Math Exam 2
Since we Halve the Exposure we know we have to Decrease by 15% so.... 60(.85) = 51 kVp 51 kVp at 4 mAs would half the exposure to the IR because decreasing kVp by 15% and keeping the same mAs will HALF THE EXPOSURE.
A Radiograph of the Elbow is produced using 4 mAs at 60 kVp. What kVp is required to halve the exposure to the IR? (mAs too)
kVp = 80(.85) = 68 <----- Decrease kVp by 15% mAs = 600mA x 1/40s(.025s) = 15 mAs 68kVp at 15 mAs will provide an IR exposure that is HALF of the original (think - when we decrease by 15% and keep mAs the same, we are reducing by half or 50%) Since 68 kVp at 15 mAs is HALF the original, we will then need to multiply our mAs by 2 (doubling the amount of mAs is will get us back to the original exposure) 68 kVp at 30 mAs will produce the same exposure as one done at 80 kVp at 15 mAs But Since we are given our time of .10 secs, we need to solve for mAs mA = 30 mAs/.10 = 300 mA So.... kVp = 68 <------ Answer mA = 300 <------ Answer
Decrease the kVp using the 15% rule but with Equal Exposure Given: Original Technique 600 mA 1/40s 80 kVp New technique mA - ________ Sec - .10 kVp - ________
Increase kVp by 15% because that in itself will double the exposure.
For 15% rule questions, if you are increasing the kVp by 15% but wanting to double the exposure you wont have to do ANYTHING ELSE BUT:
I(2)/I(1) = D(1)^2/ D(2)^2 <--- Inverse Square Law x/5mR = 40^2/36^2 <--- plug in x/5mR = 1600/1296 <--- Exponents 1296x = 8000mR <--- Cross Multiply x = 6.17 mR <--- Divide 6.17 mR<--- Answer
For the Following, Determine the new radiation dose the pt receives after a change in distance. Original: Distance: 40 Radiation Dose (mR): 5 New: Distance: 36 Radiation Dose (mR): ????
I(2)/I(1) = D(1)^2/D(2)^2 <---- Inverse Square Law x/8mR = 48^2/72^2 <--- plug in x/8mR = 2304/5184 <--- Exponents 5184x = 18432 mR <--- Cross multiply x = 3.56mR <--- Divide 3.56 mR <--- Answer
For the Following, Determine the new radiation dose the pt receives after a change in distance. Original: Distance: 48 Radiation Dose (mR): 8 New: Distance: 72 Radiation Dose (mR): ????
I(2)/I(1) = D(1)^2/ D(2)^2 <--- Inverse Square Law x/2mR = 72^2/60^2 <--- plug in x/2mR = 5184/3600 <--- Exponents 3600x = 10368mR <--- Cross Multiply x = 2.88mR <--- Divide 2.88 mR <---Answer
For the Following, Determine the new radiation dose the pt receives after a change in distance. Original: Distance: 72 Radiation Dose (mR): 2 New: Distance: 60 Radiation Dose (mR): ????
mAs with grid/mAs without grid
Grid Conversion Factor (GCF) = (---)/(---)
mAs2/mAs1 = D(2)^2/D(1)^2 <---- Direct Square Law x/10mAs = 60^2/40^2 <--- Plug In x/10mAs = 3600/1600 <--- Exponents 1600x = 36000 mAs <--- Cross Multiply x = 22.5 mAs <--- Divide 22.5 mAs <--- Answer
If 10 mAS was used to produce a satisfactory radiograph at a 40 inch distance, what mAs would be required at a distance of 60 inches?
mAs1 / mAs2 = d1² / d2² <---- Direct Square Law 10 mAs/ X = 40 SID²/ 60 SID² <--- Plug in Info 10 mAs/X = 1600/ 3600 <--- Exponents 1600x = 36000 <--- Cross Multiply x = 36000/1600 <--- Divide x = 22.5 mAs mAs2 = 22.5 mAs <--- Answer
If 10 mAs was used to produce a satisfactory radiograph at 40 Inch Distance, what mAs would be required at a distance of 60 inches?
GCF = mAs with the grid/ mAs without the grid mAs with grid = 18 GCF (16:1 grid) = 6 (from GCF table) 6= 18/mAs without the grid [GCF = mAs with grid/ mAs without the grid] [divide each side by 6] 3 mAs without the grid <---- Answer
If A radiographer produces a knee image using a 16:1 ratio grid and 18 mAs, and on the next exposure wants to use a nongrid exposure, what mAs should be used to produce an image with comparable quality?
Object Size = Image Size/ Magnification Factor <--- Object Size formula O = 5"/1.02 O = 4.9" <--- Answer
If Projected Image measures 5" and the magnification factor is 1.02, what is the size of the actual object?
Because SOD is not Supplied, it must be found using the formula: SID = SOD + OID 40" = SOD + 2" SOD = 40" - 2" SOD = 38" then M = SID/SOD <--- Magnification Factor Formula M = 40"/38" M = 1.05 <---- Answer
If SID is 40" and OID is 2" what is the magnification factor?
GCF = mAs with the grid/mAs without the grid mAs without grid = 2 GCF (8:1 grid) = 4 from GCF table 4= mAs with the Grid/2 [GCF = mAs with grid/ mAs without the grid] [multiply each side by 2] 8 mAs with the Grid <---- Answer
If a Radiographer produces a knee image with a non-grid exposure using 2 mAs and next wants to use an 8:1 ratio grid, what mAs should be used to produce a comparable quality image?
mAs1/mAs2 = GCF1/GCF2 Exposure 1: 30 mAs, 6:1 grid, GCF = 3 Exposure 2: x mAs, 12:1 grid, GCF = 5 30/ mAs2 = 3/5 [mAs1/mAs2 = GCF1/GCF2] [Cross multiply] mAs2 = 50 <----- Answer
If a radiographer performs a routine portable abdomen examination using 30 mAs with a 6:1 ratio grid, what mAs should be used if a 12:1 ratio grid is substituted?
mAs1/mAs2 = GCF1/GCF2 Exposure 1: 40 mAs, 8:1 Grid, GCF = 4 Exposure 2: x mAs, 5:1 Grid, GCF = 2 40/mAs2 = 4/2 [mAs1/mAs2 = GCF1/GCF2] [Cross multiply] mAs2 = 20 <------ Answer
If a radiographer uses 40 mAs with an 8:1 ratio grid, what mAs should be used with a 5:1 ratio grid to maintain the same exposure to the IR?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 1.6/0.1 Grid Ratio = 16 Grid Ratio = 16:1
If the height of the lead strips is 1.6mm and the width between the strips is 0.1mm, the ratio of the grid is:
I1/I2 = D2^2/D1^2 <--- Inverse Square Law Formula Intensity 1 = 400 Intensity 2 = X Distance 1 = 40 Inches Distance 2 = 72 Inches 400 mR/x = 72^2/ 40^2 <---- Plug in to formula 400 mR/x = 5184/1600 <---- Exponents 640,000 mR = 5184x <---- Cross Multiply 123.5 mR = X <---- Divide Both Sides 123.5 mR <--- Answer
If the intensity of radiation at an SID of 40 Inches is equal to 400 mR, what is the intensity of radiation when the distance is increased to 72 inches?
Original kVp multiplied by 1.15 Original mAs Reduced by 1/2
If the kVp must be increased by 15% but the problem states "maintain same exposure" what should be done with kVp? mAs? [15% rule]
Original kVp multiplied by .85 Original mAs doubled
If the kVp must be reduced by 15% but the problem states "maintain same exposure" what should be done with kVp? mAs? [15% rule]
mAs1 / mAs2 = d1² / d2² <---- Direct Square Law 25mAs/X = 40 SID²/ 72 SID² <--- Plug in Information 25mAs/X = 1600/5184 <---- Exponents 1600x = 129,600 <---- Cross Multiply x = 129600/1600 <----- Divide x = 81 mAs mAs2 = 81 mAs <---- Answer
If the mAs at an SID of 40 inches is equal to 25 mAs what is the mAs when the distance is increased to 72 inches?
I(2)/I(1) = D(1)^2/ D(2)^2 <--- Inverse Square Law x/10mR = 40^2/30^2 <--- Plug in x/10mR = 1600/900 <--- Exponents 900x = 16,000mR <--- Cross Multiply x = 17.77 mR <--- Divide 17.77 mR <--- Answer
If the radiation dose to a patient was 10 milliroentgens(mR) at a 40 inch distance, what would the dose be if the distance were reduced to 30 inches?
mAs2/mAs1 = D(2)^2/D(1)^2 <---- Direct Square Law x/400 mAs = 60^2/40^2 <--- Plug in x/400 mAs = 3600/1600 <--- Exponents 1600x = 1,440,000 mAs <--- Cross Multiply x = 900 mAs <--- Divide 900 mAs <--- Answer
If the technique chart states that you should use a particular mAs for a procedure at a specified distance and you must change the distance, what new mAs would be required at the new distances listed in the problems below? Original: Distance: 40 mAs: 400 New: Distance: 60 mAs: ???
mAs2/mAs1 = D(2)^2/D(1)^2 <---- Direct Square Law x/240 mAs = 40^2/44^2 <---- Plug In x/240 mAs = 1600/1936 <--- Exponents 1936x = 384,000 mAs <--- Cross Multiply x = 198.35 <--- Divide 198.35 <--- Answer
If the technique chart states that you should use a particular mAs for a procedure at a specified distance and you must change the distance, what new mAs would be required at the new distances listed in the problems below? Original: Distance: 44 mAs: 240 New: Distance: 40 mAs: ???
kVp = 60(1.15) = 69 <---- Increase kVp by 15% mAs = 100mA(.50s) = 50 mAs A Radiograph taken at 69 kVp and 50 mAs will provide us with double the exposure already; when we increase by 15% we are ALREADY increasing the exposure by double So... kVp = 69 <----- Answer mAs = 50 <----- Answer
Increase the kVp using the 15% rule with double the exposure given: Original Technique 100 mA .50s 60 kVp New Technique mAs - ??? kVp - ???
No Grid - 1 5:1 - 2 6:1 - 3 8:1 - 4 12:1 - 5 16:1 - 6
Name the Grid Conversion Factor for Each: 1. No Grid 2. 5:1 3. 6:1 4. 8:1 5. 12:1 6. 16:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 10 (3/5) New mAs = 6
Solve For New mAs Original mAs: 10 Original Grid: 10:1 New Grid: 6:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 12.5 (1/5) New mAs = 2.5
Solve For New mAs Original mAs: 12.5 Original Grid: 12:1 New Grid: No Grid
New mAs = Old mAs (New GCF/Old GCF) New mAs = 16 (2/6) New mAs = 5.3
Solve For New mAs Original mAs: 16 Original Grid: 16:1 New Grid: 5:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 16 (4/2) New mAs = 32
Solve For New mAs Original mAs: 16 Original Grid: 5:1 New Grid: 8:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 20 (5/4) New mAs = 25
Solve For New mAs Original mAs: 20 Original Grid: 8:1 New Grid: 12:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 25 (6/5) New mAs = 30
Solve For New mAs Original mAs: 25 Original Grid: 12:1 New Grid: 16:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 40 (1/6) New mAs = 6.7
Solve For New mAs Original mAs: 40 Original Grid: 16:1 New Grid: No Grid
New mAs = Old mAs (New GCF/Old GCF) New mAs = 50 (1/5) New mAs = 10
Solve For New mAs Original mAs: 50 Original Grid: 10:1 New Grid: No Grid
New mAs = Old mAs (New GCF/Old GCF) New mAs = 6.4 (6/1) New mAs = 38.4
Solve For New mAs Original mAs: 6.4 Original Grid: No Grid New Grid: 16:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 8 (5/3) New mAs = 13.3
Solve For New mAs Original mAs: 8 Original Grid: 6:1 New Grid: 10:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 80 (2/3) New mAs = 53.3
Solve For New mAs Original mAs: 80 Original Grid: 6:1 New Grid: 5:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 10 (6/1) New mAs = 60
Solve for New mAs Original mAs: 10 Original Grid: No Grid New Grid: 16:1
New mAs = Old mAs (New Grid Ratio Multiplier/Old Grid Ratio Multipier) New mAs = 20 (5/2) New mAs = 50
Solve for New mAs Original mAs: 20 Original Grid: 5:1 New Grid: 12:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 4 (4/1) New mAs = 16
Solve for New mAs Original mAs: 4 Original Grid: No Grid New Grid: 8:1
New mAs = Old mAs (New GCF/Old GCF) New mAs = 8 (2/4) New mAs = 4
Solve for New mAs Original mAs: 8 Original Grid: 8:1 New Grid: 5:1
Exposure 1: 32 mAs, 8:1 Grid, GCF = 4 Exposure 2: x mAs, 16:1 Grid, GCF = 6 mAs1/mAs2 = GCF1/GCF2 32/mAs2 = 4/6 Cross Multiply New mAs = 48 <---- Answer
Solve for New mAs: Original mAs: 32 Original Gird: 8:1 New Grid: 16:1
mAs1 = Original mAs mAs2 = New mAs d1² = Old Distance Squared d2² = New Distance Squared
The Exposure Maintenance Formula/ Direct Square Law is shown in the image. What do each of the variables represent? mAs1 = mAs2 = d1² = d2² =
I1/I2 = D2^2/D1^2 <--- Inverse Square Law Formula Intensity 1 = 4 Intensity 2 = X Distance 1 = 90 cm Distance 2 = 180 cm 4 mgy/x = 180^2/90^2 <----- Plug in using formula 4 mGy/x = 32,400/8,100 <--- Exponents/SID 32,400X = 32,400 <---- Cross Multiply X = 1 mGy <---- Divide 1 mGy <--- Answer
The Exposure from an X-ray Tube Operated at 70 kVp, 200 mAs is 4 mGya at 90 cm. What will the Exposure be at 180 cm?
New mAs = Old mAs(New Grid Ratio Multiplier/Old Grid Ratio Multiplier) New mAs = 10 (5/3) New mAs = 16.7 <----Answer
The Original Technique Factors are 10 mAs, 75 kVp, with the use of a 6:1 ratio grid. A change is made to a 12:1 ratio grid. Calculate the new mAs with this grid ratio change.
I1: Old Intensity I2: New Intensity D1^2: Old Distance Squared D2^2:New Distance Squared
The formula for the Inverse Square Law is shown in the image what do each of the variables represent? I1: I2: D1^2: D2^2:
kVp: 110(.85) = 93.5 <--- multiply kVp by .85 to penetrate less mAs: 120(2) = 240 <--- Double mAs to maintain exposure since we decreased kVp by 15% Seconds: 240/400 = .6 <--- Divide mAs by new mA Seconds = .6 <--- Answer kVp = 93.5 <--- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that has less penetration, but equal IR Exposure. Original Technique: 120 mAs 400 mA 110 kVp New Technique: mA= 400 Sec= ??? kVp = ???
kVp: 120(.85) = 102 <--- multiply kVp by .85 to penetrate less mAs: 200(.06) = 12 <--- multiply sec and mA However, to have equal exposure we must DOUBLE mAs since we DECREASED kVp by 15% mAs: 12(2) = 24 <---Multiply by 2 kVp: 102 <---- Answer mAs: 24 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that has less penetration, but equal IR Exposure. Original Technique: 200 mA .06 sec 120 kVp New Technique: mAs = ??? kVp = ???
kVp: 90(.85) = 76.5 <--- multiply kVp by .85 to penetrate less mAs: 25(.5) = 12.5 <---multiply sec and mA However, to have EQUAL EXPOSURE we must DOUBLE mAs since we DECREASED kVp by 15% mAs: 12.5(2) = 25 <---- Multiply by 2 Seconds: 25mAs/25mA = 1 <--- Divide mAs by new mA Seconds = 1 <--- Answer kVp = 76.5 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that has less penetration, but equal IR Exposure. Original Technique: 25 mA 0.5 sec 90 kVp New Technique: mA = 25 sec = ??? kVp = ???
kVp: 85(.85) = 72.25 <--- multiply kVp by .85 to penetrate less mAs: 500(.07) = 35 <---- multiply mA by Sec However, to have EQUAL EXPOSURE we must double the mAs since we decreased kVp by 15% mAs: 35(2) = 70 <--- Double mAs kVp: 72.25 <--- Answer mAs: 70 <--- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that has less penetration, but equal IR Exposure. Original Technique: 500 mA 0.07 sec 85 kVp New Technique: mAs: ??? kVp: ???
kVp: 95(1.15) = 109.25 <----multiply kVp by 1.15 to penetrate more mAs: 100(.02) = 2 <--- multiply mA and Sec However, to have Equal Exposure we must HALF the mAs since we INCREASED kVp by 15% mAs: 2/2 = 1 <--- Half mAs mA: 1/.03 = 33.33 <---- Divide mAs by New Seconds mA: 33.33 <--- Answer kVp: 109.25 <--- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that penetrates more, but has equal IR Exposure. Original Technique: 100 mA .02 sec 95 kVp New Technique: Sec = 0.03 mA = ??? kVp = ???
kVp: 60(1.15) = 69 <-- Multiply kVp by 1.15 to penetrate more mAs: 300mA(.15sec) = 45 <-- (multiply secs and mA) However, to have Equal Exposure we must HALF the mAs since we INCREASED kVp by 15% mAs: 45/2 = 22.5 <--- Half mAs kVp: 69 <---- Answer mAs: 22.5 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that penetrates more, but has equal IR Exposure. Original Technique: 300 mA 0.15 sec 60 kVp New Technique: mAs = ??? kVp = ???
kVp: 100(1.15) = 115 <-- Multiply kVp by 1.15 to penetrate more mAs: 50/2 = 25 <--- Half mAs since we increased kVp by 15% to maintain equal exposure Seconds: 25/500 = .05 <--- Divide mAs by new mA to find S kVp = 115 <---- Answer Sec = .05 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that penetrates more, but has equal IR Exposure. Original Technique: 50 mAs 500 mA 100 kVp New Technique: mA = 500 sec = ??? kVp = ???
kVp: 80(1.15) = 92 <--Multiply kVp by 1.15 to penetrate more mAs = 600(.02) = 12 <---- Multiply secs and mA However, to have Equal Exposure we must HALF the mAs since we INCREASED kVp by 15% mAs: 12/2 = 6 <--- Half mAs Seconds: 6/600 = .01 <---- Divide mAs by new mA to find S kVp = 92 <---- Answer Sec = .01 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that penetrates more, but has equal IR Exposure. Original Technique: 600 mA .02 sec 80 kVp New Technique: mA = 600 Sec = ??? kVp = ???
kVp: 75(1.15) = 86.25 <---- multiply kVp by 1.15 to penetrate more mAs: 800(.035) = 28 <----Multiply sec and mA However, to have Equal Exposure we must HALF the mAs since we INCREASED kVp by 15% mAs: 28/2 = 14 <---- Half mAs kVp: 86.25 <---- Answer mAs: 14 <---- Answer
Using the 15% rule for changing kVp, fill in the missing information to produce a radiograph that penetrates more, but has equal IR Exposure. Original Technique: 800 mA .035 sec 75 kVp New Technique: mAs: ??? kVp: ???
kVp = 60(1.15) = 69 <----- Increase kVp by 15% mAs: 400mA x .07s = 28 mAs <--- solve for mAs Our Exposure of 69 kVp and 28 mAs will provide DOUBLE the exposure in regard to 60 kVp at 28 mAs. But the question asks for HALF the Exposure mAs: 28 mAs/2 = 14 mAs <-- Half mAs (Equal Exposure) So Now we have an exposure of 69 kVp at 14 mAs. This Exposure is EQUAL EXPOSURE to 60 kVp at 28 mAs. But we WANT TO HALF THE EXPOSURE OF THE ORIGINAL. So we will need to HALF mAS AGAIN. mAs: 14 mAs/2 = 7 mAs <---- Half mAs again (Half Exposure) A Radiograph of 69 kVp at 7 mAs will be HALF the exposure of 60 kVp at 28 mAs, but with more penetration. kVp: 69 <---- Answer mAs: 7 <---- Answer
What New mAs and kVp are needed to produce a radiograph with an increase of 15% kVp (more penetration) and half the exposure if the original technique called for 60 kVp at 400 mA for .07 sec?
mAs1 / mAs2 = d1² / d2²
What is the Exposure Maintenance Formula (Direct Square Law)?
SID (Source to Image Receptor Distance) = SOD (Source to Object Distance) + OID (Object to Image Receptor Distance) SID = SOD + OID
What is the Formula for SID?
H/D (Height of lead strips/ distance between the lead strips)
What is the Formula used to find Grid Ratio?
Object Size = Image Size/Magnification Factor O = I/M
What is the Formula used to find Object Size?
Grid Ratio = h/D (height of lead strips/distance btwn lead strips) 3.2/0.2 = 16 or 16:1
What is the Grid Ratio when the lead strips are 3.2 mm high and separated by 0.2 mm
Magnification Factor = Source to Image Receptor Distance/ Source to Object Distance M = SID/SOD
What is the Magnification Factor Formula?
I1/I2 = D2^2/D1^2
What is the formula for the inverse square law?
mAs1/mAs2 = GCF1/GCF2
What is the formula used for converting different grid ratios?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 0.5/0.1 Grid Ratio = 5 Grid Ratio = 5:1
What is the ratio of a grid if the height of the lead strips is 0.5 mm and the distance between them is 0.1?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 0.6/0.1 Grid Ratio = 6 Grid Ratio = 6:1
What is the ratio of a grid if the height of the lead strips is 0.6 mm and the distance between them is 0.1 mm?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 0.8/0.1 Grid Ratio = 8 Grid Ratio = 8:1
What is the ratio of a grid if the height of the lead strips is 0.8 mm and the distance between them is 0.1 mm?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 1.0/0.1 Grid Ratio = 10 Grid Ratio = 10:1
What is the ratio of a grid if the height of the lead strips is 1.0 mm and the distance between them is 0.1 mm?
Grid Ratio = Height of lead strips (h)/ Distance btwn strips (d) Grid Ratio = 1.2/0.1 Grid Ratio = 12 Grid Ratio = 12:1
What is the ratio of a grid if the height of the lead strips is 1.2 mm distance between them is 0.1 mm?
kVp: 80(.85) = 68 <---- Multiply kVp by .85 to decrease by 15% mAs: 300(3/20) = 300(.15) = 45 <--- multiply mA by Sec ATM this answer provides HALF the Exposure, so we must Double the mAs mAs: 45(2) = 90 <-- Double mAs ATM this answer provides EQUAL EXPOSURE so we must double mAs AGAIN! mAs: 90(2) = 180 <--- Double mAs ATM this Answer provides DOUBLE THE EXPOSURE so we gud. kVp: 68 <--- Answer mAs: 180 <--- Answer
What new mAs and kVp are needed to produce a radiograph with a decrease of 15% kVp and double the exposure of the original technique called for 80 kVp at 300 mA for 3/20 sec?
kVp: 100(.85) = 85 <--- multiply kVp by .85 to decrease kVp by 15% mAs: 600(.035) = 21 <--- Multuiply mA and Sec Since we want to half the exposure, NOTHING else needs to be done since decreasing kVp by 15% already HALVES the Exposure. kVp = 85 <--- Answer mAs = 21 <--- Answer
What new mAs and kVp are needed to produce a radiograph with a decrease of 15% kVp and half the exposure if the original technique called for 100 kVp at 600 mA for 0.035 sec?
kVp: 70(1.15) = 80.5 <--- Multiply kVp by 1.15 to increase by 15% mAs: 200(0.01) = 2 <-- Multiply mA and Seconds Since we already increased kVp by 15%, NOTHING ELSE needs to be done, the exposure has already been DOUBLED. kVp: 80.5 <--- Answer mAs: 2 <--- Answer
What new mAs and kVp are needed to produce a radiograph with an increase of 15% kVp and double the exposure if the original technique called for 70 kVp at 200 mA for 0.01 sec?
