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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?

Short Answer

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

Step by step solution

01

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.
02

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 \( 7.00 \, \text{mA} = 0.007 \, \text{A} \).
03

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.
04

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 \( 560 \, \text{W} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

An insulating belt moves at speed \(30 \mathrm{~m} / \mathrm{s}\) and has a width of \(50 \mathrm{~cm}\). It carries charge into an experimental device at a rate corresponding to \(100 \mu \mathrm{A}\). What is the surface charge density on the belt?

A certain brand of hot-dog cooker works by applying a potential difference of \(120 \mathrm{~V}\) across opposite ends of a hot dog and allowing it to cook by means of the thermal energy produced. The current is \(10.0 \mathrm{~A}\), and the energy required to cook one hot dog is \(60.0 \mathrm{~kJ} .\) If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously?

A block in the shape of a rectangular solid has a crosssectional area of \(3.50 \mathrm{~cm}^{2}\) across its width, a front-to-rear length of \(15.8 \mathrm{~cm}\), and a resistance of \(935 \Omega\). The block's material contains \(5.33 \times 10^{22}\) conduction electrons \(/ \mathrm{m}^{3}\). A potential difference of \(35.8 \mathrm{~V}\) is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

An \(18.0 \mathrm{~W}\) device has \(9.00 \mathrm{~V}\) across it. How much charge goes through the device in \(4.00 \mathrm{~h}\) ?

A common flashlight bulb is rated at \(0.30 \mathrm{~A}\) and \(2.9 \mathrm{~V}\) (the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(1.1 \Omega\), what is the temperature of the filament when the bulb is on?

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