Question

An energy meter makes 600 revolutions per kWh. If it makes 300 revolutions in 30 minutes, then the load power is

• A. 0.5 kW
• B. 1 kW
• C. 2 kW
• D. 4 kW

Hint:

Think of it proportionally: The meter requires 600 revolutions for 1 kWh. It completed 300 revolutions, which means exactly half a unit of energy (0.5 kWh) was consumed. Since this 0.5 kWh was used over exactly half an hour (30 minutes), the constant power load must be 1 kW.

Answer 1

The Correct Option is B

Concept: The bridge relationship between the rotation profile of an induction energy meter disc and the electricity consumed is defined by the meter constant (\(K\)): \[ K = \frac{\text{Number of Revolutions}}{\text{Energy Consumed in kWh}} \] Energy consumed (\(E\)) can also be related to the load power (\(P\), in kW) and time (\(t\), in hours) by the simple equation: \[ \text{Energy } (E) = \text{Power } (P) \times \text{Time } (t) \]

Step 1: Calculate total energy consumed during the period. We are given:
• Meter Constant (\(K\)) = 600 revolutions/kWh
• Observed revolutions (\(N\)) = 300 revolutions Rearranging the formula to find energy consumption: \[ E = \frac{N}{K} = \frac{300}{600} = 0.5\text{ kWh} \]

Step 2: Convert the operating time interval into hours units. The time duration provided is: \[ t = 30\text{ minutes} = \frac{30}{60}\text{ hours} = 0.5\text{ hours} \]

Step 3: Calculate the load power from the energy and time. Using the energy equation: \[ E = P \times t \quad \Rightarrow \quad 0.5\text{ kWh} = P \times 0.5\text{ hours} \] Solving for power \(P\): \[ P = \frac{0.5\text{ kWh}}{0.5\text{ hours}} = 1\text{ kW} \] Hence, the total power delivery requirement is exactly 1 kW, which matches option (B).