Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne X-Ray and Heat
Name:  Rick M.
Status: other
Age: 40s
Location: N/A
Country: N/A
Date: 2000-2001


Question:
How many calories are generated at the plate of an x-ray tube when 30,000 volts at 0.1 ma are applied?


Replies:
Energy (joules) = voltage (volts) X current (amperes) = 30,000 X 0.1x10^-3 = 3 joules. 1calorie = 4.184 joules, so the energy in calories is 0.717 cal.

Vince Calder


Power generated (or used) by a current I = q/t going through a voltage drop V (= Ed) is given by P = IV. Here I = 0.0001 Amperes and V = 30,000 volts, so P = IV = 3 W.

Here I is the current in amperes, q is the charge in coulombs transferred by the current in t seconds, E is the average electric field in newtons/coulomb acting on the charge over a distance of d meters.

If you know that power is work done per unit time (P = W/t), that work is given by force times distance (W = Fd), that the force exerted on a charge of q coulombs by an electric field of E newtons per coulomb is given by F = qE, and that E = V/d, this is easy to derive. Otherwise ignore the next paragraph!

W = Fd = (qE)d; P = W/t = Fd/t = qEd/t = q(V/d)d/t = (q/t)V = IV

Best, Dick Plano... Richard J. Plano


Rick,

Actually, the question is incorrect. The number of calories produced will depend on the length of time that the current flows through the tube. The calorie is a unit of energy. If the tube is operating for twice as much time, twice as much energy is produced. The milliamp,(1/1000 amperes), however, is a rate: how many milliCoulombs of electric charge flow through the tube per second. I will show you how to find the rate at which energy is produced, often called "power".

The standard energy rate unit is the Watt: 1 Watt = 1 Joule per second. 1 calorie = 4.18 Joules. As for relating electric current and voltage to energy, the product of the two is correct. (1 Volt)*(1 Ampere) = 1 Watt.

(30,000 Volts)*(0.1 x 10^-3 Amperes) = 3.0 Watts = 3.0 Joules/second 3.0 Joules/second *[1 calorie/4.18 Joules] = 0.72 cal/sec

In each second of current flow, 0.72 calories of electrical energy are converted into x-ray radiation, heat, and possibly visible light. It may not seem like a great deal of energy, but you must consider the source. X-ray radiation interacts with individual cells. A cell can be destroyed with just a little bit of energy, if that energy is in the right form.

Dr. Ken Mellendorf
Illinois Central College


(30,000 V) x (0.1 mA) = 3 Watts

This is equal to 3 Joules / Sec

This is 0.72 cals per sec (or 7.2 x 10-4 Cals / sec)

Hope this helps

-Wil Lam



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory