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RAD 121

Imaging 1 Problem Solving

QuestionAnswer
Name the three formulas for mAs, mA, or Time when two of the three are known. mAs = mA x Time; mA = mAs / Time; Time = mAs / mA
List the steps to find Percentage difference in mAs or kVp 1. Find the difference (high - low) 2. Divide the difference by the original 3. Move the decimal two places to right (x 100) (highest - lowest)/original x 100
Application of kVp 15% rule: Double receptor exposure, all other factors constant original kVp x 1.15 = new kVp
Application of kVp 15% rule: Halve receptor exposure, all other factors constant original kVp x .85 = new kVp
Application of kVp 15% rule: to increase contrast and maintain receptor exposure decrease kVp by 15% and 2x mAs
Application of kVp 15% rule: to decrease contrast and maintain receptor exposure increase kVp by 15% and 1/2 mAs
Maintain receptor exposure with mAs for changes in SID: (mAs - distance compensation formula) DIRECT SQUARE LAW mAs(1) / mAs(2) = (SID(1))squared / (SID(2))squared
Exposure rate with changes in distance from source: INVERSE SQUARE LAW I(2) = I(1) X (SID(1))squared / (SID(2))squared or I(1) / I(2) = (SID(2))squared / (SID(1))squared
The time required to maintain a given receptor exposure is directly proportional to the square of the SID. t(1) / t(2) = (d(1))squared / (d(2))squared
The mA required to maintain a given receptor exposure is directly proportional to the square of the SID mA(1) / mA(2) = d(1)squared / d(2)squared
The mAs required to maintain a given receptor exposure is directly proportional to the square of the SID mAs(1) / mAs(2) = (SID(1))squared / (SID(2))squared DIRECT SQUARE LAW
True/False; For a given receptor exposure, mA and Time are inversely proportional to each other. Give formula. TRUE; mA(1) / mA(2) = t(2) / t(1)
When calculating with mAs, ALWAYS use DIRECT SQUARE LAW; mAs(1) / mAs(2) = (d(1))squared / (d(2))squared
What factor should always be squared when working problems? SID distance
Problem Solving What is the mAs? 500mA .25 sec 125 mAs
Problem Solving What is the mA? time 1/5 mAs 120 600 mA
Problem Solving What is the Time? mA 800 mAs 280 .35 sec
Problem Solving What is the mAs? 1200 mA 100 millisec 120 mAs
Problem Solving What is the mA? time .16 mAs 256 1600 mA
Problem Solving What is the Time? mA 200 mAs 50 1/4 or .25 secs
Problem Solving What is the mA? 2/15 time 80 mAs 600 mA
Problem Solving What is the mAs? mA 300 time 2 600 mAs
Problem Solving What is the time? 450 mA 180 mAs .4 secs
Problem Solving What is the mAs? mA 1600 Time 1/20 80 mAs
Problem Solving What is the missing factor? 68 mAs was used to produce an image. If the mAs is increased to 92, what is the percent of increase? 35.29%
Problem Solving What is the missing factor? 2 mAs is used to produce an image. If you are instructed to increase the mAs by 40%, what new mAs will you use? 2.8 mAs
Problem Solving What is the missing factor? An image is made using .06 sec and 600 mA. If the time is changed to .03 sec, what mA will be required to maintain receptor exposure? 1200 mA
Problem Solving What is the new mAs? 60" and 10 mAs to 30" and _____mAs 2.5 mAs
Problem Solving What is the new mAs? 40" and 200 mAs to 28" and _____mAs 98 mAs
Problem Solving What is the mAs? 36" and 40 mAs to 72" and _____mAs 160 mAs
Problem Solving What is the new mAs? 48" and 80 mAs to 72" and _____mAs 180 mAs
Problem Solving What is the new mAs? 56" and 128 mAs to 21" and _____mAs 18 mAs
Problem Solving An image is made using 58 kVp and 3 mAs. If the mAs is decreased to 1.5 mAs, what kVp must be used to maintain receptor exposure within 1 kVp accuracy? 67 kVp
Problem Solving An image is made using 48 mAs and 79 kVp. If kVp remains at 79, what mAs must you use in order to decrease receptor exposure by 50% 79 kVp and 24 mAs
Problem Solving What is the missing factor? 1/5 sec and 40" to _____ sec and 60" .45 sec
Problem Solving What is the missing factor? 1 sec and 72" to _____ sec and 60" .69 sec
Problem Solving What is the missing factor? .75 sec and 40" to .5sec and _____" 32.66"
Problem Solving What is the missing factor? 1/10 sec and 36" to 1/40 sec and _____" 18"
Problem Solving What is the missing factor? 1/2 sec and 48" to _____ and 36" .28 sec
Problem Solving What is the missing factor? 200 mA and 40" to _____ mA and 50" 312.5 mA
Problem Solving What is the missing factor? 360 mA and 72" to _____ mA and 48" 160 mA
Problem Solving What is the missing factor? 100 mA and 40" to 200 mA and _____" 56.57 " round up to 57"
Problem Solving What is the missing factor? 300mA and 40" to 200 mA and _____" 32.66" round up to 33"
Problem Solving What is the missing factor? 900 mA and 36" to _____mA and 24" 400 mA
Problem Solving What is the missing factor? 50 mAs and 48" to _____ mAs and 60" 78.125 mAS
Problem Solving What is the missing factor? 300 mAs and 40" to _____ mAs and 60" 675 mAs
Problem Solving What is the missing factor? 72 mAs and 42" to _____ mAs and 32" 41.8 mAs
Problem Solving What is the missing factor? 125 mAs and 20" to _____ mAs and 32" 320 mAs
Problem Solving What is the missing factor? 128 mAs and 72" to _____ mAs and 64" 101.14 mAs
Problem Solving What is the missing factor? 400 mA and 1/20 sec to _____ mA and 1/15 sec Workbook p 130 300 mA
Problem Solving What is the missing factor? 600 mA and 1/4 sec to _____ mA and 2/3 sec 225 mA
Problem Solving What is the missing factor? 300 mA and .05 sec to 600 mA and _____ sec .025 sec
Problem Solving What is the missing factor? 100 mA and .75 sec to 600 mA and _____ sec .125 sec
Problem Solving What is the missing factor? 800 mA and 1/20 sec to 100 mA and _____ sec .4 sec
Problem Solving What is the percent change from first exposure to second? Exposure 1 at 60 mAs to Exposure 2 at 80 mAs 33.3%
Problem Solving What is the percent change? 66 mAs to 59.4 mAs 10%
Problem Solving What is the percent change from first exposure to second? Exposure 1 at 15 mAs to Exposure 2 at 21 mAs 40%
Problem Solving What is the percent change from first exposure to second? Exposure 1 at 80 kVp to Exposure 2 at 92 kVp 15%
Problem Solving What is the percent change from first exposure to second? Exposure 1 at 60 kVp to Exposure 2 at 51 kVp 15%
Problem Solving State the new exposure factor needed to make the change given. An exposure is made using 18mAs and 72kVp. The exposure should be repeated with the mAs decreased by 1/3 and kVp left the same. 12 mAs and 72 kVp
Problem Solving State the new exposure factor needed to make the change given. An exposure is made using 48mAs and 82kVp. The exposure should be repeated with the mAs increased by 100% and kVp reduced 15%. Round your numbers. 96 mAs and 70kVp
Problem Solving State the new exposure factor needed to make the change given. An exposure is made using 120mAs and 90kVp. The exposure should be repeated with the mAs decreased by 40% and kVp left the same. 72 mAs and 90 kVp
Problem Solving State the new exposure factor needed to make the change given. An exposure is made using 14mAs and 64kVp. The exposure should be repeated with the mAs increased by 50% and kVp increased 8%. 21 mAs and 69 kVp
What is the formula for mAs, mA or Time when 2 of the 3 are known? a) mA x Time = mAs; b) mAs / Time = mA; c) mAs/mA = Time
How do you find the percentage difference in mAs or kVp? A) find the difference; B) Divide the difference by the original; C) move decimal 2 places to the right (x100)
State the 15% rule with kVp. Give all possibilities. 2x IR exposure, AOFC: Orig kVp x 1.15 = new kVp; 1/2 IR exposure, AOFC: orig kVp x .85 = new kVp; increase contrast and maintain IR exposure, decrease kVP by 15% and 2x mAs; decrease contrast and maintain IR exposure, increase kVp by 15% and 1/2 mAs
State the direct square law. mAs1 / mAs2 = (SID1)2 divided by (SID2)2; maintains receptor exposure with mAs changes for SID
State the mAs-distance compensation formula. mAs1 / mAs2 = (SID1)2 divided by (SID2)2; maintains receptor exposure with mAs changes for SID
How do you maintain receptor exposure with mAs if SID changes? mAs1 / mAs2 = (SID1)2 divided by (SID2)2; maintains receptor exposure with mAs changes for SID
State the Inverse Square Law. I1 / I2 = (SID2)2 / (SID1)2; exposure rate with changes in distance from source
Higher kVp creates what scale of contrast? long scale/low contrast - think CHEST x-ray at 125kVp - allows differentiation between different tissues
Lower kVp creates what scale of contrast? short scale/high contrast - think HAND x-ray at 60kVp - primarily interested in bone only, not soft tissue
Maintaining receptor exposure with a change in grid ratio mAs1/mAs2 = GF1/GF2
What effect does Focal Spot Size (FSS) have on receptor exposure? NO EFFECT
What effect does Filtration have on receptor exposure? inverse (increase filtration, decrease Receptor exposure)
What effect does mAs have on receptor exposure? direct (increase mAs, increase receptor exposure)
What effect does kVp have on receptor exposure? direct (increase kVp, increase receptor exposure)
What effect does OID have on receptor exposure? inverse (increase OID, decrease receptor exposure)
What effect does Field Size have on receptor exposure? direct (increase field size, increase receptor exposure)
What effect does SID have on receptor exposure? inverse (increase SID, decrease receptor exposure)
What effect does Grid have on receptor exposure? inverse (increase Grid, decrease receptor exposure)
What effect does Beam Restriction have on receptor exposure? inverse (increase BEAM RESTRICTION, decrease receptor exposure)
What effect does Patient Motion have on receptor exposure? NO EFFECT
What effect does CR Angle have on receptor exposure? inverse (increase CR Angle, decrease receptor exposure)
What effect do the Prime Factors have on Brightness? NO EFFECT - because brightness is a quality of the monitor. It provides VISIBILITY
What affects tube current? mA - increasing tube current increases filament current, the quantity of electrons produced at the filament
What affects the POTENTIAL difference in exposure rate? kVp - increasing kVp allows more electrons move to target with more energy creating stronger x-rays that can penetrate better
What is photographic Property? Workbook p26 Visibility of detail - includes receptor exposure and image contrast - the quality of the picture and what it contains (like a photo)
What is Geometric Property? workbook p26 Sharpness of detail - includes spatial resolution and distortion (how clear is the anatomy, is there motion or elongation/ foreshortening)
What is the best way to change receptor exposure? mAs
What does the tech need to change to maintain receptor exposure for a full expiration chest compared to full inspiration? increase mAs by 30-35%
If same mAs is used for supine and erect imaging of the abdomen, the erect image may be _____. underexposed because the abdomen is thicker in erect vs supine positions
How does OID affect receptor exposure? INVERSELY - the air-gap allows more scatter, reducing receptor exposure - and magnifying image
What changes to kVp for post-reduction wet plaster cast? small - medium + increase 5-7kVp; Large - increase 8-10kVp
What changes to kVp for followup dry cast? small - medium + increase 5-7kVp; Large - increase 8-10kVp
What changes to kVp for fiberglass cast? Increase kVp 3-5
Why is mAs the controlling factor for changing receptor exposure? Because it has a DIRECTLY PROPORTIONAL effect - kVp has a greater effect, but since it is not proportional there is too much possible variation
Created by: Larobbins
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