Question

1.An x-ray tube operates with a tube current of 20.0 mA. The tube voltage is 140.0 kV. If both the tube current and tube voltage are constant, at what rate (j/sec) is energy delivered to the anode?

2.patient is radiographed using 60 kVp and 10 mAs, resulting in an x-ray exposure of 28 mR. If the technique is changed to 55 kVp and 10 mAs,what is the new x-ray exposure?

3.An x-ray film image is taken at 68 kVp and 20 mAs. The optical density is fine, but the contrast is poor. If the kVp is increased to 78 kVp, what should be the new mAs setting?

4.In an x-ray tube, suppose the apparent focal spot size is 2.0 mm and the true focal spot size is 6.0 mm. What is the target angle?

Answer 1

Solution of Question 1:

Let us go to the basics first.

This question is based on the concept of

1. Energy delivered to x ray anode

2. Rate of energy I.e. power

The detailed solution is described below:

Rate of energy

= Voltage * current

= 140 kV * 20mA

= (140*1000)* (20*10^-3)

= 2800 J/sec (Answer)

Thanks!!!