To achieve the greatest radiographic exposure, which combination of kVp, mA, and seconds is preferred?

To achieve the greatest radiographic exposure, the choice that includes a combination of kVp, mA, and time (seconds) must effectively maximize the overall exposure to the film or detector. The correct option, featuring 90 kVp, 200 mA, and 0.07 seconds, represents a favorable balance among these three variables.

Using 90 kVp increases the penetration power of the X-rays, which is essential for obtaining a clear image by penetrating tissues effectively. The moderate mA of 200 provides a substantial increase in the number of X-rays produced, thereby enhancing the dose without excessively increasing the exposure time. The exposure time of 0.07 seconds strikes a balance that minimizes motion blur, while still allowing for sufficient radiation to achieve the desired exposure level.

Together, this combination results in an optimal exposure, taking into account the need for sufficient penetration and quantity of X-rays produced in a brief timeframe to minimize patient exposure while ensuring image quality. This consideration makes it the most suitable option compared to the others, which either have lower kVp, higher or lower mA, or exposure times that do not match the effective balance found in the correct choice.