Energy Meters Questions and Answers

Answer: A

Ten revolutions represent \( 10/1200 = 0.00833 \) kWh, which at 1.5 kW takes \( 0.00833/1.5 = 0.00556 \) hour, that is 20 seconds. A convenient test interval is one long enough for the stopwatch error to be small, so a slow meter is given more revolutions.

Answer: B

The error is \( (5.2 - 5.0)/5.0 = +0.04 \), that is four per cent fast. Referring the difference to the registered value instead would give 3.85 per cent, which is the commoner mistake, but error is always referred to the true value.

Answer: C

Correct answer: plus 2.5 per cent

The meter records \( 41/800 = 0.05125 \) kWh in one minute, which is a power of \[ 0.05125 \times 60 = 3.075\ \mathrm{kW}. \] Against the true 3 kW that is \[ \frac{3.075-3}{3} = +2.5\%. \] The sign matters commercially: the meter is running fast and would over-charge the consumer.

Answer: D

One revolution is \( 1/1200 = 8.33 \times 10^{-4} \) kWh, and over a sixth of an hour that is \( 8.33 \times 10^{-4} \times 6 = 5 \times 10^{-3} \) kW, or five watts. Over a month such creeping would register more than three units the consumer has not used.

Answer: A

The product is \( 5 \times 8 \times 30 = 1200 \) kilowatt hours. This is the quantity the register accumulates, and it is quite independent of how the demand was distributed within each day, which is why a separate demand charge exists.

Answer: B

Load factor is the average power divided by the maximum demand. The average is \( 6000/720 = 8.33 \) kW, and \( 8.33/50 = 0.167 \). A low load factor means plant sized for a peak that is seldom approached, which is expensive to provide.

Answer: C

A kilowatt is a thousand joules per second, and an hour is 3600 seconds, so the product is \( 3.6 \times 10^{6} \) joules. The kilowatt hour survives as a unit because it is a convenient size for billing, not because it is coherent with the rest of the system.

Answer: D

Correct answer: 2.4 kW

Five revolutions represent \( 5/375 = 0.013333 \) kWh, delivered in 20 seconds, which is \( 20/3600 \) of an hour: \[ \frac{0.013333}{0.005556} = 2.4\ \mathrm{kW}. \] The check on any such calculation is the units - revolutions divided by the constant give kilowatt hours, and dividing by the time in hours gives kilowatts.

Answer: A

The current ratio is forty and the voltage ratio one hundred, so the energy indicated is a four-thousandth of the true energy. Such a meter is often supplied with its register already geared to allow for the transformers, in which case no multiplier is applied.

Answer: B

Driving torque is proportional to power and braking torque to speed, so the steady speed is proportional to power. That linear relation is the whole reason the revolutions counted are proportional to the energy consumed.