Chapter 11 Lecture Notes

The Prime Factors

Factors Affecting X-Ray Quantity

**Milliampere-seconds (mAs):**

**Example 1**

A certain technique calls for 100 kVp, 25 mAs, which results in an x-ray intensity of 28 mR at the position of the patient. If the mAs is increased to 35 mAs, what will the x-ray intensity be?

First we will use the formula above and plug in the values.

Now use algebra and solve for by first cross multiplying.

Then divide by 25 mAs to isolate the variable.

Now simplify. Notice that the units will cancel to have mR as the final units.

Example 2

A certain technique calls for 75 kVp, 60 mAs, which results in an x-ray intensity of 12 mR at the position of the patient. If the mAs is decreased to 25 mAs, what will the x-ray intensity be?

First we need to put the values in the formula.

Then solve for the variable.

Now divide both sides by 60 mAs.

Simplify.

**Kilovolt Peak (kVp):**

**Example 3**

A certain technique calls for 75 kVp, 50 mAs, which results in an x-ray intensity of 20 mR at the position of the patient. If the kVp is increased to 125 kVp, what will the x-ray intensity be?

First, plug in the values in the formula.

Now solve for the variable.

And solve for the variable.

Simplify equation.

Example 4

A certain technique calls for 150 kVp, 50 mAs, which results in an x-ray intensity of 40 mR at the position of the patient. If the x-ray intensity decreased to 22 mR, what is the new kVp?

Plug in the values into the formula.

Now solve for the variable, notice that the variable will be squared.

Use algebra to get the variable by itself.

Simplify the right hand side of the equation.

Then take the square root of both sides of the equation.

**Distance:** Notice that distance
and x-ray exposure are an inverse square law.

**Example 5**

A certain technique calls for 95 kVp/20 mAs at a SID of 150 cm. The exposure at that the image receptor is 15 mR. If the same technique is used at a SID of 100 cm, what will be the x-ray exposure?

Plug the values into the formula.

Solve for the variable.

And simplify.

Example 6

A certain technique calls for 120 kVp/35 mAs at a SID of 125 cm. The exposure at that the image receptor is 15 mR. If you need an exposure of 6 mR, where must the patient be placed?

Plug in the values into the formula.

In this case it is not necessary to cross multiply. Multiply both sides by to isolate the variable.

Source-to-image receptor distance (SID)

When SID is increased, mAs must be increased by to maintain constant optical density (OD).

**Example 7**

A certain technique calls for 75 kVp/25 mAs at a SID of 175 cm. The exposure at that the image receptor is 15 mR. If the SID is changed to 100 cm, what will the new mAs need to be?

Plug in the values into the formula.

Solve for the variable.

Example 8

A certain technique calls for 125 kVp/35 mAs at a SID of 110 cm. The exposure at that the image receptor is 20 mR. If the mAs is changed to 75 mAs, what will the SID need to be to keep the same exposure?

First plug in the values into the formula.

Solve for the variable.

The Entrance Skin Exposure (ESE) can be calculated by using the following equation.

**Example 9**

A certain technique calls for 85 kVp/40 mAs and results in 120 mR. What is the expected ESE if the kVp is increased to 100 kVp and the mAs is increased to 50 mAs?

This formula is a little different. But we will solve it the same way. First plug in the values.

Cross multiply.

Solve for the variable.

Example 10

A certain technique calls for 110 kVp/50 mAs and results in 110 mR. What is the expected ESE if the kVp is decreased to 95 kVp and the mAs is increased to 60 mAs?

Plug in the values into the formula.

Cross multiply.

Solve for the variable.

The End