Question

using the 15 percent rule, calculate the kvp that will produce the same image receptor exposure as the original set of factors.

	1st exposure	2nd exposure
b	25 mas 80 kvp	50 mas type your answer..
c	60 mas 75 kvp	120 mas type your answer..
d	200 mas 90 kvp	type your answer.. 104 kvp
e	10 mas 50 kvp	type your answer.. 58 kvp

Explanation:

Step1: Recall 15 - percent rule formula

The 15 - percent rule states that changing the kVp by 15% will have the same effect on image receptor exposure as doubling or halving the mAs. If the mAs is doubled, the kVp should be decreased by 15% to maintain the same exposure, and if the mAs is halved, the kVp should be increased by 15%. The formula to increase kVp by 15% is $kVp_{new}=kVp_{old}(1 + 0.15)$ and to decrease it is $kVp_{new}=kVp_{old}(1 - 0.15)$.

Step2: Solve for part a

The mAs is doubled from 150 mAs to 300 mAs. So we need to decrease the kVp by 15%. Given $kVp_{old}=60$ kVp. Then $kVp_{new}=60\times(1 - 0.15)=60\times0.85 = 51$ kVp.

Step3: Solve for part b

The mAs is doubled from 25 mAs to 50 mAs. Given $kVp_{old}=80$ kVp. Then $kVp_{new}=80\times(1 - 0.15)=80\times0.85 = 68$ kVp.

Step4: Solve for part c

The mAs is doubled from 60 mAs to 120 mAs. Given $kVp_{old}=75$ kVp. Then $kVp_{new}=75\times(1 - 0.15)=75\times0.85 = 63.75$ kVp.

Step5: Solve for part d

The mAs is decreased. First, find the ratio of mAs change. Let the new mAs be $mAs_{new}$ and old mAs be $mAs_{old}$. The ratio $r=\frac{mAs_{new}}{mAs_{old}}$. Here, we need to find $mAs_{new}$ when $kVp$ changes from 90 kVp to 104 kVp. Since $104=90\times(1 + x)$, then $x=\frac{104}{90}-1\approx0.156$. Since kVp is increased by about 15%, mAs should be halved. But we are asked to find mAs when going from 200 mAs with 90 kVp to 104 kVp. Since kVp is increased by 15%, $mAs_{new}=\frac{200}{2}=100$ mAs.

Step6: Solve for part e

The kVp is increased from 50 kVp to 58 kVp. The increase factor is $\frac{58}{50}=1.16\approx1.15$. Since kVp is increased by about 15%, the mAs should be halved. Given old mAs = 10 mAs, new mAs = 5 mAs.

Answer:

a. 51 kVp
b. 68 kVp
c. 63.75 kVp
d. 100 mAs
e. 5 mAs