} = 1.587 \times 1.25 = 1.98375 \text{ A} \]

### 3. Choose a Standard MPCB Rating

MPCBs are available in standard sizes. The next standard size available for MPCBs after 1.98 A would typically be 2 A or 2.5 A. 

For practical purposes, a 2 A MPCB is commonly available and would be suitable for this motor. If there’s a slight difference in standard available ratings, choose the one closest to but not lower than the calculated value.

### Conclusion

For a 1 HP, 3-phase induction motor, a 2 A MPCB would be an appropriate choice to protect the motor adequately while allowing for startup currents.