If the nameplate rating of a 3Ø boiler is 12 KW at 240 volts, what will a meter indicate in amps?

To determine the current in amps for a 3-phase boiler with a nameplate rating of 12 kW at 240 volts, you can use the power formula for a 3-phase system, which is:

[

P = \sqrt{3} \times V \times I \times \text{Power Factor}

]

Assuming the power factor is 1 (since it's not specified), the formula simplifies to:

[

I = \frac{P}{\sqrt{3} \times V}

]

1. Substituting the values:

• ( P = 12,000 ) watts (since 12 kW = 12,000 watts)

• ( V = 240 ) volts

The calculation will be:

[

I = \frac{12,000}{\sqrt{3} \times 240}

]

2. Calculating the square root of 3:

• ( \sqrt{3} \approx 1.732 )

3. Substituting this into the formula:

[

I = \frac{12,000}{1.732 \times 240}

]