Calculations for radiography

Question 1

kV - Kilovoltage

Answer 1

controls the penetrating power

refers to the size of electrical potential difference which is applied between the cathode and the anode in x-ray tube

affects the speed that the electrons are accelerated from cathode to anode

Question 2

kV

Answer 2

is kV is too high your contrast is too low and if your kV is too low your contrast is too high

Question 3

Milliamperage mA

Answer 3

number of electrons produced at the cathode by thermionic emission

the higher the mA, the greater number of electrons produced and the greater intensity of the x-ray beam

Higher mA = the blacker the film
lower mA = the lighter the film

Question 4

Time

Answer 4

when seeting the exposure factors, we need to determine an amount of time (seconds) over which the production of X-rays will take place

mAs is the amperage and time combined to provide the mA’s

longer exposure time the blacker the film

Question 5

Film Focal Distance

Answer 5

distance from the focal spot to the film

as FFD increases, the intensity of the xray beam decreases

usually kept constant between 75cm and 100cm

Question 6

Calculating exposure factors

Answer 6

MAs = mA x s

MA = mAs / s

s= mAs/mA

If we raise the kV by 10 the mAs should be halved

If the kV is lowered by 10, the mAs should be doubled

if the FFD is changed, New mAs = old mAs x new FFDsquared = old FFDsquared