Problem 43 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring.The voltage \(V\) across a semiconductor in a computer is given by \(V=\alpha I+\beta I^{2},\) where \(I\) is the current (in A). If a 6 -V battery is conducted across the semiconductor, find the current if \(\alpha=2 \Omega\) and \(\beta=0.5 \Omega / \mathrm{A}\).

Step-by-Step Solution

Answer

The current is 2 A.

Step 1: Write the Given Equation

Given the formula for voltage across a semiconductor: \( V = \alpha I + \beta I^2 \). Substitute the known values: \( \alpha = 2 \Omega \), \( \beta = 0.5 \Omega/\mathrm{A} \), and \( V = 6 \) V. The equation becomes \( 6 = 2I + 0.5I^2 \).

Step 2: Rearrange to Form a Quadratic Equation

Rearrange the equation to standard quadratic form \( ax^2 + bx + c = 0 \). Move all terms to one side: \( 0.5I^2 + 2I - 6 = 0 \). Simplify by multiplying every term by 2 to clear the fraction: \( I^2 + 4I - 12 = 0 \).

Step 3: Factor the Quadratic Equation

Find two numbers that multiply to \(-12\) (the constant term) and add up to \(4\) (the linear coefficient). These numbers are \(6\) and \(-2\). Thus, factor the equation as \((I + 6)(I - 2) = 0\).

Step 4: Solve for the Roots

Set each factor equal to zero to find the possible solutions for \(I\): \(I + 6 = 0\) gives \(I = -6\). \(I - 2 = 0\) gives \(I = 2\). Since current cannot be negative, choose the positive root.

Key Concepts

Factoring

Factoring is like breaking down a complex expression into simpler, multipliable parts. It's a key skill in solving quadratic equations. Quadratic equations are polynomials where the highest degree is two, typically expressed in the form \( ax^2 + bx + c = 0 \). Factoring simplifies these equations by finding two binomials (expressions with two terms) whose product equals the quadratic equation. For example, if we have \( I^2 + 4I - 12 = 0 \), we need to find two numbers that multiply to \(-12\) and add to \(4\). These numbers are \(6\) and \(-2\), allowing us to rewrite the quadratic as \((I + 6)(I - 2) = 0\).

• Factoring turns a quadratic into simpler expressions.
• Prepares the equation for finding solutions, or 'roots'.
• Involves identifying numbers that satisfy the multiply-add condition.

Applying this technique allows us to find the solutions by setting each binomial equal to zero, giving us potential values for the variable.

Semiconductors

Semiconductors are materials that have the ability to conduct electricity under certain conditions, making them crucial in electronic devices. Unlike conductors, semiconductors do not carry current as freely, allowing them to control and manipulate the flow of electricity. Silicon is a common material used in semiconductors because it can be altered by doping, a process where small amounts of other elements are added to change its properties. This ability to switch between conducting and insulating states makes semiconductors the backbone of circuits and components like diodes and transistors in computers.
Semiconductors play a pivotal role in:

• Managing electrical flow in circuits.
• Forming the basis of integrated circuits and chips.
• Providing the building blocks for all modern electronics.

So, when we talk about voltage and current equations related to semiconductors, we're dealing with advanced control systems that enhance device functionalities.

Current Calculation

In electrical circuits, current calculation is essential for determining how much electric charge flows through a circuit at a given time. Current, typically denoted by \( I \), is measured in amperes (A), and it tells us how fast the electrons are moving within the circuit. Understanding these calculations is crucial for designing, troubleshooting, and maintaining electronic devices. When given a voltage equation like \( V = \alpha I + \beta I^2 \), we set up a system to find the current by balancing the equation with known variables. In the example problem, after substituting values, we have \( 6 = 2I + 0.5I^2 \). This rearranges to the quadratic form \( I^2 + 4I - 12 = 0 \). Calculating current involves factoring or using other algebraic methods to solve for \( I \).

• Understand the relation between voltage, current, and resistance.
• Use equations to solve for unknowns with given conditions.
• Apply physical laws such as Ohm's Law where necessary.

Effective current calculations ensure that devices operate within safe and efficient parameters.

Voltage Equations

Voltage equations describe the relationship between voltage, current, and resistance or other factors in electrical circuits. Voltage is the potential difference that pushes electric charges through the circuit. The equation \( V = \alpha I + \beta I^2 \) represents this relationship for a semiconductor, where \( \alpha \) and \( \beta \) are constants related to the material properties. By rearranging this equation, especially in problems involving quadratic expressions, we can solve for unknowns such as current. In this context, \( \alpha \) might represent linear resistance while \( \beta \) could represent non-linear resistance phenomena. This helps to:

• Determine how different factors influence potential difference.
• Provide predictions about current levels at certain voltage values.
• Optimize the performance of electronic components.

These equations are vital in designing circuits because they ensure that components like semiconductors function correctly under varying conditions.