A load of 5 kW runs for 8 hours a day for 30 days. The energy registered is
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.
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A consumer's maximum demand is 50 kW and the energy taken in a 720-hour month is 6000 kWh. The load factor is
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.
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One kilowatt hour expressed in joules is
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.
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A meter of constant 375 rev/kWh makes 5 revolutions in 20 seconds. The load is
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.
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A meter is used with a current transformer of ratio 200/5 and a potential transformer of 11000/110. Its register readings must be multiplied by
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.
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If the power taken by a load is doubled, the speed of the disc of a correctly adjusted energy meter
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.
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A purely inductive load draws 10 A at 230 V. Over an hour a correctly adjusted energy meter registers
Answer: C
The torque is proportional to \( VI\cos\phi \) and the cosine of ninety degrees is zero, so the disc does not turn. The load nevertheless draws current and causes losses in the network, which is why reactive energy is metered separately for industrial consumers.
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A three-element meter on a four-wire system has its elements balanced. If one phase carries 4 kW and the other two carry 2 kW each, the register advances at a rate corresponding to
Answer: D
Each element contributes a torque proportional to the power of its own phase, and since they act on a common spindle the torques add. The disc speed therefore corresponds to the total of eight kilowatts however unequally that total is divided.
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A meter tested at 5 per cent of rated current runs 6 per cent slow, and the permitted tolerance at that load is 3.5 per cent. The adjustment to make is to
Answer: A
Running slow at a small fraction of rated current is the signature of friction not fully compensated, and the friction adjustment is exactly the control provided for it. The braking magnet would shift the whole characteristic and spoil the full-load setting already made.
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True or false: an energy meter is an indicating instrument.
Answer: B
Indicating instruments show the present value of a quantity; an energy meter accumulates it over time, which makes it an integrating instrument. The distinction matters because an integrating instrument is tested over an interval and not at an instant.
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