A time of day meter registers energy in separate registers according to
Answer: C
Peak and off-peak energy have different costs to the supply, so separate registers allow separate tariffs and encourage consumers to shift load. A clock or an external signal switches the registers at the appointed hours.
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A static energy meter differs from the induction type in that it has
Answer: D
Voltage and current are sampled, multiplied and integrated electronically, and the result is shown on a display and often transmitted as well. The absence of bearings removes friction, creeping and wear at a stroke, which is why such meters have displaced the induction type.
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A static energy meter senses the load current with
Answer: A
A low-value shunt gives a voltage proportional to the current for the electronics to sample, and a current transformer does the same while providing isolation. Both are far less burdensome than the series magnet of an induction meter.
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The pulse output of a static energy meter, expressed in pulses per kilowatt hour, corresponds to the induction meter's
Answer: B
Each pulse stands for a fixed increment of energy, exactly as each revolution of the disc does, so the two constants play the same part in a test. A test set counts pulses instead of revolutions, and the arithmetic is unchanged.
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A meter marked 1200 rev/kWh makes 20 revolutions in 30 seconds. The load is
Answer: C
The energy is \( 20/1200 = 0.01667 \) kWh, and dividing by the period expressed in hours, \( 30/3600 = 0.00833 \) h, gives 2 kW. The general form is \( P = 3600n/(Kt) \) with n revolutions in t seconds and K in rev/kWh.
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A meter of constant 600 rev/kWh is on a load of 3 kW. In one minute the disc makes
Answer: D
In one minute the energy is \( 3/60 = 0.05 \) kWh, and multiplying by 600 revolutions per kWh gives 30. Counting disc revolutions over a measured interval is the quickest way to check a meter against a wattmeter reading.
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A meter of constant 1200 rev/kWh is on a 1.5 kW load. The time for 10 revolutions is
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.
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A meter registers 5.2 kWh while a substandard shows 5.0 kWh. The error of the meter is
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.
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A meter of constant 800 rev/kWh makes 41 revolutions in 60 seconds while a wattmeter reads 3 kW. The error is about
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.
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A meter of constant 1200 rev/kWh creeps one revolution in 10 minutes with no load. The equivalent power it is registering is
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.
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