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An AP radiograph of the femur was made using 300 mA, 0.03 second, 76 kV, 40-in. SID, 1.2-mm focal spot, and a 400-speed film–screen system. With all other factors remaining constant, which of the following exposure times would be required to maintain radiographic density using 87 kV, a 200-speed film–screen system, and the addition of a 12:1 grid?
The correction factor for a 12:1 grid is 5; therefore, the milliampere-seconds value (9) is multiplied by 5 to arrive at the new required milliampere-seconds value (45). Using the milliampere-seconds equation mA × s = mAs, it is determined that 0.15 second will be required at 300 mA:
An exposure was made using 600 mA and 18 ms. If the mA is changed to 400, which of the following exposure times would most closely approximate the original radiographic density?
Since 18 ms is equal to 0.018 s, and since mA × time = mAs, the original mAs was 10.8. Now it is only necessary to determine what exposure time must be used with 400 mA to provide the same 10.8 mAs (and thus the same radiographic density). Because mA × time = mAs,
400x = 10.8
x = 0.027 second (27 milliseconds)
An exposure was made at 40-in. SID using 5 mAs and 105 kVp with an 8:1 grid. In an effort to improve radiographic contrast, the image is repeated using a 12:1 grid and 90 kVp. Which of the following exposure times will bemost appropriate, using 400 mA, to maintain the original density?
seconds will be required (remember, 10 mAs is now the old mAs):
Thus, x = 12.5 mAs at 90 kVp. Now determine the exposure time required with 400 mA to produce 12.5 mAs:
The exposure factors of 400 mA, 17 ms, and 82 kV produce a milliampere-seconds value of
To calculate milliampere-seconds, multiply milliamperage times exposure time. In this case, 400 mA × 0.017 second (17 ms) = 6.8 mAs. Careful attention to proper decimal placement will help to avoid basic math errors. (Shephard, p. 170)