Problem 12
Question
ELECTRICITY For Exercises \(9-12\) , use the following information. The current \(I\) in amperes in an electrical circuit with three resistors in series is given by the equation \(I=\frac{V}{R_{1}+R_{2}+R_{3}},\) where \(V\) is the voltage in volts in the circuit and \(R_{1}, R_{2},\) and \(R_{3}\) are the resistances in ohms of the three resistors. What domain and range values are meaningful in the context of the problem?
Step-by-Step Solution
Verified Answer
The domain for the resistances \(R_1, R_2, R_3\) is \([0, \infty)\) with a sum \(>0\). The range for current \(I\) is \((0, \infty)\).
1Step 1: Analyze the Equation
The equation provided is \( I = \frac{V}{R_1 + R_2 + R_3} \). This formula shows the relationship between current \( I \), voltage \( V \), and the resistances \( R_1, R_2, R_3 \).
2Step 2: Determine Meaningful Domain Values
In the context of this problem, the resistors' resistance values \( R_1, R_2, \) and \( R_3 \) must be non-negative, as negative resistance is not physically meaningful. Therefore, the sum \( R_1 + R_2 + R_3 \) must also be positive, ensuring the denominator is non-zero. Thus, the domain for \( R_1, R_2, \) and \( R_3 \) is all non-negative real numbers \([0, \infty)\) that add up to a positive value.
3Step 3: Determine Meaningful Range Values
The current \( I \) is positive when real voltage \( V \) is divided by a positive denominator \( R_1 + R_2 + R_3 \). Hence, the possible values or range for \( I \) is also positive real values, denoted as \((0, \infty)\). Note that the current becomes smaller as resistance increases, but it cannot be zero unless the voltage is zero.
4Step 4: State the Contextual Constraints
For voltages \( V \), we assume positive values for normal operating conditions since negative or zero voltages imply different scenarios or no functionality, so \( V \) is typically in the range \((0, \infty)\). Therefore, voltage has to be a positive number to generate a physical current.
Key Concepts
Resistors in SeriesOhm's LawDomain and RangeVoltage and Current
Resistors in Series
A circuit configuration where resistors are connected end-to-end is known as a series arrangement. This implies that the current flows through each resistor one after the other, making the total resistance the sum of the individual resistances. Imagine water flowing through several pipes in sequence, where each pipe adds some resistance to the flow of water. In a similar way, each resistor contributes to the total resistance in the circuit.
This addition rule for resistors in series is crucial for calculating the total resistance and, subsequently, the current in the circuit using Ohm's Law. The combined resistance is calculated by:
This addition rule for resistors in series is crucial for calculating the total resistance and, subsequently, the current in the circuit using Ohm's Law. The combined resistance is calculated by:
- \[ R_{total} = R_1 + R_2 + R_3 \]
Ohm's Law
Ohm's Law is a fundamental principle in electricity that defines the relationship between voltage, current, and resistance. It is simply stated as:
In series circuits, applying Ohm's Law allows us to determine how current and voltage are distributed across the components. Knowing either two of these values allows you to solve for the third, making Ohm's Law an indispensable tool in solving circuit problems.
For students and engineers alike, mastering Ohm’s Law is the gateway to understanding more complex electrical phenomena.
- \[ V = I \cdot R \]
In series circuits, applying Ohm's Law allows us to determine how current and voltage are distributed across the components. Knowing either two of these values allows you to solve for the third, making Ohm's Law an indispensable tool in solving circuit problems.
For students and engineers alike, mastering Ohm’s Law is the gateway to understanding more complex electrical phenomena.
Domain and Range
When discussing electrical circuits, particularly with Ohm's Law, it's vital to identify meaningful domain and range values for the variables involved. The domain refers to all the possible values that the input (resistances and voltage in this context) can take. Given that resistance cannot be negative, the domain for resistances \( R_1, R_2, R_3 \) is the set of all non-negative real numbers,
The range is concerned with all potential output values, namely the current \( I \). Here, current values are always positive, starting just above zero and extending to higher real numbers, since current becomes effectively zero only if there's no voltage. Thus, the range of current is:
- \([0, \infty)\)
The range is concerned with all potential output values, namely the current \( I \). Here, current values are always positive, starting just above zero and extending to higher real numbers, since current becomes effectively zero only if there's no voltage. Thus, the range of current is:
- \((0, \infty)\)
Voltage and Current
Voltage and current are two core concepts in understanding electric circuits, especially when using formulas like Ohm's Law. Voltage, expressed in volts (V), is the electrical force that drives an electric current between two points. It acts like the pressure pushing water through a pipe.
Current, measured in amperes (A), is the flow of electrical charge through the circuit. The higher the current, the more charge is moving, similar to how more water flows through a pipe as you increase the flow rate.
Current, measured in amperes (A), is the flow of electrical charge through the circuit. The higher the current, the more charge is moving, similar to how more water flows through a pipe as you increase the flow rate.
- Voltage provides the necessary energy to push current through resistors.
- The amount of current that flows is influenced by the total resistance encountered along its path.
Other exercises in this chapter
Problem 12
If \(y\) varies inversely as \(x\) and \(y=5\) when \(x=10,\) find \(y\) when \(x=2\)
View solution Problem 12
Simplify each expression. $$ \frac{x}{x+1}+\frac{3}{x^{2}-4 x-5} $$
View solution Problem 12
Simplify each expression. \(\frac{12 p^{2}+6 p-6}{4(p+1)^{2}} \div \frac{6 p-3}{2 p+10}\)
View solution Problem 13
Solve each equation or inequality. Check your solutions. $$ a+1=\frac{6}{a} $$
View solution