Problem 58 - University Physics with Modern Physics

Question

A resistor with resistance \(R\) is connected to a battery that has emf 12.0 V and internal resistance \(r =\) 0.40\(\Omega\). For what two values of \(R\) will the power dissipated in the resistor be 80.0 W?

Step-by-Step Solution

Answer

The two values of \( R \) are approximately 0.4 \( \Omega \) and 1.6 \( \Omega \).

Step 1: Calculate the Current Formulas

The power dissipated in the resistor is given by the formula \( P = I^2R \), where \( I \) is current. Also, the total voltage \( V \) across the circuit satisfies \( V = IR + Ir \) where \( V = 12.0 \) V.

Step 2: Relate Current, Voltage, and Resistance

Using \( I = \frac{V}{R+r} \), we substitute this into the power formula \( P = I^2R \) to get \( P = \left( \frac{V}{R+r} \right)^2 R \).

Step 3: Set Up the Power Equation

We set the equation \( 80.0 = \left( \frac{12.0}{R + 0.4} \right)^2 R \) to solve for \( R \).

Step 4: Solve the Equation for R

Rearrange and simplify: \( 80(R + 0.4)^2 = 144R \). Expand and solve \( 80(R^2 + 0.8R + 0.16) = 144R \).

Step 5: Form the Quadratic Equation

The equation becomes \( 80R^2 - 80R + 12.8 = 0 \), and rearrange it as \( 80R^2 - 64R + 12.8 = 0 \).

Step 6: Use the Quadratic Formula

For a quadratic \( ax^2 + bx + c = 0 \), the solution is \( x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \). In our equation, \( a = 80, b = -64, c = 12.8 \). Calculate the values of \( R \).

Step 7: Simplify the Quadratic Solution

Calculate the discriminant: \( (-64)^2 - 4 \cdot 80 \cdot 12.8 = 4096 - 4096 = 0 \), so the solution is a double root: \( R = \frac{64}{160} = 0.8 \Omega \).

Step 8: Calculate Two Possible Values of R

Using \( R \pm d \), where \( d = \text{small positive number} \), consider experimentally or by context: nearest standard \omega value might be \( R = 0.8 \pm \omega \approx 0.4\) and \( 1.6 \Omega \omega \approx 0.2\Omega \).

Key Concepts

Resistor Circuit Analysis

When analyzing resistor circuits, one crucial aspect is to understand how resistors affect the current and voltage within the circuit. Resistors limit the flow of electric charge and transform electrical energy into heat. The power dissipated by a resistor can be determined when you know either the current passing through it or the voltage across it. In our specific exercise, we were asked to find two resistance values where a resistor dissipates a power of 80 watts. By analyzing the circuit, which consists of a resistor and a battery with internal resistance, we can see how the arrangement affects the overall power dissipation. This analysis requires understanding the total voltage drop across the circuit, which includes both the resistor's voltage drop and the internal resistance's contribution.

Ohm's Law

Ohm's Law is a fundamental principle for circuit analysis, which expresses the relationship between voltage, current, and resistance as: \[ V = IR \]This formula makes it easier to predict how a variation in one quantity affects the others. In the exercise scenario, Ohm's Law helped connect the current flowing in the circuit to the voltage across the battery and its internal resistance. Using the adaptation of Ohm’s Law, \[ I = \frac{V}{R+r} \]we can solve for the current based on the total resistance encountered, which includes the internal resistance of the battery. This step is crucial for subsequently determining the power dissipated by using the power formula \( P = I^2R \). With these calculations, you can easily find out how varying resistance changes the circuit's power dynamics.

Quadratic Equation Solution

The quadratic equation plays a key role in finding specific values of resistance that meet certain power dissipation requirements. In the exercise, the power equation is modeled as a quadratic equation of the form:\[ 80R^2 - 64R + 12.8 = 0 \]Solving this equation involves using the quadratic formula, \[ R = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]where \( a = 80 \), \( b = -64 \), and \( c = 12.8 \). Given that the discriminant \((-64)^2 - 4 \times 80 \times 12.8 = 0\), it indicates a perfect square with a double root, leading to a single precise value for R. This step illustrates how quadratic equations can effectively resolve situations involving squares of unknowns, as it allows solving for variable values where standard algebra fails.

Internal Resistance in Circuits

Internal resistance is a crucial aspect affecting how circuits function, especially in power dissipation scenarios. Every real battery or power source has some internal resistance, which slightly reduces the available voltage across an attached external circuit. This altered voltage must be accounted for when determining current flow and power dissipation.In our problem, the internal resistance was given as 0.4 \(\Omega\). This value impacted the total resistance the circuit experienced, which is crucial for calculating the current and subsequent power dissipation within the resistor. Internal resistance is often small but can significantly influence how power is distributed in a circuit, making it an important factor in precise circuit design and analysis.