Final answer:
The heat unit (HU) is calculated by multiplying current (mA), exposure time (seconds), and peak kilovoltage (kVp), and then applying a correction factor for single-phase generators. In this case, the result is 1120 HU, which does not match any of the provided options, indicating a possible error in the question.
Step-by-step explanation:
The heat unit (HU) of the anode in radiology can be calculated by using the formula HU = mA × seconds × kVp. To find the heat unit for given parameters, we simply multiply the current in milliamps by the exposure time in seconds and the peak kilovoltage potential (kVp) used.
So for the given parameters of 200 mA, 0.10 sec, and 80 kVp on a single-phase generator, the calculation would be:
HU = 200 mA × 0.10 sec × 80 kVp = 1600 Heat Units
However, because we are working with a single-phase generator, we must apply a 0.7 correction factor to account for the waveform produced by the generator. Thus, the actual heat units would be:
Corrected HU = 1600 HU × 0.7 = 1120 Heat Units
None of the answer choices provided matches this calculation, so there might have been a misunderstanding or typo in the question itself.