Problem 43
Question
$$ \text { Solve the given quadratic equations by factoring.}$$ The voltage \(V\) across a semiconductor in a computer is given by \(V=\alpha I+\beta I^{2},\) where \(I\) is the current (in \(\mathrm{A}\) ). If a \(6-\mathrm{V}\) battery is conducted across the semiconductor, find the current if \(\alpha=2 \Omega\) and \(\beta=0.5 \Omega / \mathrm{A}\)
Step-by-Step Solution
Verified Answer
The current is 2 A.
1Step 1: Write the Initial Equation
The voltage equation is given as \( V = \alpha I + \beta I^2 \). We need to find \( I \) when \( V = 6 \) volts, \( \alpha = 2 \) \( \Omega \), and \( \beta = 0.5 \) \( \Omega / \mathrm{A} \). Substitute these values into the equation to form \( 6 = 2I + 0.5I^2 \).
2Step 2: Rearrange into Quadratic Form
Rearrange the equation \( 6 = 2I + 0.5I^2 \) into standard quadratic form. Start by subtracting 6 from both sides: \( 0.5I^2 + 2I - 6 = 0 \). This is our quadratic equation.
3Step 3: Simplify the Equation
Multiply the entire equation by 2 to eliminate the decimal, resulting in \( I^2 + 4I - 12 = 0 \).
4Step 4: Factor the Quadratic
We need to factor the quadratic equation \( I^2 + 4I - 12 = 0 \). Look for two numbers that multiply to \(-12\) (the constant term) and add to \(4\) (the linear coefficient). These numbers are \(6\) and \(-2\). Thus, the equation factors to \((I + 6)(I - 2) = 0\).
5Step 5: Solve for the Values of I
Set each factor equal to zero: \( I + 6 = 0 \) and \( I - 2 = 0 \). Solving these gives \( I = -6 \) and \( I = 2 \).
6Step 6: Determine the Valid Solution
Given that current cannot be negative, the valid solution for the current is \( I = 2 \) A.
Key Concepts
FactoringCircuit AnalysisCurrent-Voltage RelationshipSemiconductors
Factoring
Factoring is an essential skill in algebra that involves breaking down polynomials into the product of simpler expressions. This method is particularly useful for solving quadratic equations. When you encounter a quadratic equation like \[ ax^2 + bx + c = 0, \]you can solve it by finding two numbers that multiply to the constant term \( c \) and add to the linear coefficient \( b \). This process is often referred to as finding the roots of the quadratic equation.
For example, in the equation \( I^2 + 4I - 12 = 0 \), we looked for numbers that multiply to \(-12\) and add to \(4\). The numbers \(6\) and \(-2\) satisfy these conditions, allowing us to factor the equation as \((I + 6)(I - 2) = 0\). Once factored, setting each factor equal to zero gives the potential solutions.
Factoring is not only limited to quadratics; it includes various techniques like factoring by grouping, the difference of squares, and using the quadratic formula. Each offers unique methods to simplify and solve equations, which are foundational in mathematics and many applied sciences.
For example, in the equation \( I^2 + 4I - 12 = 0 \), we looked for numbers that multiply to \(-12\) and add to \(4\). The numbers \(6\) and \(-2\) satisfy these conditions, allowing us to factor the equation as \((I + 6)(I - 2) = 0\). Once factored, setting each factor equal to zero gives the potential solutions.
Factoring is not only limited to quadratics; it includes various techniques like factoring by grouping, the difference of squares, and using the quadratic formula. Each offers unique methods to simplify and solve equations, which are foundational in mathematics and many applied sciences.
Circuit Analysis
In understanding electronics, circuit analysis is fundamental for forecasting a circuit's behavior. It involves the study of currents and voltages within an electrical network to determine the condition and performance of the circuit elements. This requires using principles from Ohm's Law and Kirchhoff’s Laws.
When you perform circuit analysis, you focus on how electrical current flows through each component and how voltage is distributed across them. In circuits containing semiconductors, it’s crucial to monitor these parameters accurately because semiconductors behave differently under varying current and voltage conditions.
When you perform circuit analysis, you focus on how electrical current flows through each component and how voltage is distributed across them. In circuits containing semiconductors, it’s crucial to monitor these parameters accurately because semiconductors behave differently under varying current and voltage conditions.
- Series and parallel connections greatly influence how voltages and currents distribute.
- Identifying equivalent resistance and applying nodal analysis are typical steps in solving circuit configurations.
Current-Voltage Relationship
The current-voltage (I-V) relationship is a fundamental concept in electronics, describing how portable charges flow in a circuit in response to voltage. This relationship is often depicted through the equation \( V = IR \), known as Ohm's Law, where \( V \) is the voltage across the conductor, \( I \) is the current through the conductor, and \( R \) is the resistance of the conductor.
In semiconductors, the I-V relationship can be nonlinear, requiring us to incorporate additional terms, such as the quadratic term in our specific example equation \( V = \alpha I + \beta I^2 \). This relationship is crucial to understanding how electrical components, like diodes and transistors, operate within circuits and respond to different voltages and currents.
Devices with nonlinear I-V characteristics can exhibit complex behaviors, such as negative resistance, which must be accounted for in advanced circuit designs. Identifying and analyzing the I-V relationship ensures devices function correctly, protecting them from potential overcurrent or overvoltage scenarios.
In semiconductors, the I-V relationship can be nonlinear, requiring us to incorporate additional terms, such as the quadratic term in our specific example equation \( V = \alpha I + \beta I^2 \). This relationship is crucial to understanding how electrical components, like diodes and transistors, operate within circuits and respond to different voltages and currents.
Devices with nonlinear I-V characteristics can exhibit complex behaviors, such as negative resistance, which must be accounted for in advanced circuit designs. Identifying and analyzing the I-V relationship ensures devices function correctly, protecting them from potential overcurrent or overvoltage scenarios.
Semiconductors
Semiconductors are materials with electrical conductivity between that of a conductor and an insulator. They are pivotal in modern electronics, forming the backbone of components like diodes, transistors, and integrated circuits. The unique property of semiconductors is their ability to conduct current under certain conditions while blocking it under others.
Each semiconductor device can be tailored through doping—adding impurities to control carrier concentration and enhance its conductivity properties. This flexibility allows semiconductors to perform complex functions like amplification, switching, and signal modulation. Common materials used as semiconductors include silicon and germanium.
Each semiconductor device can be tailored through doping—adding impurities to control carrier concentration and enhance its conductivity properties. This flexibility allows semiconductors to perform complex functions like amplification, switching, and signal modulation. Common materials used as semiconductors include silicon and germanium.
- Diodes, made of semiconductor materials, allow current to flow in one direction only, acting as electronic valves.
- Transistors can act as switches or amplifiers, crucial for processing and amplifying signals within electronic equipment.
Other exercises in this chapter
Problem 42
Solve the given problems. All numbers are accurate to at least two significant digits. Find the smallest positive integral value of \(k\) if the equation \(x^{2
View solution Problem 42
Solve the given quadratic equations by factoring.In analyzing the path of a rocket, the equation \(16 t^{2}=320 t\) is used. Solve for \(t\).
View solution Problem 43
Solve the given applied problem. The height \(h\) (in \(\mathrm{m}\) ) of a fireworks shell shot vertically upward as a function of time \(t\) (in s) is \(h=-4.
View solution Problem 43
Solve the given problems. All numbers are accurate to at least two significant digits. Solve the equation \(x^{4}-5 x^{2}+4=0\) for \(x\). (Hint: The equation c
View solution