Problem 73 - University Physics with Modern Physics

Question

A \(12.6-\mathrm{V}\) car battery with negligible internal resistance is connected to a series combination of a \(3.2-\Omega\) resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current-voltage relationship \(V=\alpha I+\beta I^{2},\) with \(\alpha=\) 3.8\(\Omega\) and \(\beta=1.3 \Omega / \mathrm{A} .\) What is the current through the \(3.2-\Omega\) resistor?

Step-by-Step Solution

Answer

The current through the 3.2-Ω resistor is approximately 1.40 A.

Step 1: Express total voltage across components

The total voltage across the series combination of the resistor and the thermistor is equal to the voltage of the battery, which is 12.6 V.

Step 2: Determine voltage across the 3.2-Ω resistor

According to Ohm's law, the voltage across the resistor can be written as \( V_R = I \cdot R \). For a 3.2 Ω resistor, this is \( V_R = I \cdot 3.2 \).

Step 3: Formulate the equation for the thermistor

The given relationship for the thermistor is \( V_T = \alpha I + \beta I^2 \). Substituting the given values, \( \alpha = 3.8 \) Ω and \( \beta = 1.3 \) Ω/A, we get \( V_T = 3.8I + 1.3I^2 \).

Step 4: Apply Kirchhoff’s voltage law to the circuit

For the entire circuit, the sum of the voltage drops equals the battery voltage: \( V_R + V_T = 12.6 \). Substituting the expressions we have: \( 3.2I + 3.8I + 1.3I^2 = 12.6 \).

Step 5: Simplify and solve the quadratic equation

Combine the linear terms for \(I\) to simplify the equation: \( 7I + 1.3I^2 = 12.6 \). Rearrange this into a standard quadratic form: \( 1.3I^2 + 7I - 12.6 = 0 \). Use the quadratic formula \( I = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1.3, b=7, c=-12.6 \).

Step 6: Calculate the roots using the quadratic formula

Calculate the discriminant \( b^2 - 4ac \):\(7^2 - 4 \cdot 1.3 \cdot (-12.6) = 49 + 65.52 = 114.52\).Solve for \(I\):\( I = \frac{-7 \pm \sqrt{114.52}}{2 \cdot 1.3} \).Calculate the two possible values for \(I\):\( I_1 = \frac{-7 + \sqrt{114.52}}{2.6} \) and \( I_2 = \frac{-7 - \sqrt{114.52}}{2.6} \).Since current cannot be negative, we focus on \( I_1 \).

Step 7: Conclusion of current value

By calculating \( I_1 \) precisely, find the positive current value that suits the physical context of the circuit: \( I \approx 1.40 \) A.

Key Concepts

Kirchhoff’s Voltage Law

In the world of electrical engineering, Kirchhoff's Voltage Law (KVL) is a fundamental principle used in circuit analysis. It states that the sum of the electrical voltage around any closed network is zero. This means that the energy provided by power sources is completely accounted for by voltage drops across circuit elements.

To understand KVL, imagine a simple loop in a circuit containing a battery and a few components like resistors and thermistors in series. The battery introduces a voltage rise, and as current moves through resistors and other elements, voltages drop across these components. According to Kirchhoff's Voltage Law, the sum of these rises and drops equals zero. Essentially, in a circuit loop, the total voltage provided by sources must be entirely converted to potential difference across the load elements.

This law is crucial when analyzing complex circuits as it helps in writing equations that describe the relations of voltage across different components. It ensures consistency in the conservation of energy principle—a core reason why circuits function as intended.

Quadratic Equation

In solving the problem involving a thermistor that does not obey Ohm’s Law directly, we encounter a quadratic equation. A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and it can have two solutions.

In our specific exercise, we arrive at a quadratic equation after applying Kirchhoff's Voltage Law to the circuit. We have variables of current \( I \) squared term due to the thermistor's behavior being defined as \( V_T = \alpha I + \beta I^2 \). We then perform a common practice of using the quadratic formula \( I = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

The quadratic formula is a powerful tool in mathematics to find unknowns in equations of this type. Here, \( a = 1.3 \), \( b = 7 \), and \( c = -12.6 \). Calculating the discriminant \( b^2 - 4ac \) helps determine the nature of solutions. If the result is positive, two real solutions exist—the scenario we observe leading to a viable \( I \) for our circuit.

Circuit Analysis

Circuit analysis involves the process of finding current and voltage across each element of an electrical circuit. It integrates laws like Kirchhoff's along with mathematical tools such as solving linear and non-linear equations.

The essence of this technique helps to predict how circuits function under different conditions. In our exercise, the analysis revolved around calculating the current through the resistor combined in series with a thermistor. Circuit analysis begins by identifying all known components and their relationships like resistive values and voltage sources.

Drawing circuit diagrams helps visualize connections. Then apply systematic approaches such as writing KVL for voltage loops and using Ohm's Law to determine current through resistors. In circuits involving non-linear components like our thermistor, specialized relationships \( V = \alpha I + \beta I^2 \) are considered which leads to solving equations to find desired quantities like current or voltage.

• Identify known and unknown parameters.
• Set up equations using core principles like KVL and Ohm's Law.
• Simplify and solve using appropriate mathematical approaches.

Mastering these steps allows not only solving academic problems but also real-world circuit design and diagnostics.