Problem 21 - Basic Technical Mathematics with Calculus

Question

The electric current in a microprocessor circuit is \(0.230 \mu\) A. How many coulombs pass a given point in the circuit in \(1.50 \mathrm{ms} ?\)

Step-by-Step Solution

Answer

3.45 x 10^-10 C

Step 1: Understanding Current and Charge

We need to calculate how much charge has passed a point in the circuit given the electric current and time. The electric current is the rate of flow of electric charge, and it is represented by the formula \( I = \frac{Q}{t} \), where \( I \) is the current in amperes (A), \( Q \) is the charge in coulombs (C), and \( t \) is the time in seconds (s).

Step 2: Convert Units

The problem provides current as \(0.230\,\mu \text{A}\) and time as \(1.50\,\text{ms}\). We need to convert these to standard units: \(\mu\text{A}\) to A and \(\text{ms}\) to s. - \(0.230\,\mu \text{A} = 0.230 \times 10^{-6}\,\text{A}\)- \(1.50\,\text{ms} = 1.50 \times 10^{-3}\,\text{s}\)

Step 3: Apply Formula to Calculate Charge

Substitute the converted values into the formula \( I = \frac{Q}{t} \) to solve for \( Q \). Rearrange the formula to get \( Q = I \times t \).- \( Q = (0.230 \times 10^{-6}\,\text{A}) \times (1.50 \times 10^{-3}\,\text{s})\)- Calculate \( Q = 345 \times 10^{-9} \)- Simplify to: \( Q = 3.45 \times 10^{-10}\,\text{C} \)

Key Concepts

Coulombs

The concept of electric charge is fundamental in understanding electricity and circuits. Electric charge is measured in units called coulombs (C). One coulomb represents a large amount of charge — approximately equivalent to the charge carried by 6.242 x 10^18 electrons. However, in most electronic devices, we deal with much smaller amounts of charge, such as picocoulombs (pC) or nanocoulombs (nC), because modern components function with minute currents and quick transactions.

When dealing with electric currents, coulombs can be defined using the relationship between current and time. Electric current (I) is defined as the flow of charge per unit of time, usually measured in amperes (A), where 1 ampere is equivalent to 1 coulomb per second. This relationship is crucial in calculating how much charge passes a point in a circuit over a given time. Using the equation \( I = \frac{Q}{t} \), where \( Q \) is charge in coulombs and \( t \) is time in seconds, you can determine the charge if you know the current and the time.

Microprocessor Circuit

Microprocessor circuits are sophisticated electronic circuits designed to execute programmed instructions. They form the central processing unit (CPU) of computers and other electronic devices, controlling everything from simple calculations to complex operations. In a microprocessor, the flow of electric current is carefully managed to ensure that instructions are processed quickly and accurately.

Electric current in these circuits is typically very small, often measured in microamperes (\( \mu A \)) or smaller. Microprocessors operate at low power to reduce heat generation and improve energy efficiency. The microprocessor's ability to switch transistors on and off efficiently, using these small electric currents, is what makes modern digital electronics possible.

• Manipulating and managing electrical signals is essential for data processing and storage.
• Efficient electric current control aids in minimizing power consumption and heat production.

Understanding the behavior of current within microprocessor circuits is essential for designing and troubleshooting electronic systems.

Unit Conversion

Unit conversion is a fundamental skill in scientific calculations, especially when dealing with different measurement scales. In electronics, correct unit conversions can significantly alter the outcome of your calculations and ensure consistency when communicating results.

For electric currents and time measurements, converting microamperes (\( \mu A \)) to amperes (A) and milliseconds (ms) to seconds (s) is a frequent necessity.

Here's how you convert these units:

• Microamperes to Amperes: Since 1 microampere is 1 x 10^-6 amperes, conversion involves multiplying the \( \mu A \) value by \( 10^{-6} \).
• Milliseconds to Seconds: Similarly, 1 millisecond is 1 x 10^-3 seconds, so you multiply the ms value by \( 10^{-3} \).

Being adept at unit conversions ensures that all your calculations remain accurate and interpretable. This competence is crucial for solving problems in physics and engineering, where unit mismatch can otherwise lead to misunderstanding or erroneous conclusions.