Question

1. an image was taken using 150 ma, 0.08 sec., 36 sid. what is the correct adjustment to maintain ir exposure if time is 0.03 sec.?

1. i’m taking a skull x-ray of a patient with parkinson’s disease. because of his uncontrollable shaking, i’m going to have to use less time. the original technique is 40” sid, 75 kvp, 200 ma and 0.04 sec. i want to maintain exposure, so what is the new factor if i use 500 ma?

1. exposure was made using 15 mas, 65 kvp, 40 sid. what is the prime factor’s value if beam quantity is double?

1. my image has motion at 50 kvp, 200 ma, 0.25 sec sid 42 inches. what would my new factor be if i increase my ma to 600 and maintain density?

1. exposure was made using 8 mas, 65 kvp, 40 sid. what is the prime factor’s value if beam quantity is double?

Explanation:

Question 1

Step1: Recall the exposure maintenance formula (mA₁ × t₁ = mA₂ × t₂)

We know \( mA_1 = 150 \), \( t_1 = 0.08 \) sec, \( t_2 = 0.03 \) sec. We need to find \( mA_2 \).
From \( mA_1 \times t_1 = mA_2 \times t_2 \), we can re - arrange to \( mA_2=\frac{mA_1\times t_1}{t_2} \)

Step2: Substitute the values into the formula

\( mA_2=\frac{150\times0.08}{0.03} \)
First, calculate the numerator: \( 150\times0.08 = 12 \)
Then, divide by the denominator: \( \frac{12}{0.03}=400 \) mA

Step1: Use the exposure maintenance formula \( mA_1\times t_1 = mA_2\times t_2 \)

We know \( mA_1 = 200 \), \( t_1 = 0.04 \) sec, \( mA_2 = 500 \). We need to find \( t_2 \)
Re - arrange the formula: \( t_2=\frac{mA_1\times t_1}{mA_2} \)

Step2: Substitute the values

\( t_2=\frac{200\times0.04}{500} \)
Calculate the numerator: \( 200\times0.04 = 8 \)
Then, divide by the denominator: \( \frac{8}{500}=0.016 \) sec

Step1: Recall that beam quantity (exposure) is related to mAs. If the beam quantity is doubled, the new mAs (\( mAs_2 \)) is twice the original mAs (\( mAs_1 \))

Original \( mAs_1=15 \) mAs, so \( mAs_2 = 2\times15=30 \) mAs
The prime factor here is mAs, and the new mAs value is 30 mAs. Also, note that kVp and SID do not affect the mAs calculation for doubling the beam quantity (assuming kVp and SID are constant, but since we are only asked about the prime factor related to beam quantity which is mAs here)

Answer:

The new mA should be 400 mA.

Question 2