Electrical Energy Audit Questions and Answers

Answer: C

Correct answer: varies widely, so the correction must be switched in steps

An automatic bank measures the power factor continuously and switches stages in and out to follow it, which is what a varying load needs - fixed capacitors sized for the peak would overcompensate at light load and drive the power factor leading, which is penalised on many tariffs and can cause overvoltage. Individual compensation at the machine suits a single large motor or a long cable run, since it relieves the cable as well.

Answer: D

The site then exports reactive power, the voltage at the busbar rises and many tariffs penalise a leading factor as they do a lagging one. It happens most often when a fixed bank sized for full load remains connected at light load.

Answer: A

The still-rotating machine and the capacitor form a self-excited generator whose voltage can rise well above nominal and damage both. Sizing the capacitor below the motor's magnetising current, and switching it with the motor, prevents it.

Answer: B

The core is magnetised whenever the transformer is energised, so the hysteresis and eddy current loss continues through the night and the weekend. On a lightly loaded transformer it is the dominant loss and the reason for switching one off.

Answer: C

Correct answer: the square of the load current

Load loss is the copper loss, \( I^2R \), so it varies with the square of the current: a transformer at half load has a quarter of its full-load copper loss. The no-load or iron loss, by contrast, is essentially constant while the unit is energised, and the interaction of those two different laws is what produces the efficiency peak examined elsewhere in this topic.

Answer: D

Differentiating the efficiency expression puts the maximum where the load-dependent loss has risen to equal the constant loss, which for most distribution transformers falls between forty and seventy per cent of rating. It is not at full load.

Answer: A

Each is paying its full iron loss for a fraction of its rating, and switching one off removes that loss entirely while the other moves closer to its efficient region. Reliability and the ability to carry the peak have to be checked before doing it.

Answer: B

Correct answer: raising the load to the full nameplate rating

Loading a transformer to its full rating maximises the copper loss, since that grows with the square of the current, so it increases the losses rather than reducing them. The other three genuinely help: a low-loss core cuts the constant iron loss, consolidating two lightly loaded units removes one unit’s iron loss entirely, and correcting the power factor at the load reduces the current the transformer must carry.

Answer: C

Halving the current quarters the loss, which is why power factor correction and load balancing pay twice over: they reduce a squared quantity. Increasing the conductor size reduces the resistance and the loss in proportion.

Answer: D

An induction motor's torque falls with the square of the voltage, so it draws more current to deliver the same output and heats itself further. Correcting the drop therefore addresses efficiency and reliability together.