AEC and Technique charts

The back-up timer terminates the exposure when the total mAs has reached: A. 100 mAs B. 300 mAs C. 600 mAs D. 1,200 mAs

. 600 mAs The back-up timer stops the exposure at 150% of the expected exposure or 600 mAs. This is only likely to happen if the technologist activates the automatic exposure control (AEC) cells within the wrong bucky in the x-ray room.

A radiograph is acquired using 450 mA and an exposure time of 0.1 seconds. What new exposure time must be used to obtain the same receptor exposure if the radiograph is repeated with 300 mA? A. 0.15 seconds B. 0.066 seconds C. 0.23 seconds D. 0.044 seconds

.15 second 450 mA x 0.1 s = 45 mAs. 45 mAs/300 mA = 0.15 seconds.

A radiograph is acquired using 200 mA and an exposure time of 200 milliseconds. What new exposure time must be used to obtain the same receptor exposure if the radiograph is repeated with 300 mA? A. 133 milliseconds B. 300 milliseconds C. 89 milliseconds D. 450 milliseconds

200 milliseconds/1000 = 0.2 seconds 200 mA x 0.2 s = 40 mAs. 40 mAs/300 mA = 0.133 seconds. 0.133 seconds x 1000 = 133 milliseconds.

A radiograph is acquired using 500 mA and an exposure time of 100 milliseconds. What new mA must be used to obtain the same receptor exposure if the radiograph is repeated using 400 milliseconds? A. 125 mA B. 31.25 mA C. 2000 mA D. 0.8 mA

500 mA x 0.1 s = 50 mAs. 50 mAs/0.4 s = 125 mA.

If a technologist receives a dose of 0.3 µGy from leakage radiation at 1 meter from the x-ray tube, estimate their dose at a distance of 5 meters from the x-ray tube. A. 0.012 µGy B. 7.5 µGy C. 1.5 µGy D. 0.06 µGy

A. 0.012 µGy This question requires using the inverse square calculation. Start with the original technologist dose (0.3), multiply by the original distance squared (12), and divide by the new distance squared (52). That's 0.3 x 12 / 52 = 0.012 µGy. Always check your math to make sure you squared the distances.. The distance INCREASED, so the new technologist dose should DECREASE.

If the patient dose is 1 mGy at a 40 inch SID, what is the patient dose at a 72 inch SID using the same technical factors? A. 0.3 mGy B. 0.55 mGy C. 3.25 mGy D. 1.8 mGy

A. 0.3 mGy This question requires using the inverse square calculation. Start with the original patient dose (1), multiply by the original distance squared (402), and divide by the new distance squared (722). That's 1 x 402 / 722 = 0.3 µGy. Always check your math to make sure you squared the distances. The distance INCREASED, so the new patient dose should DECREASE.

An x-ray exposure at 40 inches results in a beam intensity of 200 µGy. Estimate the new intensity for a distance of 50 inches. A. 128 µGy This question requires using the inverse square calculation. The basic form of the inverse square calculation is (I1)/(I2)=(D22)/(D12). Another useful form of the calculation is (I2) = (I1) x (D12)/(D22), which allows us to directly calculate the new intensity without any additional algebra or cross-multiplication. Start with the original intensity (200), multiply by the original distance squared (402), and divide by the new distance squared (502). That's 200 x 402 / 502 = 128 µGy. Always check your math to ensure you squared the distances. The distance INCREASED, so the new intensity should DECREASE. B. 160 µGy C. 312 µGy D. 250 µGy

A. 128 µGy This question requires using the inverse square calculation. The basic form of the inverse square calculation is (I1)/(I2)=(D22)/(D12). Another useful form of the calculation is (I2) = (I1) x (D12)/(D22), which allows us to directly calculate the new intensity without any additional algebra or cross-multiplication. Start with the original intensity (200), multiply by the original distance squared (402), and divide by the new distance squared (502). That's 200 x 402 / 502 = 128 µGy. Always check your math to ensure you squared the distances. The distance INCREASED, so the new intensity should DECREASE.

If an optimal radiograph is acquired at 72 inches with an air kerma of 4 mGy, what will the air kerma be if the distance is reduced to 40 inches? A. 13 mGy B. 1.23 mGy C. 4.6 mGy D. 7.2 mGy

A. 13 mGy This question requires using the inverse square calculation. Start with the original air kerma (4), multiply by the original distance squared (722), and divide by the new distance squared (402). That's 4 x 722 / 402 = 13 mGy. Always check your math to make sure you squared the distances. The distance DECREASED, so the new air kerma should INCREASE.

A radiograph is acquired using 300 mA and an exposure time of 0.15 seconds. What new mA must be used to obtain the same receptor exposure if the radiograph is repeated with an exposure time of 0.25 seconds? A. 180 mA. B. 500 mA C. 108 mA D. 833 mA

A. 180 mA The receptor exposure will remain the same when the mA is changed to 180 mA. This ensures that the new mAs is the same as the original mAs. Start by calculating the original mAs: 300 mA x 0.15 s = 45 mAs. Calculate the new mA by dividing the mAs by the new exposure time: 45 mAs/0.25 s = 180 mA.

When performing a lateral thoracic spine radiograph with an automatic exposure control (AEC)system, what technical change will increase the total exposure time without increasing the receptor exposure? A. Decrease mA B. Increase kVp C. Decrease back-up timer D. Increase density setting

A. Decrease mA When using automatic exposure control (AEC), decreasing the mA will result in increased exposure time. Decreasing the mA decreases the exposure RATE, so it takes more time to reach the desired exposure level at which the system terminates the exposure. Changes to mA affect the exposure RATE and the exposure TIME, but the total receptor exposure remains the same.

What is the advantage of a focused grid as compared to a non-focused grid? A. Decreased grid cut-off B. Increased image contrast C. Increased recorded detail D. Decrease distortion

A. Decreased grid cut-off There is less grid cut-off with the focused grid. There is less absorption of useful radiation by the grid because the lead strips are angled to match the divergence of the beam.

According to the inverse square law, if the source-to-image distance decreases the receptor exposure will: A. Increase B. Decrease C. Remain the same

A. Increase Assuming no other factors change, decreasing the source-to-image distance (SID) increases receptor exposure or the amount of radiation striking the receptor. Bringing the x-ray tube closer to the receptor results in a more concentrated x-ray beam and more photons striking the receptor.

If the source-to-image distance (SID) is decreased by a factor of 3, the x-ray beam intensity will: A. Increase by a factor of 32 B. Decrease by a factor of 32 C. Increase by a factor of 3 D. Decrease by a factor of 3

A. Increase by a factor of 32 Remember that the relationship between distance (SID) and beam intensity is both INVERSE and SQUARED. Since the SID has DECREASED by a factor of 3, the beam intensity will INCREASE by 3 SQUARED.

When using a radiographic unit with automatic exposure control (AEC), which of the following can be employed to reduce patient motion while maintaining proper receptor exposure? A. Increase mA B. Decrease kVp C. Increase back-up timer D. Decrease density setting

A. Increase mA When using automatic exposure control (AEC), a reduced patient motion may be achieved by increasing the mA. Increasing the mA increases the exposure RATE, and decreases the exposure TIME. Decreasing the exposure time helps to minimize the effect of motion blur on a radiograph.

Which of the following modifications may be used to decrease the total exposure time when using an automatic exposure control (AEC) system? A. Increase mA B. Decrease kVp C. Decrease back-up timer D. Increase density setting

A. Increase mA When using automatic exposure control (AEC), increasing the mA will result in a decreased exposure time. Increasing the mA increases the exposure RATE, so it takes less time to reach the desired exposure level at which the system terminates the exposure.

Which of the following projections may be improved by using a low-mA long-exposure time technique? A. Lateral thoracic spine B. Supine abdomen C. Decubitus chest D. AP axial cranium

A. Lateral thoracic spine The lateral thoracic spine projection may be improved by using a low-mA long-exposure time technique. This is called a "breathing technique". The exposure time is increased and the patient is instructed to breathe gently during the exposure. Breathing causes the ribs to appear blurred on the image and increases visibility of the underlying thoracic spine. When the exposure time is INCREASED, the mA must be DECREASED by the same proportion to maintain the same total mAs and the same total receptor exposure.

Which of the following statements best describes a focused grid? A. Lead strips are angled to match the divergence of the x-ray beam B. Lead strips are perpendicular to each other C. Two sets of lead strips are present at 90 degrees to each other D. Lead strips are closer together in focused grids

A. Lead strips are angled to match the divergence of the x-ray beam Lead strips are angled to match the divergence of the x-ray beam in a focused grid. The angled lead strips allow more of the useful beam to penetrate through the grid without absorption, which prevents grid cut-off.

Which of the following is an important benefit of using an automatic exposure control (AEC) system? A. Minimize patient dose B. Minimize patient motion C. Minimize image artifact D. Minimize heat accumulation in the tube

A. Minimize patient dose An essential benefit of using an automatic exposure control, or AEC, system is minimizing patient dose. Once the correct receptor exposure is achieved, the AEC system shuts down. This prevents overexposure and excess dose to the patient.

Which of the following projections may be improved by using a low-mA long-exposure time technique? A. RAO sternum B. Upright abdomen C. AP axial clavicle D. Lateral chest

A. RAO sternum The RAO sternum projection may be improved by using a low-mA long-exposure time technique. This is called a "breathing technique". The exposure time is increased and the patient is instructed to breathe gently during the exposure. Breathing causes the ribs to appear blurred on the image and increases visibility of the underlying sternum. When the exposure time is INCREASED, the mA must be DECREASED by the same proportion to maintain the same total mAs and the same total receptor exposure.

A radiograph is acquired with a source-to-skin distance (SSD) of 50 inches. How will the patient's dose change if the radiograph is repeated using the same exposure factors but decreasing the SSD to 35 inches? A. The patient's dose will increase B. The patient's dose will decrease C. The patient's dose will remain the same

A. The patient's dose will increase Decreasing the SSD from 50 inches to 35 inches will result in an increased x-ray beam intensity. Moving the x-ray tube closer to the patient results in a more concentrated x-ray beam, which is described as an increase in intensity (quantity). DECREASING distance INCREASES intensity. This relationship is governed by the inverse square law.

What is the main effect of grid cut-off? A. Underexposure of the image receptor B. Overexposure of the image receptor C. Decreased image contrast D. Reduced image spatial resolution

A. Underexposure of the image receptor Grid cut-off results in underexposure of the image receptor. Grid cut-off is the absorption of some of the useful x-ray beam, meaning that less radiation will reach the image receptor.

Which of the following situations will result in grid cut-off? 1. Incorrect SID with a focused grid 2. Upside down focused grid 3. Upside down non-focused grid A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

B. 1 and 2 only The lead strips in a focused grid are designed to match the divergence of the x-ray beam, but only when used at the correct source-to-image distance (SID). At an incorrect SID, the useful radiation can be traveling directly into the lead strips and will be absorbed. A focused grid, when used upside-down, will absorb almost all of the useful radiation, except for a small amount in the center of the x-ray field. Because the lead strips are parallel to each other in a non-focused grid, using it upside-down will have no effect.

If a technologist receives a dose of 0.3 µGy from leakage radiation at 25 inches from the x-ray tube, estimate their dose at a distance of 5 inches from the x-ray tube. A. 0.012 µGy B. 7.5 µGy C. 1.5 µGy D. 0.06 µGy

B. 7.5 µGy This question requires using the inverse square calculation. Start with the original technologist dose (0.3), multiply by the original distance squared (252), and divide by the new distance squared (52). That's 0.3 x 252 / 52 = 7.5 µGy. Always check your math to make sure you squared the distances. The distance DECREASED, so the new technologist dose should INCREASE.

According to the inverse square law, when the distance between the x-ray source and the point of measurement (patient, worker, or receptor) increases the beam quantity will: A. Increase B. Decrease The Inverse Square Law states that as the distance from the x-ray source increases, beam quantity (the number of photons) decreases. Similar to stepping away from a flashlight, the light beam becomes less intense with fewer light photons and appears less bright. C. Remain the same

B. Decrease The Inverse Square Law states that as the distance from the x-ray source increases, beam quantity (the number of photons) decreases. Similar to stepping away from a flashlight, the light beam becomes less intense with fewer light photons and appears less bright.

If the source-to-image distance (SID) is increased by a factor of 2, the x-ray beam intensity will: A. Increase by a factor of 22 B. Decrease by a factor of 22 C. Increase by a factor of 2 D. Decrease by a factor of 2

B. Decrease by a factor of 22 The relationship between distance (SID) and beam intensity is both INVERSE and SQUARED. Since the SID has INCREASED by a factor of 2, the beam intensity will DECREASE by 2 SQUARED.

What is the main advantage of the use of a moving grid over a stationary grid? A. Grid lines will be more visible with a moving grid B. Grid lines will be blurred and not visible with a moving grid C. A moving grid is more efficient at absorbing scatter radiation D. A moving grid will produce higher contrast images

B. Grid lines will be blurred and not visible with a moving grid Gridlines are blurred and not visible with a moving grid. By creating the movement of the grid in the table or upright Bucky, grid lines become blurred and are not visible on the image. A stationary grid may not be moved during the exposure so that grid lines may be visible on the image. The movement of the grid does not change the efficiency of the grid at absorbing scatter radiation.

A radiograph of an noncompliant patient demonstrates adequate receptor exposure but substantial motion blur. Which of the following technical changes will reduce motion blur in the repeat radiograph without changing the receptor exposure? A. Increase kVp and increase mA B. Increase mA and decrease exposure time C. Decrease mA and increase exposure time D. Decrease kVp and decrease mA

B. Increase mA and decrease exposure time Increasing mA and decreasing exposure time will reduce motion blur without changing the total receptor exposure. Motion blur is minimized by decreasing the exposure time. The receptor exposure is maintained by increasing the mA so that the total mAs remains the same.

When performing a PA chest radiograph using an automatic exposure control (AEC) system, the technologist engages only the center AEC ionization chamber. How might this affect the radiographic image? A. Increased exposure time and decreased receptor exposure B. Increased exposure time and increased receptor exposure C. Decreased exposure time and increased receptor exposure D. Decreased exposure time and decreased receptor exposure

B. Increased exposure time and increased receptor exposure\because the center of the lunf is spine so AEC has to increase EX.Time + increase EX. receptor This scenario will result in increased exposure time and increased receptor exposure. The technologist should have activated the outer cells under the lung fields, which are the primary structures of interest. The center cell is under the high-density spine, which will cause the automatic exposure control (AEC) cell to "fill up" more slowly and extend the exposure beyond the exposure necessary for this radiograph.

Consider an x-ray examination that uses these technical factors: 300 mA, 100 ms, 75 kVp, and a 72 inch SID. Changing the SID to 40 inches will result in: A. Increased x-ray beam energy B. Increased x-ray beam intensity C. Decreased x-ray beam energy D. Decreased x-ray beam intensity

B. Increased x-ray beam intensity Decreasing the SID from 72 inches to 40 inches will result in an increased x-ray beam intensity. Moving the x-ray tube closer to the receptor results in a more concentrated x-ray beam, which is described as an increase in intensity (quantity). DECREASING distance INCREASES intensity. This relationship is governed by the inverse square law.

What is the primary function of the back-up timer? A. Prevent over-exposure to the receptor B. Prevent over-exposure to the patient C. Ensure adequate receptor exposure D. Ensure adequate image contrast

B. Prevent over-exposure to the patient The primary function of the back-up timer with an automatic exposure control (AEC) system is to prevent over-exposure to the patient. The back-up timer stops the exposure at 150% of the expected exposure or 600 mAs. This is only likely to happen if the technologist activates the AEC cells within the wrong bucky in the x-ray room.

A radiograph is acquired using a source-to-image distance (SID) of 40 inches. How will the receptor exposure change if the radiograph is repeated using the same exposure factors but increasing SID to 60 inches? A. Receptor exposure will increase B. Receptor exposure will decrease. C. Receptor exposure will remain the same

B. Receptor exposure will decrease Increasing the SID from 40 inches to 60 inches will result in decreased x-ray beam intensity. Moving the x-ray tube farther from the receptor results in a less concentrated x-ray beam, which is described as a decrease in intensity (quantity). INCREASING distance DECREASES intensity. This relationship is governed by the inverse square law.

If the patient dose is 1 mGy at a 72 inch SID, what is the patient dose at a 40 inch SID using the same technical factors? A. 0.3 mGy B. 0.55 mGy C. 3.24 mGy. D. 1.8 mGy

C. 3.24 mGy This question requires using the inverse square calculation. Start with the original patient dose (1), multiply by the original distance squared (722), and divide by the new distance squared (402). That's 1 x 722 / 402 = 3.24 µGy. Always check your math to make sure you squared the distances. The distance DECREASED, so the new patient dose should INCREASE.

An exposure is made using 70 kVp, 300 mA, and a 0.1 second exposure time. Calculate the mAs. A. 7 mAs B. 4 mAs C. 30 mAs D. 370 mAs

C. 30 mAs The mAs is calculated by multiplying the or milliamperage (mA) by the exposure time (in seconds): 300 mA x 0.1 s = 30 mAs. The kVp is not used to calculate the mAs.

An x-ray exposure at 50 inches results in a beam intensity of 200 µGy. Estimate the new intensity for a distance of 40 inches. A. 128 µGy B. 160 µGy C. 312 µGy D. 250 µGy

C. 312 µGy This question requires using the inverse square calculation. Start with the original intensity (200), multiply by the original distance squared (502), and divide by the new distance squared (402). That's 200 x 502 / 402 = 312 µGy. Always check your math to make sure you squared the distances. The distance DECREASED, so the new intensity should INCREASE.

A radiograph is acquired with an automatic exposure control (AEC) system at 40 inches SID with a receptor exposure of 4 mGy. If the radiograph is repeated on the same system at 72 inches SID, the approximate receptor exposure will be: A. 12.9 mGy B. 1.23 mGy Your Answer C. 4.0 mGy D. 7.2 mGy

C. 4.0 mGy Correct Answer When using automatic exposure control (AEC), changing the source-to- image distance (SID) from 40 inches to 72 inches will result in the same receptor exposure of approximately 4.0 mGy. The AEC system is designed to maintain the appropriate receptor exposure, even when variables such as the SID are changed. Increasing the SID will decrease the exposure RATE and increase the exposure TIME, but the total receptor exposure will remain the same.

An exposure is made using 80 kVp, 250 mA, and a 0.25 second exposure time. Calculate the mAs. A. 20 mAs B. 3 mAs C. 63 mAs D. 330 mAs

C. 63 mAs The mAs is calculated by multiplying the mA by the exposure time (in seconds): 250 mA x 0.25 s = 62.5 mAs, which rounds to 63 mAs. The kVp is not used to calculate the mAs.

What is the primary advantage of using a grid for a radiographic procedure? A. Reduced scatter production B. Increased recorded detail C. Decreased scatter reaching the image receptor D. Decreased image contrast

C. Decreased scatter reaching the image receptor Grids decrease the amount of scatter radiation reaching the image receptor. The function of the grid is to absorb the scatter radiation coming from the patient before it hits the receptor, therefore increasing image contrast. Because the grid is located behind the patient, it will not affect the amount of scatter produced.

Which of the following errors in grid use will have the same results as an angulation error? A. Upside down focused grid B. Off-focus, or incorrect SID with a focused grid C. Off-level, or tilted grid error. D. Upside down non-focused grid

C. Off-level, or tilted grid error Off-level grids, or tilted grid error, have the same effect as angulation grid errors. The beam angled across the lead strips and a tilted grid will both result in useful radiation being directed into the lead strips instead of between the lead strips. The result of either error is an overall decrease in exposure across the entire image.

When the distance between the x-ray source and the point of measurement increases, the beam energy will: A. Increase B. Decrease C. Remain the same

C. Remain the same Changing the distance has NO EFFECT on the beam ENERGY, or quality, but only affects beam quantity, or the number of photons.

The total mAs is calculated as the product of which exposure factors? A. Tube potential and tube current B. Tube potential and exposure time C. Tube current and exposure time D. Tube current and filament current

C. Tube current and exposure time The total mAs is the product of the tube current or milliamperage (mA) and the exposure time in seconds (s). The mA multiplied by the exposure time (in seconds) equals the mAs.

The inverse square law defines the relationship between distance and: A. Beam energy B. Beam quality C. Beam penetrability D. Beam intensity

D. Beam intensity The inverse square law defines the relationship between distance and beam intensity. The x-ray beam is divergent, meaning it is spread out through space and becomes less concentrated. This decreases the number of photons in the x-ray beam, or intensity but does not affect the beam energy (a.k.a. quality or penetrability).

What technical change is required to decrease receptor exposure when using automatic exposure control (AEC)? A. Increase mA B. Decrease kVp C. Increase back-up timer D. Decrease density setting

D. Decrease density setting Decreasing the density setting is the primary way to decrease receptor exposure when using automatic exposure control (AEC). The design of the AEC system is to maintain the appropriate receptor exposure, even when variables such as the patient size, mA, and kVp are changed. Decreasing the density setting decreases the exposure level at which the system terminates the exposure. Changes to mA and kVp affect the exposure RATE and the exposure TIME, but the total receptor exposure remains the same.

Increasing the distance from the x-ray source results in: A. Increased beam energy B. Decreased beam energy C. Increased beam intensity D. Decreased beam intensity

D. Decreased beam intensity Increasing the distance from the x-ray source will cause a decrease in beam intensity.The relationship is inverse because when distance INCREASES the beam intensity DECREASES. Changes in distance affect the beam intensity, or quantity, according to the inverse square law. Changes in the distance have no effect on the beam energy or quality.

Which of the following effects will occur when using an automatic exposure control (AEC) system to image a hyposthenic patient? A. Decreased spatial resolution B. Increased contrast resolution C. Increased receptor exposure D. Decreased exposure time

D. Decreased exposure time Imaging a hyposthenic patient with an automatic exposure control (AEC) system will result in a decreased exposure time. A hyposthenic patient is a small or feeble patient. A high percentage of x-rays transmit through smaller patients, which allows the ideal receptor exposure in a shorter amount of time. The reverse would be true for larger patients. They absorb MORE radiation, resulting in less x-ray transmission and longer exposure time.

What is the effect of using an automatic exposure control (AEC) system to image a patient with a destructive pathology such as osteoporosis? A. Decreased spatial resolution B. Increased contrast resolution C. Increased receptor exposure D. Decreased exposure time

D. Decreased exposure time Imaging a patient with osteoporosis using an automatic exposure control (AEC) system will result in a decreased exposure time. This will have the same effect as imaging a smaller, hyposthenic patient. Osteoporosis is a destructive pathology, causing bone demineralization and an overall decrease in bone density. A high percentage of x-rays transmit through demineralized bone, which allows the ideal receptor exposure to achieve in a shorter amount of time. The reverse would be true for additive pathologies, such as ascites or fluid in the abdomen.

A technologist is performing a PA chest radiograph on a bariatric patient using an automatic exposure control (AEC) system. Which of the following changes will be noted if the technologist increases the mA? A. Increased patient dose B. Increased receptor exposure C. Decreased contrast D. Decreased exposure time

D. Decreased exposure time When using automatic exposure control (AEC), increasing the mA will result in a decreased exposure time. Increasing the mA increases the exposure RATE, so it takes less time to reach the desired exposure level at which the system terminates the exposure. The patient dose and receptor exposure remain the same. The changes in mA do not affect image contrast.

When performing a lateral cervical spine using an automatic exposure control (AEC) system, the technologist activates the outer ionization chambers. How might this affect the radiographic image? A. Increased exposure time and decreased receptor exposure B. Increased exposure time and increased receptor exposure C. Decreased exposure time and increased receptor exposure D. Decreased exposure time and decreased receptor exposure

D. Decreased exposure time and decreased receptor exposure This scenario will result in decreased exposure time and decreased receptor exposure. The technologist should have activated only the center automatic exposure control (AEC) cell directly under the spine. The outer cells are under the low density soft tissues of the neck, or possibly air, which will cause the AEC cells to "fill up" rapidly and terminate the exposure before adequate exposure of the cervical spine has been achieved.

When performing an AP lumbar spine using an automatic exposure control (AEC) system, the technologist inadvertently positions the patient with the active ionization chamber under a gas pocket in the abdomen. How will this affect the radiographic image? A. Decreased spatial resolution B. Decreased image contrast C. Increased distortion D. Decreased receptor exposure

D. Decreased receptor exposure Positioning a gas pocket over an active structure of interest will cause decreased receptor exposure by automatic exposure control (AEC) cells. AEC systems are designed to increase or decrease the exposure time in response to changes in patient size, density, or pathology to ensure the correct receptor exposure. However, a pocket of bowel gas allows for more x-ray transmission and causes the AEC chambers to "fill up" more rapidly, causing a decrease in exposure time and an overall decrease in receptor exposure.

The automatic exposure control (AEC) system communicates with which portion of the x-ray circuit when the optimal receptor exposure has been achieved? A. Filament circuit B. Autotransformer C. mA selector D. Exposure timer

D. Exposure timer The automatic exposure control (AEC) system communicates with the exposure timer within the x-ray circuit when the optimal receptor exposure has been achieved. The purpose of AEC is to control the receptor exposure by controlling the exposure time. The exposure timer is within the low voltage,primary, portion of the x-ray circuit.

Which grid error results in grid cut-off and loss of exposure on the outside edges of the image receptor? A. Upside down non-focused grid B. Incorrect angulation of the beam across lead strips C. Off-level, or tilted grid error D. Incorrect SID with a focused grid Using an incorrect SID with a focused grid result in grid cut-off and loss of exposure on the outside edges of the image receptor. This is also called off-focus grid error Focused grids are designed to be used with specific SID ranges as specified by the manufacturers. Grid cut-off will occur more at the outer edges of the grid when the SID is outside of this range, resulting in loss of exposure to the image receptor in those areas.

D. Incorrect SID with a focused grid Using an incorrect SID with a focused grid result in grid cut-off and loss of exposure on the outside edges of the image receptor. This is also called off-focus grid error Focused grids are designed to be used with specific SID ranges as specified by the manufacturers. Grid cut-off will occur more at the outer edges of the grid when the SID is outside of this range, resulting in loss of exposure to the image receptor in those areas.

What technical change is required to increase receptor exposure when using automatic exposure control (AEC)? A. Increase mA B. Decrease kVp C. Decrease back-up timer D. Increase density setting

D. Increase density setting Increasing the density setting is the primary way to increase receptor exposure when using automatic exposure control (AEC). The AEC system is designed to maintain the appropriate receptor exposure, even when variables such as the patient size, mA, and kVp are changed. Increasing the density setting increases the exposure level at which the system terminates the exposure. Increasing the mA increases the exposure RATE and decreases the exposure TIME, but does not change the total receptor exposure.

A radiograph of the lumbar spine is produced using automatic exposure control (AEC). The exposure results in optimal contrast but low receptor exposure. What is the ideal way to increase the receptor exposure? A. Increase mA B. Increase kVp C. Decrease grid ratio D. Increase density setting

D. Increase density setting Increasing the density setting is the primary way to increase receptor exposure when using automatic exposure control (AEC). This is because increasing the density setting increases the exposure level at which the system terminates the exposure. Increasing the mA or kVp will increase the exposure RATE and shorten the exposure TIME, but the receptor exposure will remain the same. Avoid changing the kVp because kVp is a key factor influencing image contrast.

Which of the following effects will occur when using an automatic exposure control (AEC) system to image a hypersthenic patient? A. Increased spatial resolution B. Decreased contrast resolution C. Decreased receptor exposure D. Increased exposure time

D. Increased exposure time Imaging a hypersthenic patient with an automatic exposure control (AEC)system will result in an increased exposure time. A hypersthenic patient is a larger patient with a broad body and a deep chest and abdomen. A lower percentage of x-rays will transmit through larger patients, which causes the ideal receptor exposure to be achieved in a longer amount of time. The reverse would be true for smaller patients. They absorb LESS radiation, resulting in more x-ray transmission and shorter exposure time.

A technologist is performing an AP abdomen radiograph on a geriatric patient using an automatic exposure control (AEC) system. Which of the following changes will be noted if the technologist decreases the kVp? A. Decreased patient dose B. Increased receptor exposure C. Decreased contrast D. Increased exposure time

D. Increased exposure time When using automatic exposure control (AEC), decreasing the kVp will result in increased exposure time, decreasing the kVp results in fewer photons in the beam and less penetrating ability. By decreasing the exposure RATE, it takes more time to reach the desired exposure level at which the system terminates the exposure. Changes to mA and kVp affect the exposure RATE and the exposure TIME, but the total receptor exposure remains the same.

A technologist performing an AP pelvis projection unintentionally places gonadal shielding over two of the three active automatic exposure control (AEC) ionization chambers. How will this affect the exposure time and patient dose? A. Decreased exposure time and decreased patient dose B. Decreased exposure time and increased patient dose C. Increased exposure time and decreased patient dose D. Increased exposure time and increased patient dose E. No change to exposure time and patient dose

D. Increased exposure time and increased patient dose Placing gonadal shielding over ANY portion of ANY active automatic exposure control (AEC) chamber will cause a substantial increase in exposure time and increase in patient dose. Lead shielding absorbs a significant amount of the primary beam, which will cause the ionization chamber to "fill up" more slowly. This results in an increase in exposure time and an increase in patient dose. For this reason, it is critical that lead shielding is never placed over an active AEC cell.

What component within the automatic exposure control (AEC) system is responsible for measuring the quantity of photons exiting the patient? A. Rectifier bridge B. Autotransformer C. Exposure timer D. Ionization chamber

D. Ionization chamber An automatic exposure control (AEC) system includes three or more ionization chambers that serve to measure the quantity of photons exiting the patient and striking the image receptor. When the ionization chamber(s) has detected the ideal receptor exposure, a signal is sent to the exposure timer to terminate the exposure.

When using automatic exposure control (AEC), the technologist CANNOT directly assign which of the following parameters? A. Kilovoltage potential (kVp) B. Source-to-image distance (SID) C. Object-to-image distance (OID) D. Milliamp-seconds (mAs)

D. Milliamp-seconds (mAs) The technologist cannot directly assign the milliampere-seconds (mAs) when using an automatic exposure control (AEC) system. The technologist assigns the milliamperage (mA), but the AEC system controls the exposure time in seconds (s). As a result, the total mAs is not defined until the AEC system terminates the exposure.

Which situation is most likely to result in grid errors? A. Stationary radiography at the upright bucky B. Tabletop radiography C. Stationary radiography at the table bucky D. Mobile radiography

D. Mobile radiography Mobile radiography is most likely to result in grid errors. Mobile radiography creates more difficulties in maintaining the correct alignment of the x-ray beam to the grid and image receptor.

Consider a technologist performing a lumbar spine series using an AEC system. After acquiring the lateral radiograph, the technologist angles the beam to 5o caudal and collimates to the area of the L5-S1 joint. How should the technologist change the technical parameters to ensure the same receptor exposure as the lateral view? A. Increase kVp B. Increase mA C. Decrease the SID D. No change is required

D. No change is required When the beam angle and collimation is changed while using an automatic exposure control (AEC) system, the technologist should NOT make any changes to the technical parameters. The AEC ensures an ideal receptor exposure, even when variables change, such as the beam and collimation. The increased beam angle and decreased field size will result in a decreased exposure rate to the AEC ionization chambers, and consequently, a longer total exposure time. The technologist does not control the change in the exposure time; the AEC system controls this.

In order to ensure optimal receptor exposure, which portion of the patient's anatomy should be positioned over the active automatic exposure control (AEC) chamber(s)? A. The highest density structure B. The lowest density structure C. The structure of least concern D. The structure of primary concern.

D. The structure of primary concern The structure of primary concern should always be placed over the active automatic exposure control (AEC) cell(s). AEC systems are designed to maintain optimal receptor exposure for the structures over the active AEC cells. When imaging the chest, the lungs should be positioned over the active outer cells and the center (under the spine) should be deactivated. A chest x-ray requires optimal exposure through the lung fields, not the thoracic spine.

Which of the following exposure factors is controlled by the automatic exposure control (AEC) system? A. kVp B. mA C. SID D. Time

D. Time An automatic exposure control (AEC) system only controls the exposure time, and therefore the total produced milliamperage (mA) over a set amount of time, seconds (s), or mAs. AEC uses ionization chambers to measure the radiation passing to the receptor and shuts down the exposure when there is an appropriate receptor exposure. The total exposure time depends on how long it takes to achieve the correct exposure; therefore, larger patients and larger body parts require more time. When using an AEC system, the technologist must still manually set the kVp, mA, the SID, and all other procedural variables.

What is the primary purpose of an automatic exposure control (AEC) system? A. To enhance image contrast B. To increase radiographic spatial resolution C. To automatically modulate the kVp D. To maintain optimal receptor exposure

D. To maintain optimal receptor exposure The primary purpose of automatic exposure control (AEC) is to maintain optimal receptor exposure. Setting a manual technique can be very difficult, with a wide range of patient sizes and conditions. AEC uses ionization chambers to measure the radiation passing to the receptor and shuts down the exposure when the appropriate receptor exposure is achieved.