Problem 75 - Fundamentals of Physics

Question

A certain x-ray tube operates at a current of \(7.00 \mathrm{~mA}\) and a potential difference of \(80.0 \mathrm{kV}\). What is its power in watts?

Step-by-Step Solution

Answer

The power of the x-ray tube is 560 watts.

Step 1: Understanding the Formula

To find the power (P) of the x-ray tube, we can use the formula: \[ P = VI \], where \( V \) is the potential difference in volts and \( I \) is the current in amperes.

Step 2: Convert Units

The given potential difference is \( 80.0 \) kilovolts (kV), which needs to be converted to volts (V). Similarly, the current is provided as \( 7.00 \) milliamperes (mA), which must be converted to amperes (A). Thus, \( 80.0 \, \text{kV} = 80,000 \, \text{V} \) and \( 7.00 \, \text{mA} = 0.007 \, \text{A} \).

Step 3: Calculate Power

Substitute the converted values into the power formula: \[ P = 80,000 \, \text{V} \times 0.007 \, \text{A} \]. Solve for \( P \) to find the power in watts.

Step 4: Perform the Calculation

Multiply the values: \( 80,000 \, \text{V} \times 0.007 \, \text{A} = 560 \, \text{W} \). Thus, the power of the x-ray tube is \( 560 \, \text{W} \).

Key Concepts

X-ray Tube

An X-ray tube is an essential device used in medical and industrial applications to produce X-rays. It is a type of vacuum tube that accelerates electrons using high-voltage across a cathode and an anode. When these high-speed electrons hit the anode, usually made of tungsten, the sudden deceleration causes an emission of X-ray photons.
The process involves distinct key components:

• The **cathode**, which emits electrons when heated.
• The **anode**, a target material that emits X-ray photons upon electron impact.
• A vacuum environment to facilitate uninterrupted electron flow.

Understanding X-ray tubes is fundamental in applications like medical imaging, non-destructive testing, and material analysis.
They function efficiently with high-speed electron collisions repurposed into X-ray emissions, facilitating detailed imaging through discernible contrasting radiation.

Potential Difference

Potential difference, commonly known as voltage, refers to the difference in electric potential between two points in an electric field.
It's what 'pushes' the charge through a circuit and is measured in volts (V).
In an X-ray tube, potential difference represents a crucial parameter. It influences the energy of the electrons traveling from the cathode to the anode.
Higher potential difference results in electrons accelerating more rapidly, thus producing higher energy X-rays when they collide with the anode.
This energy transfer is directly related to the effective resolution and penetrating ability of the X-rays produced.

• In the problem here, we convert potential difference provided in kilovolts to volts for calculation purposes.
• A potential difference of 80 kV converts to 80,000 V.

This conversion aligns units for power calculations to be performed seamlessly.
Maintaining awareness of potential difference helps in modifying and controlling the X-ray tube's output effectively.

Current Conversion

When dealing with electrical calculations, it's vital to use consistent units.
Current, measured in amperes (A), indicates the rate of electric charge flow in a circuit. Often, especially in smaller circuits or components, current may be denoted in milliamperes (mA).
To perform power calculations in standard international units, conversions may be necessary:

• Convert milliamperes to amperes by dividing by 1000. Thus, 7.00 mA becomes 0.007 A.
• Ampere values allow direct substitution in power formulas like \( P = VI \).

Such conversions ensure accuracy in power estimation and other relevant computations.
Always verifying unit consistency simplifies mathematical steps and prevents errors when working with mixed electrical units in various formulas.

Power Calculation

Power in electrical systems refers to the rate of energy conversion or transfer utilized by a component like an X-ray tube.
It is measured in watts (W) and calculated using the formula \( P = VI \).
Where

• \( V \) is the potential difference in volts.
• \( I \) is the current in amperes.

For the given X-ray tube, substituting the converted values \( V = 80,000 \, \text{V} \) and \( I = 0.007 \, \text{A} \) into the formula provides:
\[ P = 80,000 \, \text{V} \times 0.007 \, \text{A} = 560 \, \text{W} \]
Thus, the power of this X-ray tube is 560 watts.
Understanding this concept ensures proper assessment of energy usage and requirements for X-ray tube operation.
It also helps in evaluating the system’s efficiency and performance based on its power rating.